Chapter 15 Heating and chemical effect of current free study material by TEACHING CARE online tuition and coaching classes
Joules Heating.
When some potential difference V is applied across a resistance R then the work done by the electric field on charge q to flow through the circuit in time t will be W = qV = Vit = i^{2}Rt = V 2 t Joule .
R
This work appears as thermal energy in the resistor.
Heat produced by the resistance R is
H = W
= Vit = i ^{2} Rt =
V ^{2} t
Cal. This relation is called joules heating.
J 4 × 2
4 × 2
4 × 2R
Some important relations for solving objective questions are as follow :
Electric Power.
The rate at which electrical energy is dissipated into other forms of energy is called electrical power i.e.
P = W = Vi = i ^{2} R = V 2
t R
 Units : It’s S.I. unit is Joule/sec or Watt
Bigger S.I. units are KW, MW and HP, remember 1 HP = 746 Watt
 Rated values : On electrical appliances (Bulbs, Heater ……. etc.)
Wattage, voltage, ……. etc. are printed called rated values e.g. If suppose we have a bulb of 40 W, 220 V then rated power (P_{R}) = 40 W while rated voltage (V_{R}) = 220 V. It means that on operating the bulb at 220 volt, the power dissipated will be 40 W or in other words 40 J of electrical energy will be converted into heat and light per second.
 Resistance of electrical appliance : If variation of resistance with temperature is neglected then
V 2
resistance of any electrical appliance can be calculated by rated power and rated voltage i.e. by using
R = R
PR
e.g.
Resistance of 100 W, 220 volt bulb is
R = 220 ´ 220 = 484 W
100
 Power consumed (illumination) : An electrical appliance (Bulb, heater, …. etc.) consume rated power (P_{R}) only if applied voltage (V_{A}) is equal to rated voltage (V_{R}) e. If V_{A} = V_{R} so P_{consumed} = P_{R}. If V_{A} < V_{R} then
Pconsumed
2

^{ } ^{A} also we have
R
V 2

R ^{R} so
PR
æ 2 ö


Pconsumed (Brightness) = ç A ÷. PR
V
è R ø
æ 110 ö ^{2}
e.g. If 100 W, 220 V bulb operates on 110 volt supply then
Pconsumed
= ç 220 ÷
´ 100 = 25 W
Note : @If V_{A}
< V_{R}
è ø
then % drop in output power = (PR – Pconsumed ) ´ 100
PR
@ For the series combination of bulbs, current through them will be same so they will consume power in the ratio of resistance i.e., P µ R {By P = i^{2}R) while if they are connected in parallel i.e. V
is constant so power consumed by them is in the reverse ratio of their resistance i.e.
P µ 1 .
R
 Thickness of filament of bulb : We know that resistance of filament of bulb is given by
V 2
R = ^{R} ,
PR
also
R = r l , hence we can say that
A
A
(Thickness )
µ PR
µ 1
R
i.e. If rated power of a bulb is more, thickness of it’s filament
is also more and it’s resistance will be less.
1 V ^{2}
If applied voltage is constant then
P(consumed) µ R (By
P = ^{A} ). Hence if different bulbs (electrical appliance)
R
operated at same voltage supply then
Note : @Different bulbs
Pconsumed
µ PR
µ thickness µ 1
R
25W
220V
100W
220V
1000W
220V
Þ Resistance R_{25} > R_{100} > R_{1000}
Þ Thickness of filament t_{1000} > t_{100} > t_{40}
Þ Brightness B_{1000} > B_{100} > B_{25}
 Long distance power transmission : When power is transmitted through a power line of resistance R, powerloss will be i ^{2} R
Now if the power P is transmitted at voltage V
P = Vi
i.e.
i = (P / V)
So,
Power loss = P 2 ´ R
V 2
Now as for a given power and line, P and R are constant so Power loss µ (1 / V ^{2} )
So if power is transmitted at high voltage, power loss will be small and viceversa. e.g., power loss at 22 kV
is 10^{–4} times than at 220 V. This is why long distance power transmission is carried out at high voltage.
 Time taken by heater to boil the water : We know that heat required to raise the temperature Dq of any substance of mass m and specific heat S is H = S.Dq
Here heat produced by the heater = Heat required to raise the temp. Dq of water.
i.e. p ´ t = J ´ m.S.Dq Þ
t = J(m.S.Dq )
p
{J = 4.18 or 4.2 J/cal)
for m kg water
t = 4180 ( or 4200)m Dq
p
{S = 1000 cal/kg^{o}C)
Note : @ If quantity of water is given n litre then t = 4180(4200)n Dq
p
Electricity Consumption.
 The price of electricity consumed is calculated on the basis of electrical energy and not on the basis of electrical
 The unit Joule for energy is very small hence a big practical unit is considered known as kilowatt hour
(KWH) or board of trade unit (B.T.U.) or simple unit.
 1 KWH or 1 unit is the quantity of electrical energy which dissipates in one hour in an electrical circuit when the electrical power in the circuit is 1 KW thus 1 KW = 1000 W ´ 3600 sec = 6 ´ 10^{6} J.
 Important formulae to calculate the of consumed units is n = Total watt ´Total hours
1000
Example: 1 The approximate value of heat produced in 5 min. by a bulb of 210 watt is (J = 4.2 joule/calorie)
[MP PET 2000; MNR 1985]
(a) 15,000 (b) 1,050 (c) 63,000 (d) 80,000
Solution : (a) By using
H = P ´ t = 210 ´ 5 ´ 60 = 15000 Cal
4.2 4.2
Example: 2 A heater coil is cut into two parts of equal length and one of them is used in the heater. The ratio of the heat produced by this half coil to that by the original coil is [NCERT 1972]
(a) 2 : 1 (b) 1 : 2 (c) 1 : 4 (d) 4 : 1
Solution : (a) If suppose resistance of the coil is R so resistance of it’s half will be R . Hence by using H =
2
V ^{2}t R
Þ H µ 1
R
Þ HHalf = RFull = R = 2
HFull
RHalf
R / 2 1
Note : @ In general if coil is divided in n equal parts then heat produced by each part will be n times of the heat produced by coil it self i.e. H¢ = nH
Example: 3 If current in an electric bulb changes by 1%, then the power will change by [AFMC 1996]
(a) 1% (b) 2% (c) 4% (d) 1 %
2
Solution : (b) By using P = i^{2}R Þ P µ i^{2} Þ
DP = 2 Di
Þ change in power = 2%
P i
Example: 4 A constant voltage is applied on a uniform wire, then the heat is produced. The heat so produced will be doubled, if [IITJEE 1980; MP PET 1987; SCRA 1994; MP PMT 1999]
 The length and the radius of wire are halved (b) Both length and radius are doubled
(c) Only the length is doubled (d) Only the radius is doubled
V ^{2}t
l r l
V ^{2}tp r ^{2} r ^{2}
Solution : (b) By using H = R
doubled.
and
R = r A = p r 2 Þ
H = r l Þ H µ l ; on doubling both r and l heat will be
Example: 5 An electric heater of resistance 6 ohm is run for 10 minutes on a 120 volt line. The energy liberated in this period of time is [MP PMT 1996]
(a)
7.2 ´ 10^{3} J
(b)
14.4 ´ 10^{5} J
(c)
43.2 ´ 10^{4} J
(d)
28.8 ´ 10^{4} J
Solution : (b) By using
H = V 2 t Þ H =
R
(120)^{2} ´ 10 ´ 60
6
= 14.4 ´ 10^{5} J
Example: 6 An electric bulb of 100 W is designed to operate on 220 V. Resistance of the filament is
[EAMCET 1981, 82; MP PMT 1993, 97]
(a) 484 W (b) 100 W (c) 22000 W (d) 242 W
Solution : (a) By using
P = V 2
R
Þ R = V 2
P
= (220)^{2}
100
= 484 W
Example: 7 An electric bulb is rated 220 V and 100 W. Power consumed by it when operated on 110 volt is
[AFMC 2000; MP PMT 1986, 94; CPMT 1986]
(a) 50 W (b) 75 W (c) 90 W (d) 25 W
æ V ö ^{2}
æ 110 ö ^{2}
Solution : (d) By using
Pconsumed
= ç^{ } ^{A} ÷
 PR Þ PConsumed = ç
÷ ´ 100 = 25 W
è VR ø
è 220 ø
Example: 8 A 500 watt heating unit is designed to operate from a 115 Volt line. If the line voltage drops to 110 volt, the percentage drop in heat output will be [ISM Dhanbad 1994]
(a) 10.20% (b) 8.1% (c) 8.6% (d) 7.6%
æ V ö ^{2}
æ 110 ö ^{2}
Solution : (c) By using
Pconsumed
= ç^{ } ^{A} ÷
 PR Þ PConsumed = ç
÷ ´ 500 = 456.6 Watt
è VR ø
è 115 ø
So % drop in heat output = PActual – PConsumed ´ 100 = (500 – 456.6) ´ 100 = 8.6%
PActual
500
Example: 9 An electric lamp is marked 60 W, 230 V. The cost of 1 kilowatt hour of power is Rs. 1.25. The cost of using this lamp for 8 hours is [KCET 1994]
(a) Rs. 1.20 (b) Rs. 4.00 (c) Rs. 0.25 (d) Rs. 0.60
Solution : (d) By using consumed unit (n) or KWH = Total Watt ´ Total time
Þ n = 60 ´ 8 = 12
So cost = 12 ´ 1.25 = 0.60 Rs
25
1000
1000 25
Example: 10 How much energy in Kilowatt hour is consumed in operating ten 50 watt bulbs for 10 hours per day in a month (30 days) [Pb PMT 2000; CPMT 1991; NCERT 1978]
(a) 1500 (b) 15.000 (c) 15 (d) 150
Solution : (d) By using n = Total Watt ´ Total time
1000
Þ n = (50 ´ 10) ´ (10 ´ 30) = 150
1000
Example: 11 An immersion heater is rated 836 watt. It should heat 1 litre of water from 20^{o} C to 40^{o} C in about
(a) 200 sec (b) 100 sec (c) 836 sec (d) 418 sec
Solution : (b) By using t = 4180 ´ n ´ Dq
P
Þ t = 4180 ´ 1 ´ (40 – 20) = 100 sec
836
Example: 12 The power of a heater is 500 watt at 800^{o} C. What will be its power at 200^{o} C if a = 4 ´ 10 ^{–}^{4}
(a) 484 W (b) 672 W (c) 526 W (d) 611 W
per °C
2 V 2
1 P_{1}
R2 (1 + a t 2 )
500
(1 + 4 ´ 10 ^{–}^{4} ´ 200)
Solution : (d) By using
P = i R =
R Þ P µ R Þ
=
P2 R1
= Þ
(1 + a t1 ) P2
= (1 + 4 ´ 10 ^{–}^{4} ´ 800)
Þ 500 = 1.08 Þ 611W
P2 1.32
Example: 13 A heater of 220 V heats a volume of water in 5 minute time. A heater of 110 V heats the same volume of water in [AFMC 1993]
 5 minutes (b) 8 minutes (c) 10 minutes (d) 20 minutes
V ^{2}t
Solution : (d) By using H = R . Here volume of water is same. So same heat is required in both cases. Resistance is also
1 t æ V ö 2 5 æ 110 ö ^{2} 1
constant so V^{2}t = constant Þ
t µ Þ ^{ } ^{1} = ç 2 ÷ Þ = ç ÷ = Þ t _{2} = 20 min
V 2 t 2
è V1 ø
t 2 è 220 ø 4
Example: 14 Water boils in an electric kettle in 15 minutes after switching on. If the length of the heating wire is decreased to 2/3 of its initial value, then the same amount of water will boil with the same supply voltage in [MP PMT 1994]
 15 minutes (b) 12 minutes (c) 10 minutes (d) 8 minutes
Solution : (c) By using
H = V ^{2}t
where
R = r l Þ H = V 2 t A . Since volume is constant so H is also constant so t µ l
R
t l t
A r l
2 l1
which gives ^{ } ^{2} = ^{2} Þ ^{ } ^{2} = 3 Þ t 2 = 10 min
t1 l1 15 l1
Combination of Bulbs (or Electrical Appliances).
Some Standard Cases for Series and Parallel Combination.

 If n identical bulbs first connected in series so PS= n
and then connected in parallel. So P_{P}
= nP hence
PP = n^{2} .
PS
 To operate a bulb on voltage which is more then it’s rated voltage, a proper resistance is connected in series with e.g. to glow a bulb of 30 W, 6 V with full intensity on 126 volt required series resistance calculated as follows
Bulb will glow with it’s full intensity if applied voltage on it is 6 V i.e. 120 V appears across the series resistance
R current flows through bulb = current flows through resistance
i = 30 = 5 amp
6
Hence for resistance V = iR i.e. 120 = 5 ´ R Þ 5 ´ R Þ R = 24 W
æ Voperating – VR ö
Note : @ If you want to learn Short Trick then remember Series resistance = ç
è PR
÷ ´ VR
ø
 An electric kettle has two coils when one coil is switched on it takes time t_{1} to boil water and when the second coil is switched on it takes time t_{2} to boil the same
1  = 1 +  1 
HS / t S H1 / t1  H 2 / t 2 
 If three identical bulbs are connected in series as shown in figure then on closing the switch S. Bulb C short circuited and hence illumination of bulbs A and B increases
Reason : Voltage on A and B increased.
 If three bulbs A, B and C are connected in mixed combination as shown, then illumination of bulb A
decreases if either B or C gets fused
Reason : Voltage on A decreases.
 If two identical bulb A and B are connected in parallel with ammeter A and key K as shown in It should be remembered that on pressing key reading of ammeter becomes twice.
Reason : Total resistance becomes half.
Example: 15 An electric kettle has two heating coils. When one of the coils is connected to ac source the water in the kettle boils in 10 minutes. When the other coil is used the water boils in 40 minutes. If both the coils are connected in parallel, the time taken by the same quantity of water to boil will be [CBSE PMT 2003]
 4 min (b) 25 min (c) 15 min (d) 8 min
Solution : (d) By using the formula (as we discussed in theory) Þ
Note : @ In this question if coils are connected in series then the time taken by the same quantity of water to boil will be t_{s} = t_{1} + t_{2} = 10 + 40 = 50 min
Example: 16 If a 30 V, 90 W bulb is to be worked on a 120 V line, a resistance of how many ohms should be connected in series with the bulb [KCET 2003; MP PMT 2002]
(a) 10 ohm (b) 20 ohm (c) 30 ohm (d) 40 ohm
æ Voperating – VR ö
(120 – 30)
Solution : (c) By using Series resistance R = ç
è PR
÷ ´ VR
ø
(As we discussed in theory) Þ R =
90 ´ 30 = 30 W
Example: 17 In the circuit shown in figure, the heat produced in 5 ohm resistance is 10 calories per second. The heat produced in 4 ohm resistance is [BHU 2003; RPMT 1994; IIT 1981]
 1 cal/sec
 2 cal/sec
 3 cal/sec
 4 cal/sec
Solution : (b) Ratio of currents i1
= 10 = 2 by using H = i ^{2} Rt
i 4W 6W
i2 5 1
H æ i ö2 R 10 æ 2 ö^{2} 5
^{2} Line (2)
Þ 1 = ç 1 ÷ ´ 1 Þ
= ç ÷ ´
Þ H_{2} = 2cal / sec
Line (1)
H2 ç i2 ÷ R2 H2 è 1 ø 4
è ø i1
Example: 18 Two heater wires of equal length are first connected in series and then in parallel. The ratio of heat produced in the two cases is [MP PET 2002, 1999; MP PMT 2001, 2000, 1996; AIIMS 2000; MNR 1987; DCE 1997, 94]
(a) 2 : 1 (b) 1 : 2 (c) 4 : 1 (d) 1 : 4
Solution : (d) Both the wires are of equal length so they will have same resistance and by using H =
V ^{2}t R
Þ H µ 1
R
Þ Hs
= RP ;
Þ Hs
= R / 2 = 1
HP Rs HP 2R 4
Example: 19 If two bulbs of wattage 25 and 100 respectively each rated at 220 volt are connected in series with the supply of 440 volt, then which bulb will fuse [MP PET 2000; MNR 1988]
 100 watt bulb (b) 25 watt bulb (c) None of them (d) Both of them
Solution : (b) In series VA µ 1
PR
i.e. voltage appear on 25W bulb will be more then the voltage appears on 100 W bulb. So
bulb of 25 W will gets fused.
Example: 20 Three equal resistors connected in series across a source of e.m.f. together dissipate 10 watt. If the same resistors are connected in parallel across the same e.m.f., then the power dissipated will be
[KCET 1999; DCE 1998; CBSE 1998; MP PAT 1996]
(a) 10 W (b) 30 W (c) 10/3 W (d) 90 W
Solution : (d) In series consumed power Ps = P
n
while in parallel consumed power P_{p} = nP Þ P_{P} = n^{2}. P_{s}
Þ PP = (3)^{2} ´ 10 = 90W
Example: 21 Forty electric bulbs are connected in series across a 220 V supply. After one bulb is fused, the remaining 39 are connected again in series across the same supply. The illumination will be [Haryana CEE 1996; NCERT 1972]
 More with 40 bulbs than with 39 (b) More with 39 bulbs than with 40
 Equal in both the cases (d) In the ratio of 40^{2} : 39^{2}
V 2
Solution : (b) Illumination = P_{Consumed} =
. Initially there were 40 bulbs in series so equivalent resistance was 40 R, finally
R
39 bulbs are in series so equivalent resistance becomes 39 R. Since resistance decreases so illumination increases with 39 bulbs.
Example: 22 Two bulbs of 100 watt and 200 watt, rated at 220 volts are connected in series. On supplying 220 volts, the consumption of power will be
(a) 33 watt (b) 66 watt (c) 100 watt (d) 300 watt
Solution : (b) In series PConsumed =
P1P2 P1 + P2
Þ PConsumed
= 100 ´ 200 = 66W
300
Example: 23 Two wires ‘A’ and ‘B’ of the same material have their lengths in the ratio 1 : 2 and radii in the ratio 2 : 1. The two wires are connected in parallel across a battery. The ratio of the heat produced in ‘A’ to the heat produced in ‘B’ for the same time is [MNR 1998]
(a) 1 : 2 (b) 2 : 1 (c) 1 : 8 (d) 8 : 1
l l R
l æ r ö ^{2} R
1 æ 1 ö ^{2} 1
Solution : (d) Resistance
R = r
Þ R µ
Þ A = A ´ ç B ÷
Þ A =
´ ç ÷ =
By using
p r ^{2} r ^{2} RB
H = V ^{2}t Þ H_{A} = R_{B} = 8
l B è rA ø
RB 2
è 2 ø 8
R HB RA 1
Example: 24 A heating coil is labelled 100 W, 220 V. The coil is cut in half and the two pieces are joined in parallel to the same source. The energy now liberated per second is [CBSE PMT 1995]
(a) 200 J (b) 400 J (c) 25 J (d) 50 J
Solution : (b) Let resistance of the heating coil be R, when coil cut in two equal parts, resistance of each part will be R .
2
When these two parts are corrected in parallel,
Req = R
4
i.e. resistance becomes, so according to
P µ 1 ; Power becomes 4 times i.e. P¢ = 4P = 400 J/sec
R
Example: 25 Two identical electric lamps marked 500 W, 220 V are connected in series and then joined to a 110 V line.
The power consumed by each lamp is
(a)
125 W
4
(b)
25 W
4
(c)
225 W
4
 125 W
Solution : (a) Both bulbs are identical so voltage across each bulb will be 55V.
æ V ö ^{2}
æ 55 ö ^{2}
125
Hence power consumed by each bulb is ç ^{A} ÷
´ P_{R} = ç ÷ ´ 500 = W
è VR ø
è 220 ø 4
If two wires of different metals are joined at their ends so as to form two junctions, then the resulting arrangement is called a “Thermo couple”.
Seebeck Effect.
 Definition : When the two junctions of a thermo couple are maintained at different temperatures, then a current starts flowing through the loop known as thermo electric current. The potential difference between the junctions is called thermo electric emf which is of the order of a few microvolts per degree temperature difference (mV/^{o}C).
 Origin of thermo emf : The density of free electrons in a metal is generally different from the density of free electrons in another metal. When a metal is brought into intimate contact (say by soldering) with other metal, the electrons tend to diffuse from one metal to another, so as to equalise the electron As an illustration, when copper is brought into intimate contact with iron, the electrons diffuse from iron to copper. But this diffusion cannot go on continuously because due to diffusion, the potential of copper decreases and the potential of iron increases. In other words, iron becomes positive with respect to copper. This is what stops further diffusion. In the case of thermocouple whose junctions are at the same temperature, the emf’s at the junctions will be equal in magnitude but opposite in direction. So, the net emf for the whole of thermocouple will be zero.
Let us now consider the case when the temperature of one junction of the thermocouple is raised. Raising the temperature of one junction will affect the electron density in the two metals differently. Moreover, the transfer of electrons at the junction will be easier than the transfer of electrons at the cold junction. Due to both these reasons, the emf’s at the two junctions will be different. This produces a net emf in the thermocouple. This emf is known as Seebeck emf.
 Seebeck series : The magnitude and direction of thermo emf in a thermocouple depends not only on the temperature difference between the hot and cold junctions but also on the nature of metals constituting the thermo
Seebeck arranged different metals in the decreasing order of their electron density. Some of the metals forming the series are as below.
Sb, Fe, Ag, Au, Sn, Pb, Cu, Pt, Ni, Bi
 About magnitude thermo emf : Thermo electric emf is directly proportional to the distance between the two metals in series. Farther the metals in the series forming the thermo couple greater is the thermo emf. Thus maximum thermo emf is obtained for Sb–Bi thermo
 Direction of thermo electric current : If a metal occurring earlier in the series is termed as A and the metal occurring later in the series is termed as B, then the rule for the direction of conventional current in thermocouple made of elements A and B is ABC. That is, at the cold junction current will flow from A to B. g. in Fe–Cu thermocouple, at the cold junction current flows from A to B that is from Fe to Cu. At the hot junction, the current flows from Cu to Fe. This may be remembered easily by the hot coffee.
(4) Law of thermoelectricity
 Law of successive temperature : If initially temperature limits of the cold and the hot junction are t_{1} and
t_{2}, say the thermo emf is
E ^{t}2 .

1
When the temperature limits are t_{2} and t_{3}, then say the thermo emf is
t3 then


2
E^{t}2 + E^{t}3
= E^{t}3
where
E t3
is the thermo emf when the temperature limits are
E t3
t1 t2 t1 t1 t1
 Law of intermediate metals : Let A, B and C be the three metals of Seebeck series, where B lies
between A and C. According to this law, E^{B} + E^{C} = E^{C}
A B A
When tin is used as a soldering metal in Fe–Cu thermocouple then at the junction, two different thermo couples are being formed. One is between iron and tin and the other is between tin and copper, as shown in figure (i)
Now iron is thermoelectrically more positive as compared to tin and tin is more positive with respect to copper (the element which occurs earlier in the seebeck series gets positively charged on losing the electrons at the junction), so as clear from the figure below, the thermo emf’s of both the thermocouples shown in the figure (ii) are additive
 (ii)
\ If soldering metal in a thermocouple is an intermediate metal in the series then thermo emf will not be affected. It is also clear from the above discussions that if the soldering metal does not lie between two metals (in
Seebeck series) of thermocouple then the resultant emf will be subtractive.
 Effect of temperature on thermo emf : In a thermocouple as the temperature of the hot junction increases keeping the cold junction at constant temperature (say 0^{o}C). The thermo emf increases till it becomes maximum at a certain
 Thermo electric emf is given by the equation
E = a t + 1 b t ^{2}
2
where a and b are thermo electric constant having units are volt/^{o}C and volt/^{o}C^{2} respectively (t = temperature of hot junction).
 The temperature of hot junction at which thermo emf becomes maximum is called neutral temperature (t_{n}). Neutral temperature is constant for a thermo couple (g. for Cu–Fe, t_{n} = 270^{o}C)
 Neutral temperature is independent of the temperature of cold junction.
 If temperature of hot junction increases beyond neutral temperature, thermo emf start decreasing and at a particular temperature it becomes zero, on heating slightly further, the direction of emf is reversed. This temperature of hot junction is called temperature of inversion (t_{i}).
 Graphical representation of thermo emf
 tn
= ti + tc
2
 Graph is parabolic
 For E to be maximum (at t = t_{n})
dE = 0
dt
i.e. a + b t_{n} = 0 Þ th
= – a
b
 Thermo electric power : The rate of change of thermo emf with the change in the temperature of the hot junction is called thermoelectric
It is also given by the slope of parabolic curve representing the variation of thermo emf with temperature of the hot junction, as discussed in previous section.
It is observed from the above graph that as temperature of hot junction increases from that of the cold junction to the neutral junction, though the thermo emf is increasing but the slope of the graph, that is the rate of change of thermo emf with temperature of hot junction is decreasing. Note that, at the neutral temperature, the thermo emf is maximum but the slope i.e. the thermoelectric power is zero.
The thermo electric power
æ dE ö
ç dt ÷
is also called Seebeck coefficient. Differentiating both sides of the
è ø
equation of thermo emf with respect to t, we have
P = dE = d (a t + 1 b t ^{2} ) Þ
P = a + b t
dt dt 2
The equation of the thermo electric power is of the type as shown below.
y = mx + c,
so the graph of thermo electric power is
Peltier Effect.
 If a current is passed through a junction of two different metals, the heat is either evolved or absorbed at the This effect is known as Peltier effect. It is the reverse of Seebeck effect. Before going into the detailed explanation, we will first revise an important concept about absorption and evolution of energy when electric charge is made to pass through two points having some potential difference.
When a positive charge flows from high potential to low potential, it releases energy and when positive charge flows from low potential to high potential it absorbs energy.
 Explanation of Peltier effect : In the light of above statement it can be seen that if current is made to flow in FeCu thermocouple by connecting it to a battery then the junction at which current goes from Fe to Cu becomes hot because here positive charge is flowing from high potential to low potential, so energy is released. Remember that, in ironcopper thermocouple, the polarity of the contact potential at each junction is such iron is at higher Similarly the junction where current flows from Cu to Fe becomes colder because at this junction
current is flowing from negative to positive potential, so energy is absorbed. Thus it is observed that on application of potential difference in a thermocouple temperature difference is automatically created. The amount of heat absorbed at cold junction is equal to the heat released at hot junction.
 Peltier coefficient (p) : Heat absorbed or liberated at the junction is directly proportional to the charge passing through the junction e. H µ Q Þ H = pQ ; where p is called Peltier coefficient. It’s unit is J/C or volt.
 If Q = 1 then
H = p
i.e. Peltier coefficient of a junction is defined as heat absorbed or liberated at the
junction when a unit quantity of electric charge flows across the junction (H is also known as Peltier emf).
 Relation between p and absolute temperature : Suppose the temperature of the cold junction is T
and that of the hot junction is T + dT and let dE be the thermo emf produced, then it is found that
p = T dE = T ´ S ; where T is in Kelvin and
dT
dE = P =
dT
Seebeck coefficient S
 pdepends on : (a) Temperature of junction (b) Difference in electron density of the two metal used in
 Comparison between Joule and Peltier effect
Thomson’s Effect.
 Definition : In Thomson’s effect we deal with only metallic rod and not with thermocouple as in Peltiers effect and Seebeck’s effect. (That’s why sometimes it is known as homogeneous thermo electric effect. When a current flows thorough an unequally heated metal, there is an absorption or evolution of heat in the body of the This is Thomson’s effect.
(2)
Types of Thomson’s effect
 Thomson’s coefficient : In Thomson’s effect it is found that heat released or absorbed is proportional
to QDq i.e.
H µ QDq
Þ H = θQΔθ
where s = Thomson’s coefficient. It’s unit is Joule/coulomb^{o}C or volt/^{o}C and
Dq = temperature difference.
 If Q = 1 and Dq = 1 then
□ = H
so the amount of heat energy absorbed or evolved per second between
two points of a conductor having a unit temperature difference, when a unit current is passed is known as Thomson’s coefficient for the material of a conductor.
 It can be proved that Thomson coefficient of the material of conductor
s = –T d 2 E
dT ^{2}
also Seebeck co
efficient S = dE
so dS = d ^{2} E
hence s
= –Tæ dS ö = T ´ b ; where b = Thermo electric constant = dS
dT dT
dT ^{2}
ç dT ÷ dt
è ø
Application of Thermo Electric Effect.
 To measure temperature : A thermocouple is used to measure very high (2000^{o}C) as well as very low (– 200^{o}C) temperature in industries and laboratories. The thermocouple used to measure very high temperature is called
 To detect heat radiation : A thermopile is a sensitive instrument used for detection of heat radiation and measurement of their It is based upon Seebeck effect.
A thermopile consists of a number of thermocouples of SbBi, all connected in series.
This instrument is so sensitive that it can detect heat radiations from a match stick lighted at a distance of 50
metres from the thermopile.
 Thermoelectric refrigerator : The working of thermoelectric refrigerator is based on Peltier effect. According to Peltier effect, if current is passed through a thermocouple, heat is absorbed at one junction and is evolved at the other junction of the If on the whole, the heat is absorbed, then the thermocouple acts as thermoelectric refrigerator. It’s efficiency is small in comparison to conventional refrigerator.
 Thermoelectric generator : Thermocouple can be used to generate electric power using Seebeck effect in remote It can be achieved by heating one junction in a flame of kerosene oil lamp and keeping the other junction at room or atmospheric temperature. The thermo emf so developed is used to operate radio receivers or even radio transmitters.
Example: 26 The smallest temperature difference that can be measured with a combination of a thermocouple of thermo
e.m.f. 30 mV per degree and a galvanometer of 50 ohm resistance capable of measuring a minimum current
of 3 × 10^{–7} ampere is [MP PET 2000]
(a) 0.5 degree (b) 1.0 degree (c) 1.5 degree (d) 2.0 degree
Solution : (a) By using E = aq Þ i R = aq Þ 3 ´ 10^{–7} ´ 50 = 30 ´ 10^{–6} ´ q Þ q = 0.5 degree
Example: 27 The expression for thermo e.m.f. in a thermocouple is given by the relation
q 2
E = 40 q – 20 , where q is the
temperature difference of two junctions. For this, the neutral temperature will be [AMU (Engg.) 2000]
(a) 100^{o} C (b) 200^{o} C (c) 300^{o} C (d) 400^{o} C
Solution : (d) Comparing the given equation of thermo e.m.f. with E = a t +
1 b t ^{2}
2
we get a = 40
and
b = –
1 . By
10

using t = – a
b
Þ tn = 400^{o} C .
Example: 28 One junction of a certain thermoelectric couple is at a fixed temperature T_{r} and the other junction is at
temperature T. The thermo electromotive force for this is expressed by
E = K(T –
é 0 – 1 (T +
ù at


T ) T
ë 2
T )ú
û

temperature T =
1 T_{0} , the thermo electric power will be [MP PMT 1994]
2
(a)
1 KT0
(b)
KT (c)
1 KT ^{2}
(d)
1 K(T
– T )^{2}
2 0 2 0 2 0 r
Solution : (a) As we know thermo electric power
T = 1 T0 we get S = 1 KT0 .
S = dE . Hence by differentiating the given equation and putting
dT
2 2
Example: 29 The cold junction of a thermocouple is maintained at 10^{o} C. No thermo e.m.f. is developed when the hot junction is maintained at 530^{o} C. The neutral temperature is [MP PMT 1994]
(a) 260^{o} C (b) 270^{o} C (c) 265^{o} C (d) 520^{o} C
Solution : (b) Given tc
= 10 ^{o} C and ti
= 530 ^{o} C
hence by using tn
= ti + tc
2
Þ tn
= 270 ^{o} C
Example: 30 The thermo emf develops in a Cu–Fe thermocouple is 8.6 mV/^{o}C. It temperature of cold junction is 0^{o}C and temperature of hot junction is 40^{o}C then the emf obtained shall be
(a) 0.344 mV (b) 3.44 mV (c) 3.44 V (d) 3.44 mV
Solution : (a) By using thermo emf e = aq where a = 8.6 mV
o C
So e = 8.6 ´ 10^{–6} ´ 40 = 344 mV = 0.344 mV.
and q = temperature difference = 40^{o}C
Example: 31 A thermo couple develops 200 mV between 0^{o}C and 100^{o}C. If it develops 64 mV and 76 mV respectively between (0^{o}C – 32^{o}C) and (32^{o}C – 70^{o}C) then what will be the thermo emf it develops between 70^{o}C and 100^{o}C
(a) 65 mV (b) 60 mV (c) 55 mV (d) 50 mV
Solution : (b) By using e^{100} = e ^{32} + e ^{70} + e^{100} Þ 200 = 64 + 76 + e^{100} Þ e^{100} = 60mV
0 0 32 70 70 70
Example: 32 A thermo couple is formed by two metals X and Y metal X comes earlier to Y in Seebeck series. If temperature of hot junction increases beyond the temperature of inversion. Then direction of current in thermocouple will so
 X to Y through cold junction (b) X to Y through hot junction
(c) Y to X through cold junction (d) Both (b) and (c)
Solution : (d) In the normal condition current flows from X to Y through cold. While after increasing the temperature of hot junction beyond temperature of inversion. The current is reversed i.e. X to Y through hot junction or Y to X through cold junction.
Example: 33 Peltier coefficient of a thermo couple is 2 nano volts. How much heat is developed at a junction if 2.5 amp
current flows for 2 minute
(a) 6 ergs (b) 6 ´ 10^{–7} ergs (c) 16 ergs (d) 6 ´ 10^{–3} erg
Solution : (a)
H = p i t = (2 ´ 10 ^{–}^{9} ) ´ 2.5 ´ (2 ´ 60) = 6 ´ 10 ^{–}^{7} J = 6 erg
Example: 34 A thermo couple develops 40 mV/kelvin. If hot and cold junctions be at 40^{o}C and 20^{o}C respectively then the emf develops by a thermopile using such 150 thermo couples in series shall be
(a) 150 mV (b) 80 mV (c) 144 mV (d) 120 mV
Solution : (d) The temperature difference is 20^{o}C = 20 K. So that thermo emf developed E = aq = 40 mV ´ 20K = 800mV.
K
Hence total emf = 150 ´ 800 = 12 ´ 10^{4} mV = 120 mV
Current can produce or speed up chemical change, this ability of current is called chemical effect (shown by dc
not by ac).
When current is passed through an electrolyte, it dissociates into positive and negative ions. This is called chemical effect of current.
Important Terms Related to Chemical Effect.
 Electrolytes : The liquids which allows the current to pass through them and also dissociates into ions on passing current through them are called electrolytes g. solutions of salts, acids and bases in water, etc.
Note : @ These liquids which do not allow current to pass through them are called insulators (e.g. vegetable oils, distilled water etc.) while the liquids which allows the current to pass through them but do not dissociates into ions are called good conductors (e.g. Hg etc.)
@ Solutions of cane sugar, glycerin, alcohol etc. are examples of nonelectrolytes.
 Electrolysis : The process of decomposition of electrolyte solution into ions on passing the current through it is called electrolysis.
Electric current
Note : @Practical applications of electrolysis are Electrotyping, extraction of metals from the ores, Purification of metals, Manufacture of chemicals, Production of O_{2} and H_{2}, Medical applications and electroplating.
@ Electroplating : It is a process of depositing a thin layer of one metal over another metal by the method of electrolysis. The articles of cheap metals are coated with precious metals like silver and gold to make their look more attractive. The article to be electroplated is made the cathode and the metal to be deposited in made the anode. A soluble salt of the precious metal is taken as the electrolyte. (If gold is to be coated then auric chloride is used as electrolyte).
 Electrodes : Two metal plates which are partially dipped in the electrolyte for passing the current through the electrolyte.
Anode : Connected to positive terminal of battery
Cathode : Connected to negative terminal of battery
 Voltameter : The vessel in which the electrolysis is carried out is called a It contains two electrodes and electrolyte. It is also known as electrolytic cell.
 Equivalent weight : The ratio of the atomic weight of an element to its valency is defined as it’s equivalent
 Types of voltameter : Voltameter is divided mainly in following types
Cuvoltameter Ag voltameter Water voltameter
In copper voltameter, electrolyte is solution of copper e.g. CuSO_{4}, CuCl_{2}, Cu(NO_{3})_{2} etc. Cathode may be of any material, but anode must be of copper.
CuSO_{4} in water dissociates as follows

CuSO_{4} ® Cu^{++} + SO ^{–} ^{–}
Cu^{++} moves towards cathode and takes 2 electron to become neutral and deposited on cathode
Cu^{++} + 2e ® Cu
SO_{4}^{–} ^{–} moves towards anode and looses 2 electrons their. Copper is deposited on the cathode and an equivalent amount of copper is lost by the anode, but the concentration of copper sulphate solution remains the same. In this process, two
In silver voltameter electrolyte is a solution of silver, e.g. AgNO_{3}. Cathode may be of any metal but anode must be of silver. The dissociation reaction is as follows
AgNO_{3} ® Ag^{+} + NO_{3}^{–}
The silver dissolves from the anode gets deposited on the cathode. During this process, the concentration of the electrolyte remains unchanged. In this process one electron per reaction is active and valence of Ag atom is also one.
In water voltameter the electrolyte used is acidic water, because it is much more conducting than that of pure water. So acid CH_{2}SO_{4} increases the concentration of free ions in the solution. The electrodes are made of platinum, because it does not dissolve into electrolyte and does not react with the products of electrolysis. When current flows through the electrolyte, hydrogen gas is collected in the tube placed over the cathode (– ve electrode) and oxygen is collected in the tube placed over the anode (+ve electrode).
Hydrogen and oxygen are liberated in the proportional in which they are found in water i.e. the volume ratio of hydrogen and oxygen is 2 : 1.
electrons per reaction are active and valence of copper atom is also two.
R_{h}
 – +
R_{h}
Anode
AgNO_{3} solution
Ag
O_{2} H_{2}
Cu lost
A Cu C
plates
Cu solution
Cu deposited
Cathode
+ –
– A +
R_{h}
Faraday’s Law of Electrolysis.
 First law : It states that the mass of substance deposited at the cathode during electrolysis is directly proportional to the quantity of electricity (total charge) passed through the
Let m be the mass of the substance liberated, when a charge q is passed through the electrolyte. Then,
according to the Faraday’s first law of electrolysis m µ q or m = zq, where the constant of proportionality z is called
electrochemical equivalent (E.C.E.) of the substance. If a constant current i is passed through the electrolyte for time
t, then the total charge passing through the electrolyte is given by q = i t
Therefore we have m = zit . If q = 1 coulomb, then we have m = z ´ 1 or z = m
Hence, the electrochemical equivalent of substance may be defined as the mass of its substance deposited at the cathode, when one coulomb of charge passes through the electrolyte.
S.I. unit of electrochemical equivalent of a substance is kilogram coulomb^{–1} (kgC^{–1}).
 Second law : If same quantity of electricity is passed through different electrolytes, masses of the substance deposited at the respective cathodes are directly proportional to their chemical
Let m be the mass of the ions of a substance liberated, whose chemical equivalent is E. Then, according to
Faraday’s of electrolysis, m µ E
or m = constant ´ E or
m = constant
E
Note : @ Chemical equivalent E also known as equivalent weight in gm i.e.
E = Atomic mass (A)
Valance (V)
 Relation between chemical equivalent and electrochemical equivalent : Suppose that on passing same amount of electricity q through two different electrolytes, masses of the two substances liberated are m_{1} and
m_{2}. If E_{1}
and E_{2}
are their chemical equivalents, then from Faraday’s second law, we have m1
m2
= E1 E2
Further, if z_{1} and z_{2} are the respective electrochemical equivalents of the two substances, then from Faraday’s
first law, we have m_{1}
= z1q and m2
= z2 q Þ
m1 = z1 m2 z 2
So from above equation
z1 = E1 z 2 E2
Þ z µ E
Þ z 2
= z1
 E2
E1
 Faraday constant : As we discussed above
E µ z
Þ E = Fz
Þ z = E = A
F VF
‘F’ is proportionality constant called Faraday’s constant.
As z = E and z = m (from I law) so E = m
hence if Q = 1 Faraday then
E = m
i.e. If electricity supplied to
F Q F Q
a voltameter is 1 Faraday then amount of substance liberated or deposited is (in gm) equal to the chemical equivalent. e.g. to deposit 16 gm O_{2}; 2 Faraday electricity is required.
Note : @Remember Number of gm equivalent = given mass ´ valency
atomic mass
@ 1 Faraday = 96500 C
@ Also F = Ne {where N = Avogrado number)
Electro Chemical Cell.
It is an arrangement in which the chemical energy is converted into electrical energy due to chemical action taking place in it. The total amount of energy that can be provided by this cell is limited and depends upon the amount of reactants. Electro chemical cells are of two types.
 Primary cell : Is that cell in which electrical energy is produced due to chemical energy. In the primary cell, chemical reaction is This cell can not be recharged but the chemicals have to be replaced after a long use examples of primary cells; Voltaic cell, Daniel cell, Leclanche cell and Dry cell etc.
 Voltaic cell
Polarisation
Cu Zn
+ –
Electrolyte Dil. H_{2}SO_{4}
Local action
Positive electrode – Cu rod
Negative electrode – Zn rod
Electrolyte – dil. H_{2}SO_{4}
Emf – 1.08 V
Main chemical reactions

H SO ® 2H^{+} + SO ^{–} ^{–}
2 4 4
 Daniel cell
Cu
Crystals
 Lechlanche cell
C
– Zn
Rod
A +
Cu Pot
CuSO_{4} Solution (Depolariser)
Porous pot Dil. H_{2}SO_{4}
(electrolyte)
Zn ® Zn^{++} + 2e^{–}
Positive electrode – Cu rod
Negative electrode – Zn rod
Electrolyte – dil. H_{2}SO_{4}
Emf – 1.1 V
Main chemical reactions
Zn + H_{2}SO_{4} ® ZnSO_{4} + 2H^{+} + 2e^{–} 2H^{+} + CuSO_{4} ® H_{2}SO_{4} + Cu^{++}
Positive electrode – Carbon rod
Zn rod C A
– +
Graphite rod Porous pot
Negative electrode – Zn rod
Electrolyte – NH_{4}Cl solution
Glass pot
 Dry cell
C + A
–
MnO_{2} + charcoal dust (depolariser)
NH_{4}Cl Solution (Electrolyte)
Graphite rod
Zn pot Porous pot
MnO_{2} + charcoal dust (depolariser)
NH_{4}Cl Solution (Electrolyte)
Emf – 1.45 V
Main chemical reactions
Zn + 2NH_{4}Cl ® 2NH_{3} + ZnCl_{2} + 2H^{+} + 2e^{–}
2H^{+} + 2MnO_{2} ® Mn_{2}O_{3} + H_{2}O + 2 unit of positive
charge
Positive electrode – Carbon rod with brass cap
Negative electrode – Zn vessel
Electrolyte – Paste of NH_{4}Cl
and saw dust
Emf – 1.5 V
Main chemical reactions – Similar to
Leclanche cell
 Secondary cell : A secondary cell is that cell in which the electrical energy is first stored up as a chemical energy and when the current is taken from the cell, the chemical energy is reconverted into electrical energy. In the secondary cell chemical reaction are reversible. The secondary cells are also called storage cell or accumulator. The commonly used secondary cells are
In charged Lead accumulator Alkali accumulator
+ –
Glass vessel
PbO_{2}
Pb
dil. H SO
+ – Ni(OH)_{2} Fe(OH)_{2}
Perforated steel grid
KOH 20%
2 4 + Li(OH), 1%
Positive electrode Perforated lead plates coated with PbO_{2} Perforated steel plate coated with Ni(OH)_{4}
Negative electrode Perforated lead plates coated with pure lead Perforated steel plate coated with Fe
During charging Chemical reaction
At cathode : PbSO_{4} + 2H^{+} + 2e^{–} ® Pb + H_{2}SO_{4}
At anode :
Chemical reaction
At cathode :
Fe(OH)_{2} + 2OH^{+} – 2e^{–} ® Ni(OH)_{4}

PbSO_{4} + SO ^{–} ^{–}
+ 2H_{2}O – 2e^{–}
® PbO_{2} + 2H_{2}SO_{4}
At anode :
During discharging
Specific gravity of H_{2}SO_{4} increases and when specific gravity becomes 1.25 the cell is fully charged.
Emf of cell : When cell is full charged them E = 2.2 volt
Chemical reaction
At cathode : Pb + SO_{4}^{–} ^{–} – 2e^{–} ® PbSO_{4}
At anode :
PbO_{4} + 2H^{+} – 2e^{–} + H_{2}SO_{4} ® PbSO_{2} + 2H_{2}O
Specific gravity of H_{2}SO_{4} decreases and when specific gravity falls below 1.18 the cell requires recharging.
Emf of cell : When emf of cell falls below 1.9 volt
the cell requires recharging.
Fe(OH)_{2} + 2K^{+} + 2e^{–} ® Fe + 2KOH
Emf of cell : When cell is fully charge then E
= 1.36 volt Chemical reaction At cathode :
Fe + 2OH^{–} – 2e^{–} ® Fe(OH)_{2}
At anode :
Ni(OH)_{4} + 2K^{+} + 2e^{–} ® Ni(OH)_{2} + 2KOH
Emf of cell : When emf of cell falls below 1.1 V
it requires charging.
Efficiency 80% 60%
 Defects In a primary cell : In voltaic cell there are two main defects
Local action : It arises due to the presence of impurities of iron, carbon etc. on the surface of commercial Zn rod used as an electrode. The particles of these impurities and Zn in contact with sulphuric acid form minute voltaic cell in which small local electric currents are set up resulting in the wastage of Zn even when the cell is not sending the external current.
Removal : By amalgamating Zn rod with mercury (i.e. the surface of Zn is coated with Hg).
Polarisation : It arises when the positive H_{2} ions which are formed by the action of Zn on sulphuric acid, travel towards the Cu rod and after transferring, the positive charge converted into H_{2} gas atoms and get deposited in the form of neutral layer of a gas on the surface of Cu rod. This weakens the action of cell in two ways.
Removal : Either by brushing the anode the remove the layer or by using a depolariser (i.e. some oxidising agent MnO_{2}, CuSO_{4} etc which may oxidise H_{2} into water).
Note : @The end point voltage of dry cell is 0.8 V.
Example: 35 In an electroplating experiment, m gm of silver is deposited when 4 ampere of current flows for 2 minute. The amount (in gm) of silver deposited by 6 ampere of current for 40 second will be [MNR 1991; MP PET 2002] (a) 4 m (b) m/2 (c) m/4 (d) 2m
Solution : (b) By using m = zit Þ m1 = i1t1 Þ m
= 4 ´ 2 ´ 60 Þ m2 = m / 2
m2 i2t2 m2
6 ´ 40
Example: 36 A current of 16 ampere flows through molten NaCl for 10 minute. The amount of metallic sodium that appears at the negative electrode would be [EAMCET 1984] (a) 0.23 gm (b) 1.15 gm (c) 2.3 gm (d) 11.5 gm
Solution : (c) By using m = zit = A
VF
it Þ m = 23 ´ 16 ´ 10 ´ 60 = 2.3 gm
1 ´ 96500
Example: 37 For depositing of 1 gm of Cu in copper voltameter on passing 2 amperes of current, the time required will be (For copper Z = 0.00033 gm/C)
(a) Approx. 20 minutes (b) Approx. 25 minutes (c) Approx. 30 minutes (d) Approx. 35 minutes Solution : (b) By using m = zit Þ 1 = 0.00033 ´ 2 ´ t Þ t = 1515.15 sec » 25 min.
Example: 38 Two electrolytic cells containing CuSO_{4} and AgNO_{3} respectively are connected in series and a current is passed through them until 1 mg of copper is deposited in the first cell. The amount of silver deposited in the second cell during this time is approximately (Atomic weights of copper and Silver are respectively 63.57 and 107.88) [MP PMT 1996]
(a) 1.7 mg (b) 3.4 mg (c) 5.1 mg (d) 6.8 mg
Solution : (b) By using m1
m2
= E1 Þ 1
E2 m2
= 63.57 / 2
107.88 / 1
= 31.7
107.88
Þ m2
= 3.4 mg
Example: 39 When a copper voltameter is connected with a battery of emf 12 volts, 2 gms of copper is deposited in 30 minutes. If the same voltameter is connected across a 6 volt battery, then the mass of copper deposited in 45 minutes would be [SCRA 1994]
(a) 1 gm (b) 1.5 gm (c) 2 gm (d) 2.5 gm
Solution : (b) By using m = zi t = z V t
R
Þ m1 m2
= V_{1}t_{1} Þ 2
V2 t 2 m2
= 12 ´ 30 Þ m

6 ´ 45
= 1.5 gm
Example: 40 Silver and copper voltameter are connected in parallel with a battery of e.m.f. 12 V. In 30 minutes, 1 gm of silver and 1.8 gm of copper are liberated. The energy supplied by the battery is (Z_{Cu} = 6.6 × 10^{–4} gm/C and Z_{Ag} = 11.2 × 10^{–4} gm/C) [IIT 1975]
(a) 24.13 J (b) 2.413 J (c) 0.2413 J (d) 2413 J


Solution : (a) By using m = z i t, for Ag voltameter 1 = 11.2 ´ 10^{–4} ´ i_{1} ´ 30 ´ 60 Þ i_{1} = 0.5 amp.
1

For Cu voltameter 1.8 = 6.6 ´ 10^{–4} ´ i_{2} ´ 30 ´ 60 Þ i_{2} = 1.5 amp
Main current i = i_{1}
+ i_{2}
= 1.5 + 0.5 = 2A.
i i2
12 V
So energy supplied = Vi = 12 ´ 2 = 24 J
Example: 41 Amount of electricity required to pass through the H_{2}O voltameter so as to liberate 11.2 litre of hydrogen will be
 1 Faraday (b)
1 Faraday
2
 2 Faraday (d) 3 Faraday
Solution : (a) Mass of hydrogen in 11.2 litres of hydrogen = æ 11.2 ö ´ M = æ 11.2 ö ´ 2 = 1gm
ç 22.4 ÷ ç 22.4 ÷
è ø è ø
We know that 1 gm of hydrogen is equal to 1 gm equivalent wt. of hydrogen. It means that 11.2 litre of
hydrogen at NTP represents 1 gm equivalent of hydrogen, so for liberation it requires 1 Faraday electricity.
Example: 42 Amount of electricity required to liberate 16 gm of oxygen is
 1 Faraday (b) 2 Faraday (c)
1 Faraday
2
 3 Faraday
Solution : (b) Number of gm equivalent =
Given mass
gm equivalent weight
= 16
16 / 2
= 2. Hence 2 Faraday electricity is needed.
Example: 43 Total surface area of a cathode is 0.05 m^{2} and 1 A current passes through it for 1 hour. Thickness of nickel deposited on the cathode is (Given that density of nickel = 9 gm/cc and it’s ECE = 3.04 ´ 10^{–}^{4} gm/C)
 4 m (b) 2.4 cm (c) 2.4 mm (d) None of these
Solution : (c) Mass deposited = density ´ volume of the metal
m = r ´ A ´ x …… (i)
Hence from Faraday first law m = Zit ……(ii)
So from equation (i) and (ii) Zit = r ´ Ax Þ x = Zit = 3.04 ´ 10 4 ´ 10 3 ´ 1 ´ 36 ´………………….. = 2.4 ´ 10 ^{–}^{6} m = 2.4mm
rA 9000 ´ 0.05
Example: 44 Resistance of a voltameter is 2 W, it is connected in series to a battery of 10 V through a resistance of 3 W. In a certain time mass deposited on cathode is 1 gm. Now the voltameter and the 3W resistance are connected in parallel with the battery. Increase in the deposited mass on cathode in the same time will be
(a) 0 (b) 1.5 gm (c) 2.5 gm (d) 2 gm
Solution : (b) Remember mass of the metal deposited on cathode depends on the current through the voltameter and not on the
current supplied by the battery. Hence by using m = Zit, we can say
Hence increase in mass = 2,5 – 1 = 1.5 gm
3W
mParallel
mSeries
= iParallel Þ m
iSeries
Parallel
= 5 ´ 1 = 2.5gm .
2
i1 i1 = 10 = 2A
5
i2 = 10 = 5 A
2