Electrochemistry Important Questions

Download CBSE Class 12th Standard CBSE Chemistry question papers, sample papers, important questions, and previous year solved papers in PDF format. Get free study materials, NCERT solutions, and exam preparation resources for Class 12th Standard CBSE Chemistry

Questions:

Generic 2

5 x 2 = 10

Write the cell reaction of a lead storage battery when it is discharged. How does the density of the electrolyte change when the battery is discharged?

State the products of electrolysis obtained on the cathode and the anode in the following cases :
(i) A dilute solution of H₂ SO₄ with platinum electrodes
(ii) An aqueous solution of AgNO₃ with silver electrodes.

The same quantity of electrical charge deposited 0.583 g of Ag when passed through AgNO_3, AuCl₃ solution. Calculate the weight of gold formed.
(At weight of Au = 197 g mol^-1).

Ezpress the relation among cell constant, resistance of the solution in the cell and conductivity of the solution. How is molar conductivity of solution related to its conductivity.

How many Faradays of charge are required to convert:
1 mole of ${ MnO }_{ 4 }^{ - }$ to Mn²⁺ion,

Generic 3

10 x 3 = 30

One half-cell in a voltaic cell is constructed from a silver wire dipped in silver nitrate solution of unknown concentration. The other half-cell consists of a zinc electrode in a 0.10 M solution of Zn(N0₃)₂. A voltage of 1. 48 V is measured for this cell.Use this information to calculate the concentration of silver nitrate solution
[ Given $E^{ 0 }_{ Zn2+/zn }=-0.763V\quad E^{ 0 }_{ Ag+/ag }=+0.80V]$ ]

Write the reactions taking place at cathode and anode in lead storage battery when the battery is in use. What happens on charging the battery?

Calculate the emf of the following cell at 25° C:
$Fe|{ Fe }^{ 2+ }(0.01M)\parallel { H }^{ + }\left( 0.01M \right) { H }_{ 2 }(g)(1bar)|Pt(s)$
${ E }^{ ° }\left( { Fe }^{ 2+ }|Fe \right) =-0.44V{ E }^{ ° }\left( { H }^{ + }|{ H }_{ 2 } \right) =0.00V$

Conductivity of $2.5\times { 10 }^{ -4 }$M methanoic acid is $5.25\times { 10 }^{ -5 }S{ cm }^{ -1 }$. Calculate its molar conductivity and degree of dissociation.
Given: ${ \lambda }^{ ° }\left( { H }^{ + } \right) $ = 349.5 S cm^-2mol^-1 and ${ \lambda }^{ ° }\left( { HCOO }^{ - } \right) $ = 50.5 S cm²mol^-1

10.

Depict the galvanic cell in which the reaction
$Zn(s)+2{ Ag }^{ + }(aq)\longrightarrow { Zn }^{ 2+ }(aq)+2Ag(s)$ takes place. Further, show
(i) Which of the electrodes is negatively charged ?
(ii) The carriers of the current in the cell.
(iii) Individual reaction at each electrode.

11.

Cr₂O^2-₇+ 14 H⁺ + 6 e^- $\longrightarrow$ Cr⁺⁺⁺+7 H₂O, E^o = 1.33 V; 3 $\times$ [2 I^- $\longrightarrow$ I₂ + 2 e^-], E^o= -0.54 V Find out the value of the equilibrium constant and Gibbs free energy change in the reaction given above.

12.

At what pH of HCI solution will hydrogen gas electrode show electrode potential of -0.118 V? H₂ gas is bubbled at 298 K and 1 atm pressure.

13.

A solution of CuSO₄ is electrolysed for 10 minutes with a current of 1-5 amperes. What is the mass of copper deposited at the cathode? (Molar mass of Cu = 63.5 g mol^-1)

14.

How long will it take an electric current of 0.15 A to deposit all the copper from 500 ml of 0.15 M copper sulphate solution?

15.

A strip of nickle metal is placed in a 1 molar solution of Ni(NO₃)₂ and a strip of silver metal is placed in a 1-molar solution of AgNO₃. An electrochemical cell is created when the two solutions are connected by a salt bridge and the two strips are connected by wires to a voltameter.
(i) Write the balanced equation for the overall reaction occuring in the cell and calculate the cell potential.
(ii) Calculate the cell potential, E, at 25 ^oC for the cell if the initial concentration of Ni(NO₃)₂ is 0.100 molar and the initial concentration of AgNO₃ is 1.00 molar.
[${ E }_{ { Ni }^{ 2 }/Ni }^{ o }=-0.25V;{ E }_{ { Ag }^{ + }/Ag }^{ o }=0.80V;{ log10 }^{ -1 }=-1$]

Generic 5

7 x 5 = 35

16.

(a) State Faraday's first law of electrolysis. How much charge in terms of Faraday's is required for the reduction of 1 mole of Cu²⁺ to Cu.
(b) Calculate emf of the following cell at 298 K.
Mg(s) IMg²⁺(0.1 M) II Cu²⁺(0.01) I Cu(s)
Given, E⁰_cell = +2.71 V,
1F = 96500 C mol^-1

17.

Calculate the standard electrode potential of Ni²⁺ II Ni electrode if emf of the cell, Ni (s)1 Ni²⁺(0.01M) IICu²⁺(0.1 M) ICu(s) is 0.059 V. (Given $E^{ 0 }_{ cu2+/cu }=+0.34V)$

18.

(a) Define the following terms:
(i) Molar conductivity
(ii) Secondary batteries
(iii) Fuel cell
(b) State the following laws
(i) Faraday first law of electrolysis
(ii) Kohlrausch's law of independent migration of ions

19.

Rahul visited the house of his friend Shyam and found that all the water taps were rusted. On enquiry, he came to know that these were iron taps. Rahul advised his friend to use either chrome plated or nickel plated taps. Shyam accepted his advice.
On the basis of above passage give the answer of following questions.
(i) Why did iron tap get rusted?
(ii) What was the purpose of chrome plating or nickel plating?
(iii) What is the value associated with this?

20.

Two students use same stock solution of ZnSO₄ and a solution of CuSO₄ . The e.m.f. of one cell is 0.03 V higher than the other. The concentration of CuSO₄ in the cell with higher e.m.f. value is 0.5 M. Find out the concentration of CuSO₄ in the other cell (2.303 RT/F = 0.06)

21.

Calculate the standard electrode potential of Cu⁺/Cu half cell. Given that the standard reduction potentials of Cu²⁺/Cu and Cu²⁺/Cu⁺ are 0.337 V and 0.153 V respectively.

22.

(a) What type of a battery is lead storage battery? Write the anode and cathode reactions and the overall cell reaction occuring in the operation of a lead storage battery.
(b) Calculate the potential for half-cell containing 0.10 M K₂Cr₂O₇(aq),0.20 M Cr³⁺(aq) and $1.0\times { 10 }^{ -4 }M{ H }^{ +\quad }(aq)$ The half-cell reaction is
${ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }(aq)+14{ H }^{ + }(aq)+{ 6e }^{ - }\rightarrow { 2Cr }^{ 3+ }(aq)+{ 7H }_{ 2 }O(l)$ and the standard electrode potential is given as E^o= 1.33 V.

Multiple Choice Question

12 x 1 = 12

23.

Which of the following are false ?

(a)

Saline water slows down rusting

(b)

In Daniell cell, if concentrations of the solutions are doubled, the emf of the cell is also doubled.

(c)

EMF of a cell is an intensive quantity whereas free energy change, $\Delta G$ is extensive.

(d)

Galvanized iron sheets remain protected from rusting even if a crack is developed

24.

The metal that cannot be obtained by electrolysis of an aqueous solution of its salt is

(a)

(b)

(c)

(d)

25.

In a cell that utilizes the reaction $Zn(s)+{ 2H }^{ + }(aq)\longrightarrow { Zn }^{ 2+ }(aq)+{ H }_{ 2 }(g)$ addition of H₂SO₄ to the cathode compartment will

(a)

lower the E and shift equilibrium to the left

(b)

lower the E and shift equilibrium to the right

(c)

increase the E and shift equilibrium to the right

(d)

increase the E and shift equilibrium to the left

26.

A button cell used in watches functions as follows :
$Zn(s)+{ Ag }_{ 2 }O(s)+{ H }_{ 2 }O(l)\rightleftharpoons 2Ag(s)+{ Zn }^{ 2+ }(aq)+2{ OH }^{ - }(aq)$
The half-cell potentials are
${ Zn }^{ 2+ }(aq)+{ 2e }^{ - }\longrightarrow Zn(s);{ E }^{ 0 }=-0.76V$
${ Ag }_{ 2 }O(s)+{ H }_{ 2 }O(l)+{ 2e }^{ - }\longrightarrow 2Ag(s)+2{ OH }^{ - }(aq);{ E }^{ 0 }=0.34V$
The cell potential will be

(a)

1.10 V

(b)

0.42 V

(c)

0.84 V

(d)

1.34 V

27.

The sequence of ionic mobility in the aqueous solution is

(a)

K⁺ > Na⁺ > Rb⁺ > Cs⁺

(b)

Cs⁺> Rb⁺ > K⁺ > Na⁺

(c)

Rb⁺ > K⁺ > Cs⁺> Na⁺

(d)

Na⁺ > K⁺ > Rb⁺ > Cs⁺

28.

Two faradays of electricity are passed through a solution of CuSO₄. The mass of copper deposited at the cathode (at mass of Cu = 63.5 amu)

(a)

2 g

(b)

127 g

(c)

0 g

(d)

63.5 g

29.

${ E }_{ cell }^{ \circleddash }=1.1V$ for Daniell cell. Which of the following expressions are correct description of state of equilibrium in this cell?

(a)

1.1 = K_c

(b)

$\frac { 2.303RT }{ 2F } \log { { K }_{ c } } =1.1$

(c)

$\log { { K }_{ c } } =\frac { 2.2 }{ 0.059 } $

(d)

$\log { { K }_{ c } } =1.1$

30.

The cell constant of a conductivity cell ------------

(a)

changes with change of electrolyte

(b)

changes with change of concentration of electrolyte

(c)

changes with temperature of electrolyte

(d)

remains constant for a cell

31.

Which of the following statement is correct ?

(a)

E_cell and ${ \Delta }_{ r }G$ of cell reaction both are extensive properties

(b)

E_cell and ${ \Delta }_{ r }G$ of cell reaction both are intensive properties

(c)

E_cell is an intensive property while ${ \Delta }_{ r }G$ of cell reaction is an extensive property

(d)

E_cell is an extensive property while ${ \Delta }_{ r }G$ of cell reaction is an intensive property

32.

For the reduction of silver ions with copper metal, the standard cell potential was found to be + 0.46 V at 25°C. The value of standard Gibbs energy, ${ \Delta G }^{ ° }$ will be (F = 96500 C mol^-1)

(a)

- 98.0 kJ

(b)

- 89.0 kJ

(c)

- 89.0 J

(d)

- 44.5 kJ

33.

A hypothetical electrochemical cell is$\overset { \circleddash }{ A } |{ A }^{ + }\left( xM \right) \parallel { { B }^{ + } }\left( yM \right) |\overset { \oplus }{ B } $ The emf measured is + 0.20 V. The cell reaction is

(a)

The cell reaction cannot be predicted

(b)

$A+{ B }^{ + }\longrightarrow { A }^{ + }+B$

(c)

${ A }^{ + }+B\longrightarrow A+{ B }^{ + }$

(d)

${ A }^{ + }+{ e }^{ - }\longrightarrow A;{ B }^{ + }+{ e }^{ - }\longrightarrow B$

34.

The time required to liberate one gram equivalent of an element by passing one ampere current through its solution is

(a)

6.7 hrs

(b)

13.4 hrs

(c)

19.9 hrs

(d)

26.8 hrs

Fill up / 1 Marks

1 x 1 = 1

35.

Electrolysis of an aqueous solution of copper sulphate using platinum electrodes produces --------- at the cathode and ------------- at the anode.

Visit QB365-SLMS - https://www.qb365.in/slms/

Answers Key:

Generic 2

5 x 2 = 10

$Pb+Pb{ O }_{ 2 }+2{ H }_{ 2 }{ SO }_{ 4 }\longrightarrow 2Pb{ SO }_{ 4 }+2{ H }_{ 2 }O$
Density of sulphuric acid solution decreases as water is formed and sulphuric acid is consumed during discharge.

${ (i) \ H }_{ 2 }{ SO }_{ 4 }(aq)\longrightarrow { 2H }^{ + }(aq)+{ { SO }_{ 4 } }^{ 2- }(aq); \ \ \ { H }_{ 2 }O\rightleftharpoons { H }^{ + }+{ OH }^{ - }$
At cathode: ${ H }^{ + }+e\longrightarrow H,H+H\longrightarrow { H }_{ 2 }(g)$
At Anode: ${ OH }^{ - }\longrightarrow OH+{ e }^{ - },4OH\longrightarrow 2{ H }_{ 2 }O(l)+{ O }_{ 2 }(g)$
Thus, H₂ is liberated at the cathode and 02 at the anode
${ (ii) \ Ag }{ NO }_{ 3 }(s)+aq\longrightarrow { Ag }^{ + }(aq)+{ { NO }_{ 3 } }^{ - }(aq);$
At cathode. Ag⁺ ions have lower discharge potential than H⁺ ions. Hence, Ag⁺ ions will be deposited on the cathode
$\left( { Ag }^{ + }+{ e }^{ - }\longrightarrow Ag \right) $
At anode. Ag anode will be attacked by NO³ ions. Hence, Ag anode will dissolve to form Ag⁺ ions in the solution
$\left( Ag\longrightarrow { Ag }^{ + }+{ e }^{ - } \right) $

$\frac { { W }_{ Ag } }{ Eq.wt \ of \ Ag } =\frac { { W }_{ Au } }{ Eq.wt \ of \ Au } $
Eq.wt. of Ag = 108, Eq.wt. of Au = $\frac{197}{3}$
$\frac{0.583}{108}$= $\frac { { W }_{ Au } }{ \frac { 197 }{ 3 } } $, W_Au= $\frac { 0.583\times 197 }{ 108\times 3 } $= 0.354 g

$k=\frac { 1 }{ R } \times \frac { l }{ a } $, where 'k' is conductivity of solution, 'R' is resistance, $\frac { l }{ a } $ is cell constant. ${ \Lambda }_{ m }=\frac { 1000k }{ M } $, where 'k' is conductivity in S cm^-1, 'M' is molarity of solution. ${ \Lambda }_{ m }$ is molar conductivity in S cm²mol^-1.

$\mathrm{MnO}_4^{-} \rightarrow \mathrm{Mn}^{2+}, \mathrm{Mn}^{7+}+5 e^{-} \rightarrow \mathrm{Mn}^{2+} \mathrm{MnO}_4^{-}+5 e^{-}+8 \mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O} \text { i.e. }$ when 1 mole of ${M n O}_{4}^{-}$ $\mathrm{MnO}_4^{-} \text {changes to } \mathrm{Mn}^{2+}$ 5 Faradays of charge is required.

Generic 3

10 x 3 = 30

According to given information cell can be represented as
Zn(s)|Zn²⁺(aq) (1M)||ag⁺(aq)|Ag(s)
Given $E^{ 0 }_{ cell }=E^{ 0 }_{ cathode }-E^{ 0 }anode$
= 0.80-(-0.76) = 1.56V
By using Nernst equation, we get
$E=_{ cell }E^{ 0 }_{ cell }=\frac { 0.0591 }{ 2 } log\frac { \left[ Zn^{ 2+ } \right] }{ \left[ Ag^{ + } \right] ^{ 2 } } $
$1.48=1.56-\frac { 0.0591 }{ 2 } log\frac { 1 }{ \left[ Ag^{ + } \right] ^{ 2 } } $
0.08 + 0.0295log $\frac { 1 }{ \left[ Ag^{ + } \right] ^{ 2 } } $
$\frac { 0.08 }{ 0.0295 } =2log\left[ Ag^{ + } \right] $
$\frac { 0.08 }{ 0.0295\times 2 } =log\left[ Ag^{ + } \right] $
log[Ag⁺] = -1.356
Ag⁺= Antilog[-1.356]
= 0.0441 = 4.4 x 10^-2M

The cell reactions when the battery is in use are given below:
Pb_(s)+SO₄^2-_(aq)$\rightarrow$PbSO₄_(s)+2e^-
Cathode:
PbO₂_(s)+SO₄^2-_(aq)+4H⁺_(aq)+2e^-$\rightarrow$4H⁺_(aq)+2H₂O_(l)
i.e., Overall cell reaction consisting of cathode and anode reactions is
Pb_(s)+PbO_2(s)+2H₂SO_4(aq)$\rightarrow$2PbSO₄_(s)+2H₂O_(l)
On charging the battery, the electrode reactions are reverse of those that occur during discharge.

Cell reaction is
${ Fe }_{ (s) }+{ 2{ H }^{ + } }_{ (aq) }\rightarrow { { Fe }^{ 2+ } }_{ (aq) }+{ H }_{ 2(g) }$
${ { E }^{ ° } }_{ cell }=0.00-(-0.44)=0.44V$
${ E }_{ cell }={ { E }^{ ° } }_{ cell }-\frac { 0.0591 }{ 2 } log\frac { \left[ { Fe }^{ 2+ } \right] }{ { \left[ { H }^{ + } \right] }^{ 2 } } $
$=0.44-\frac { 0.0591 }{ 2 } log\frac { 0.001 }{ { \left( 0.01 \right) }^{ 2 } } $
=0.44 - 0.02955 = 0.41045 V

K = 5.25 x 10^-5 S cm^-1, M = 2.5 x 10^-4 M.
$\lambda_{\mathrm{HCOOH}}^{0}=\lambda_{\left(\mathrm{HCOO}^{-}\right)}^{0}+\lambda_{\mathrm{H}}^{0}+$
= 349.5 + 50.5 : 400 S cm² mol^-1.
$\Lambda_{m}=\frac{1000 \kappa}{\mathrm{M}}=\frac{1000 \times 5.25 \times 10^{-5}}{2.5 \times 10^{-4}}$
= $\frac{1000 \times 525}{10 \times 2.5 \times 100}$
$\Rightarrow \quad \Lambda_{m}=\frac{525}{2.5}=\frac{5250}{25}=210 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$
$\alpha=\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{0}}=\frac{210}{400}=\frac{21}{40}=0.525$
$\Rightarrow \ \alpha$ = 0.525 x 100% = 52.5%

10.

The cell will be represented as:
Zn (s) I Zn²⁺(aq) II Ag²⁺ (aq) I Ag (s)
(i) Anode, i.e., zinc electrode will be negatively charged.
(ii) The current will flow from silver to copper in the external circuit.
(iii) At Anode: Zn (s)⇾ Zn2+(aq) + 2 e^-
At Cathode: Ag⁺ (aq) + e ⇾ Ag

11.

K = 1.596 $\times 10^{80},\triangle G ^{o}$ = 457.41 kJ mol^-1.

12.

pH = 2.

13.

0.296 g.

14.

500 ml of 0.15 M CuSO₄ solution contains $\frac { 500\times 0.15 }{ 1000 } $
Mass of Cu = 0.075 x 63.5 = 4.7625 g; Eq. wt. of Cu2⁺ =$\frac{63.5}{2}$=31.75
m = Z $\times$ l $\times$ t
4.7625 = $\frac{31.75}{96500}$$\times$0.15 $\times$ t
t= $\frac { 4.7625\times 96500 }{ 31.75\times 0.15 } $ = 96500 sec = $\frac { 96500 }{ 60\times 60 } $ = 26.80 hours.

15.

$Λ_m={1000\times k\over m}$
$138.9\ S\ cm^2 mol^{-1} ={1000\times k\over 1.5}$
$k={138.9\times1.5\over 1000}={208.05\over 1000}\times10^{-1}\ S \ cm^{-1}$

Generic 5

7 x 5 = 35

16.

(a) Faraday's first law of electrolysis It states that the amount of chemical reaction occurring at an electrode by passing current is proportional to the quantity of electricity passed through the electrolyte. Charge required for the reduction of 1 mole of Cu²⁺ to Cu = 2F
(b) $Cu^{ 2+ }Mg\rightarrow Mg^{ 2+ }+Cu$
Given, E^_O _Cell= +271 V
By using Nernst equation,
$E_{ cell }=E^{ 0 }_{ cell }-\frac { 0.059 }{ n } log\frac { Mg^{ 2+ } }{ Cu^{ 2+ } } $
Here, n = 2 and E⁰ _cell = + 2.71 V
$\therefore E_{ cell }=2.71-\frac { 0.059 }{ 2 } log\frac { 0.1 }{ 0.01 } $
$=2.71-\frac { 0.059 }{ 2 } log10$
= 2.71- 0.0295
$[\therefore $ log₁₀ = 1]
= 2.68 V

17.

(i) First, find E⁰_cell
(ii) Then find $E^{ 0 }_{ anode }$ by using the formula
$E^{ 0 }_{ cell }=E^{ 0 }-_{ cathode }E^{ 0 }_{ anode }$
Given E_cell = 0.059V; E⁰_cu2+/cu = +0.34V
[Ni²⁺] = 0.01M and [cu²⁺] = 0.1M
E_cell = E^₀_cell - $\frac { 0.059 }{ 2 } log\frac { \left[ Ni^{ 2+ }(aq) \right] }{ \left[ Cu^{ 2+ }(aq) \right] } $
$0.059=E^{ 0 }_{ cell }-\frac { 0.059 }{ 2 } log\frac { \left( 0.01 \right) }{ 0.1 } \left( \because n=2 \right) $
$0.059=E^{ 0 }_{ cell }-\frac { 0.059 }{ 2 } log\frac { 1 }{ 10 } \left[ \because log10^{ -1 }=-1 \right] $
$\because 0.059=E^{ 0 }_{ cell }+0.0295 \times 1$
$E^{ 0 }_{ cell }=0.059-0.0295$
= 0.0295 V = 0.03 V
$E^{ 0 }_{ cell }=E^{ 0 }_{ cathode }E^{ 0 }_{ anode }$
0.03 = 0.34 - E⁰_anode

or$E^{ 0 }_{ anode }=E^{ 0 }_{ Ni2+/Ni }=0.34-0.03$
_{= 0.31 V}

18.

(a) (i) Molar conductivity of a solution at a given concentration is the conducting power of all the ions produced by 1 mol of an electrolyte.
(ii) Secondary battery can be recharged by passing current through it in opposite direction so that it can be used again.
(iii) Galvanic cells that are designed to convert the energy of combustion of fuels like hydrogen, methane, methanol, etc. directly into electrical energy are called fuel cells.
(b) (i) Faraday's first law of electrolysis states that the amount of chemical reaction which occurs at any electrode during electrolysis by current is proportional to the quantity of electricity passed through the electrolyte (solution or melt).
(ii) According to Kohlrausch law of independent migration of ions limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte

19.

(i) Iron is p e to rust. It gets rusted when kept in open because it gets oxides in presence air and moisture.
(ii) The purpose of depositing a layer of chromium or nickel on the surface of iron is to check rusting. These metals are not affected by air or moisture.
(iii) (a) General awareness
(b) Friendship
(c) Skill of applying knowledge of chemistry

20.

The two cells may be represented as
Zn I Zn²⁺(cone = c) II Cu²⁺(c = ?) I Cu, EMF = E₁
Zn I Zn²⁺(cone = c) II Cu²⁺(0·5 M) I Cu, EMF = E₂
The cell reaction is : Zn + Cu²⁺⇾ Zn²⁺+ Cu
$E_1=E^0-{2.303\ RT\over 2F}log{c\over [Cu^{2+}]}$
$E_2=E^0-{2.303\ RT\over 2F}log{c\over 0.5}$
$E_2-E_1={2.303\ RT\over 2F}\left( log{c\over [Cu^{2+}]}-log{c\over 0.5}\right)=0.03V$
${0.06\over 2}log{0.5\over [Cu^{2+}]}=0.03\ or\ log{0.5\over [Cu^{2+}]}=1\ or\ {0.5\over {[Cu^{2+}]}}=10\ or\ [Cu^{2+}]={0.5\over 10}=0.05M$

21.

Cu⁺ + e^-⇾ Cu; ∆G^o₃=?
(i)------(ii) gives the required result, i.e.,ΔG₃^° = ∆G⁰₁ - ΔG₂^° = [-0.674 - (- 0·153)] F = - 0·521 F
$-nFE^0_{Cu^+/CU}=-0.5621F\ or\ E^0_{CU^+/Cu}=0.521V$

22.

(a) Lead storage battery: It is a secondary cell. It consists of a lead anode and a grid of lead dioxide as cathode. A solution of sulphuric acid (38% by mass) is used as electrolyte
The cell reactions when the battery is in use are given below:
Anode : Pb(s) + SO₄^2-(aq) ⇾ Pb504(s) + z«
Cathode: PbO₂ (s) +SO₄^2- (aq) + 4H⁺(aq) + 2e^- ⇾ PbSO₄+2H₂O(I) i.e. overall cell reaction consisting of cathode and anode reactions is :
Pb (s) + PbO₂ (s) + 2H₂SO₄ (oq) ⇾ 2PbSO₄ (s) + 2H₂O(I)
On recharging the battery, the reaction is reversed.
2PbSO₄ + 2H₂O ⇾ Pb(s) + PbO₂(s) + 4H⁺ + 2SO₄^2-
(b)$E_{cell}=E^0_{cell}-{0.0591V\over 6}log{[Cr^{3+}]^2\over [Cr_2O_7^{2-}][H^+]^{14}}$
$=1.33V-{0.0591V\over 6}log{(0.2)^2\over (0.1)(1.0\times10^{-4})^{14}}$
$=1.33V-{0.0591V\over 6}log{4\times 10^{-2}\over 1.0\times10^{-57}}$
$=1.3V-{0.0591V\over 6}log4\times 10^{55}$
$=1.33V-{0.0591V\over6}\times 55.6021$
$=1.33V-{3.286V\over6}=1.33V-0.5476V=0.7824V$