# Class 12 Electrochemistry Numerical Questions

The NCERT Class 12 Electrochemistry Numerical Questions for Chapter 3 – Electrochemistry has mainly been designed to help the students in preparing well and score good marks in CBSE class 12 Chemistry paper. Further, the solutions consist of well thought or structured mathematical questions along with detailed explanations to help students learn and remember concepts easily.

The students appearing for CBSE Class 12th Chemistry Examination 2021 can go through the important numerical questions and answers for Chapter 3 – Electrochemistry to prepare themselves for the upcoming examination.

Subtopics for Class 12 Chemistry Chapter 3 – Electrochemistry

1. Electrochemical Cells
2. Galvanic Cells
1. Measurement of Electrode Potential
3. Nernst Equation
1. Equilibrium Constant from Nernst Equation
2. Electrochemical Cell and Gibbs Energy of Reaction
4. The conductance of Electrolytic Solutions
1. Measurement of the Conductivity of Ionic Solutions
2. Variation of Conductivity and Molar Conductivity with Concentration
5. Electrolytic Cells and Electrolysis
1. Products of Electrolysis
6. Batteries
1. Primary Batteries
2. Secondary Batteries
7. Fuel Cells
8. Corrosion

## Class 12 Electrochemistry Numerical Questions on Galvanic Cell, Nernst Equation and Electrochemical Series

1. Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10. (NCERT)
Solution:
If pH of solution is 10 means its [H+] ion concentration is 10-10 M.
Using, Ecell = E°cell0.059n log1[H+]
Here E°cell = 0, n =2, [H+] = 10-10
∴Ecell = 0 – 0.0592 log 1 = -0.59V
2. Calculate the standard cell potentials of galvanic cells in which the following reactions take place : (NCERT)
1. 2Cr(s) + 3Cd2+(aq) → Cr3+(aq) + 3Cd
2. Fe2+(aq) +Ag+(aq) → Fe3+(aq) + Ag(s). Calculate the ∆rG°, and equilibrium constant of the reactions.
Solution:
1. E°cell = E°cathode – E°anodg
= – 0.40 -(-0.74) = +0.34V
∆G° = -nFE° = -6 x 96500 x 0.34
= -196860J = -196.86kJ
∆G° =-2.303 RT log K
– 196860 =-2.307 x 8.314 x 298 log K
or log K = 35.5014 or K
= Antilog 34.5014 = 3.19 x 1034

2. E°cell = E°cathode – E°anodg
= 0.80 – (0.77) = +0.03V
∆G° = -nFE° = -1 x 96500 x 0.03
= -2895Jmol-1
= – 2-895 kJmol-1
∆G° = -2-303 RT log K
– 2895 Jmol-1 = -2.303 x 8.314 x 298 log K or
or K = Antilog of 0.5074 = 3.22.

3. Two half-reactions of an electrochemical cell are given below :
MnO4 (aq) + 8H+ (aq) + 5e → Mn2+ (aq) + 4H2O (I), E° = 1.51 V
Sn2+ (aq) → Sn4+ (aq) + 2e, E° = + 0.15 V.
Construct the redox reaction equation from the two half-reactions and calculate the cell potential from the standard potentials and predict if the reaction is reactant or product favoured. ( CBSE All India 2010)
The reactions can be represented at anode and at cathode in the following ways :
At anode (oxidation) :
[Sn2+ → Sn4+ (aq) + 2e ] × 5 E° = + 0.15 V
At cathode (reduction) :
[MnO4(aq) + 8H+ (aq) + 5e → Mn2+ (aq) + 4H2O (I)] × 2 E° = + 1.51 V
The Net R × M = 2MnO4(aq) + 16H+ + 5Sn2+ → 2Mn2+ + 5Sn4+ + 8H2O
Now E°cell = E°cathode – E°anode
= 1.51 – 0.15 = + 1.36 V
∴ Positive value of E°cell favours formation of product.
4. Determine the values of equilibrium constant (Kc) and ΔG° for the following reaction:
Ni(s) + 2Ag+ (aq) → Ni2+ (aq) + 2Ag(s),
E° = 1.05 V
(1F = 96500 C mol-1) (Delhi 2011)
According to the formula
ΔG° = -nFE° = – 2 × 96500 ×1.05
or ΔG° = -202650 J mol-1 = -202.65 KJ mol-1
Now ΔG° ⇒ -202650 J Mol-1
R = 8.314 J/Mol/K, T = 298 K 5. The standard electrode potential (E°) for Daniel cell is +1.1 V. Calculate the ΔG° for the reaction
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
(1 F = 96500 C mol-1). (All India 2013)
We know, ΔG° = -nF E°cell
Given : E°cell = 1.1 volt
∴ ΔG° = -2 × 96500 C mol-1 × 1.1 volt
= -212300 CV
= -212300 J = -212.3 KJ
6. The standard electrode potential for Daniell cell is 1.1 V. Calculate the standard Gibbs energy for the cell reaction. (F = 96,500 C mol-1) (Comptt. Delhi 2013)
Given : E° = 1.1V, F = 96,500 C mol-5, n = 2
Zn + Cu2 ⇌ Cu + Zn2+
Using ΔG° = -nFE° = -2 × 96500 × 1.1
= 212,300 CV mol-1
7.

## Class 12 Electrochemistry Numerical Questions on Conductance of Electrolyte Solution 1.If the specific conductance of 0.02 mol.L-1 KCI solution at 298 K is 2.48 x 10-2-1cm-1, then calculate its molar conductance.
Solution:
K = 2.48 x 10-2 cm-1, C = 0.02 mol L-1
Λm = 1000k/Cm
=2.48×10-2x1000/0.02
= 124 Scm2 mol-1

2. The resistance of 0.01(N) NaCl solution at 298 K is 200 ohm. Cell constant of the conductivity cell is unity. Calculate the equivalent conductance. 3. The conductivity of a solution containing 1 gram of anhydrous BaCl, in 200 cm3 of water has been found to be 0.0058 mho cm-1. What are the molar conductivity and equivalent conductivity of the solution (At. wt. of Ba = 137 and Cl = 35.5). 4.The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500 Ω. What is the cell constant if the conductivity of 0.001 M KCl solution at 298 K is 0.146 × 10-3 S cm-1 ? (Delhi 2011) ## Class 12 Electrochemistry Numerical Questions on Kohlrausch’s Law

1. What are weak electrolytes? Give one example. Find out molar conductivity of LiBr aqueous solution infinite dilution when joint conductance of Li-1 ion and Br-1 ion are 38.7 Scm2 mol-1 and 78.40 Scm2 mol-1 respectively.
Solution:
Weak electrolytes : These are the substances which dissociate only to a small extent.
Examples: CH3COOH,NH4OH
Λ LiBr = Λ Li+ + Λ Br
Given that,
Λ Li+ = 38.7Scm2 mol-1
Λ Br = 78.40 Scm2 mol-1
Λ LiBr = 38.7 + 78.40
Λ LiBr = 117.10Scm2mol-1

2. What are strong electrolytes? Find out the molar conductivity of aqueous solution of BaCl2 at infinite dilution when ionic conductance of Ba+2 ion and Cl ion are 127.30 Scm2 mol-1 and 76.34 Scm2 mol-1 respectively.
Solution:
Strong electrolytes : These are substances which dissociate almost completely into ions under all dilutions.
Examples : NaCl,HCl,CH3COONa
Λ BaCl2 = Λ Ba2+ + 2 Λ Cl
Given that,
Λ = 127.30Scm2+mol-1
Λ Cl-1 = 76.34Scm2+mol-1
Λ BaCl2 =127.30+2(76.34)
= 127.30 + 152.68
= 279.98 Scm2mol-1

3. calculate molar conductivity at infinite dilution of H2SO4  .Given molar conductivity at infinite dilution of HCl = 426 ,  molar conductivity at infinite dilution of NaCl = 126.5 and molar conductivity at infinite dilution of Na2SO4 =260 at 25 degree celcius. (units are Scm2/mol) 4. The molar conductivity of NaCl,HCl and CH3​COONa at infinite dilution are 126.45, 426.16 and 91 Scm2mol−1 respectively. calculate molar conductivity at infinite dilution of CH3​COOH at infinite dilution                                                                                                                                    . 5. At 291 K, the molar conductivities, at infinite dilution of NH4CI, NaOH and NaCl are 129.8, 217.4 and 108.9 ohm-1cm2 respectively. If the molar conductivity of a centinormal solution of NH4OH is 9.33 ohm-1cm2 , what is the percentage dissociation of NH4OH at this dilution ? 6. The molar conductivity of Ba(OH)2, BaCl2 and NH4CL at infinite dilution are 457.6, 240.6 and 129.8 Scm2mol−1 respectively. calculate molar conductivity at infinite dilution of NH4​OH at infinite dilution ## Class 12 Electrochemistry Numerical Questions on Faraday’s Law

1.How much electricity in terms of Faradays is required to produce 20 g of calcium from molten CaCl2? (Comptt. Delhi 2013)
Answer: Ca2+ + 2e → Ca
Thus, 1 mole of Ca i.e. 40 g of Ca requires electricity = 2F
∴ 20 g of Ca will require electricity = 1F

Similar Questions: How much electricity in terms of Faraday is required to produce

• (i) 20.0 g of Fe from molten FeCl3?
• (ii) 40.0 g of Al from molten Al2O3?

2. Calculate the mass of Ag deposited at cathode when a current of 2 amperes was passed through a solution of AgNO3 for 15 minutes.
[Given: Molar mass of Ag = 108 g mol-1 1F = 96,500 C mol-1)
(ii) Define fuel cell. (Delhi 2017)
(i) Q = I × t …(Charge = Current ∝ Time)
. = 2 × 15 × 60 = 1800 C
∵ 96500 C deposit Ag = 108 g
∴ 1800 C deposit Ag = 10896500 × 1800
= 2.0145 g

3. How many electrons flow through a metallic wire if a current of 0.5 A is passed for 2 hours? (Given: 1F = 96,500 C mol-1) (All India 2017)
I = 0.5 A
t = 2 hrs. = 2 × 60 ×60 s = 7,200 s
Q = I × t = 0.5 × 7200 = 3,600 C
96,500 C electricity flows to produce
= 6.022 × 1023 electrons
∴ 1 C electricity flows to produce 4. State Faraday’s first law of electrolysis. How much charge in terms of Faraday is required for the reduction of 1 mol of Cu2+ to Cu.

According to first law of Faraday’s “the amount of chemical reaction and hence the mass of any substance deposited/liberated at any electrode is directly proportional to the quantity of electricity passed through the electrolyte.”
The quantity of charge required for reduction of 1 mol of Cu2+
= 2 faradays (∵ Cu2+ + 2e → Cu)
= 2 × 96500 C = 193000 C

Similar Questions :  How much charge is required for the following reductions:
(i) 1 mol of Al3+ to Al?
(ii) 1 mol of Cu2+ to Cu?
(iii) 1 mol of MnO4 to Mn2+?

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