Vidaara.orgClass 12 · Chemistry
CodeVID-C12-03-CH-01
Electrochemistry — Full Chapter Assignment
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- All questions are compulsory.
- Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks.
- Show all steps in numericals, write balanced electrode reactions and draw neat labelled diagrams where asked. For full solutions, raise your doubts with your teacher.
Section A — Multiple Choice Questions
6 × 1 = 6 marks
1.
In a galvanic cell the cathode is the electrode where:
- A.oxidation occurs
- B.reduction occurs
- C.no reaction occurs
- D.ions are lost
2.
The standard EMF of the Daniell cell is:
- A.$0.76\,\text{V}$
- B.$1.10\,\text{V}$
- C.$0.34\,\text{V}$
- D.$2.0\,\text{V}$
3.
Conductivity on dilution:
- A.increases
- B.decreases
- C.is unchanged
- D.doubles
4.
Kohlrausch's law gives $\Lambda_m^0$ as the sum of:
- A.resistances
- B.ionic molar conductivities
- C.cell constants
- D.concentrations
5.
One Faraday equals:
- A.$96500\,\text{C}$
- B.$6.02 \times 10^{23}$
- C.$1.6 \times 10^{-19}\,\text{C}$
- D.$9650\,\text{C}$
6.
Which is a rechargeable cell?
- A.dry cell
- B.mercury cell
- C.lead storage battery
- D.Leclanché cell
Section B — Short Answer (2 marks)
4 × 2 = 8 marks
7.
Write the Nernst equation for a single electrode at 298 K.
8.
Define molar conductivity and give its unit.
9.
Calculate the mass of copper deposited by 9650 C ($M = 63.5$, $n = 2$).
10.
Why is the salt bridge used in a galvanic cell?
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
11.
State and explain Faraday's two laws of electrolysis.
12.
Explain the electrochemical mechanism of rusting and two methods of prevention.
Section D — Long Answer (5 marks)
2 × 5 = 10 marks
13.
Describe the Daniell cell, write its cell representation and the Nernst equation for it, and calculate the EMF when $[\text{Zn}^{2+}] = 0.1\,\text{M}$ and $[\text{Cu}^{2+}] = 1\,\text{M}$ ($E^0_{cell} = 1.10\,\text{V}$).
14.
Compare strong and weak electrolytes through the variation of molar conductivity with concentration, and state Kohlrausch's law with one application.
Answer Key
Section A — Multiple Choice Questions
- (B) reduction occurs
- (B) $1.10\,\text{V}$
- (B) decreases
- (B) ionic molar conductivities
- (A) $96500\,\text{C}$
- (C) lead storage battery
Section B — Short Answer (2 marks)
- $E = E^0 - \frac{0.059}{n}\log\frac{1}{[\text{M}^{n+}]}$ (reduction).
- $\Lambda_m = \frac{\kappa \times 1000}{C}$; unit $\text{S cm}^2\,\text{mol}^{-1}$.
- $m = \frac{63.5 \times 9650}{2 \times 96500} = 3.175\,\text{g}$.
- It completes the circuit and maintains electrical neutrality of the half-cells without mixing them.
Section C — Short Answer (3 marks)
- First: $m \propto Q$ ($m = ZIt$). Second: for equal charge, masses $\propto$ equivalent masses; $1\,F$ deposits one gram-equivalent.
- Anode $\text{Fe}\rightarrow\text{Fe}^{2+}+2e^-$, cathode reduces $\text{O}_2$; rust is $\text{Fe}_2\text{O}_3\cdot x\text{H}_2\text{O}$. Prevent by galvanising and cathodic protection.
Section D — Long Answer (5 marks)
- Zn anode, Cu cathode, salt bridge; $\text{Zn}\mid\text{Zn}^{2+}\parallel\text{Cu}^{2+}\mid\text{Cu}$; $E = 1.10 - \frac{0.059}{2}\log\frac{0.1}{1} = 1.10 + 0.0295 = 1.13\,\text{V}$.
- Strong: $\Lambda_m = \Lambda_m^0 - A\sqrt{C}$, extrapolates to $\Lambda_m^0$; weak: steep rise near zero C. Kohlrausch: $\Lambda_m^0 = \nu_+\lambda_+^0 + \nu_-\lambda_-^0$; gives $\Lambda_m^0$ of weak acids and $\alpha$.
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