Question Set C

Electro Chemistry

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Questions:

Multiple Choice Question

12 x 1 = 12

The feasibility of a redox reaction can be predicted with the help of ________.

(a)

reduction potential

(b)

oxidation potential

(c)

electrochemical series

(d)

standard emf

The emf of the cell is measured in _______.

(a)

ohm

(b)

amperes

(c)

volts

(d)

coulomb

In Daniel cell, the charges developed by Zn/Zn²⁺ and Cu / Cu²⁺ are _______.

(a)

positive, positive

(b)

negative, negative

(c)

positive, negative

(d)

negative, positive

Calculate the standard emf of the cell, provided the standard reduction potentials of cathode and anode are -0.763 V and 0.80V.

(a)

- 1.563V

(b)

0.037V

(c)

- 0.610V

(d)

None of these

The important use of Kohlrausch's law is deducing the ______.

(a)

λ_∞ value of weak electrolyte.

(b)

λ_∞ value of strong electrolyte.

(c)

λ_∞ value of weak electrolyte.

(d)

λ_∞ value of weak electrolyte

When one coulomb of electricity is passed through an electrolytic solution the mass deposited on the electrode is equal to ______.

(a)

equivalent weight

(b)

molecular weight

(c)

electrochemical equivalent

(d)

one gram

Recharging of lead storage battery involves ______.

(a)

anode is reduced to pb

(b)

cathode is reduced to pb

(c)

cathode is oxidised to pb

(d)

anode is oxidised to pbO₂

Which among the following is the strongest reducing agent?

(a)

F₂

(b)

Cl₂

(c)

(d)

Cell equation: A + 2B^- \(\rightarrow\)A²⁺+ 2B; A²⁺ + 2e^- \(\rightarrow\)A E^o = +0.34V and log₁₀ K = 15.6 at 300K for cell reactions find E^o for B⁺ + e⁻ \(\rightarrow\) B

(a)

0.80

(b)

1.26

(c)

-0.54

(d)

-10.94

10.

A conductivity cell has been calibrated with a 0.01M, 1:1 electrolytic solution (specific conductance (k = 1.25 x 10^-3cm^-1) in the cell and the measured resistance was 800 Ω at 25^oC. The cell constant is ______.

(a)

10⁻¹ cm⁻¹

(b)

10¹ cm⁻¹

(c)

1 cm⁻¹

(d)

5.7 x 10^-12

11.

While charging lead storage battery _______.

(a)

PbSO₄ on cathode is reduced to Pb

(b)

PbSO₄ on anode is oxidised to PbO₂

(c)

PbSO₄ on anode is reduced to Pb

(d)

PbSO₄ on cathode is oxidised to Pb

12.

Faraday constant is defined as_______.

(a)

charge carried by 1 electron

(b)

charge carried by one mole of electrons

(c)

charge required to deposit one mole of substance

(d)

charge carried by 6.22 ×10¹⁰ electrons

Generic 2

5 x 2 = 10

13.

The standard reduction potentials of Fe³⁺ / Fe and Fe²⁺ / Fe electrode system are -0.035 V and -0.44 V respectively. Predict which of the two oxidations is easy. Fe³⁺/ Fe and Fe²⁺ / Fe

14.

Give the expression that relates molar conductivity and degree of dissociation.

15.

Define conductance. Give its unit

16.

The resistance of a conductivity cell is measured as 190 Ω using 0.1M KCl solution (specific conductance of 0.1M KCl is 1.3 Sm^-1). When the same cell is filled with 0.003 M sodium chloride solution, the measured resistance is 6.3KΩ. Both these measurements are made at a particular temperature. Calculate the specific and molar conductance of NaCl solution.

17.

Why does conductivity of a solution decrease on dilution of the solution.

Generic 3

4 x 3 = 12

18.

Draw a neat diagram of Leclanche cell and mark the parts.

19.

What is the oxidation and reduction half cell in a Daniel cell?

20.

Ionic conductance at infinite dilution of Al³⁺ and SO₄^2- are 189 and 160 mho cm² equiv^-1. Calculate the equivalent and molar conductance of the electrolyte Al₂(SO₄)₃ at infinite dilution.

21.

Arrange the following solutions in the decreasing order of specific conductance.
i) 0.01M KCl
ii) 0.005M KCl
iii) 0.1M KCl
iv) 0.25M KCl
v) 0.5M KCl

Generic 5

2 x 5 = 10

22.

Write an account on cell terminology.

23.

Specific conductance of 1M KNO₃ solution is oberved to be 5.55 x 10^-3mho cm². What is the equivalent conductance of KNO₃ when one litre of the solution is used?

Assertion and reason

2 x 2 = 4

24.

Assertion: The relationship between ΔG and emf of cell is ΔG = -nFE
Reason: In the Nest equation, 'n' denotes number of protons.
Codes:
a) (A) and (R) are true and (R) is the correct explanation of (A)
b) Both (A) and (R) are true but (R) does not explain (A)
c) (A) is true but (R) is false
d) Both (A) and (R) are false

25.

Assertion: Increase in temperature, decreases conductance
Reason: This is due to the increase in the attractive forces between the oppositely charged ions.
Codes:
a) (A) and (R) are true and (R) is the correct explanation of (A)
b) Both (A) and (R) are true but (R) does not explain (A)
c) (A) is true but (R) is false
d) Both (A) and (R) are false

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Answers Key:

Multiple Choice Question

12 x 1 = 12

(d)
negative, positive

(a)
- 1.563V

(a)
λ_∞ value of weak electrolyte.

(a)
anode is reduced to pb

(d)
Li

(a)
0.80

10.

R = ρ.l/A
Cell constant = R/ρ
\(= k.R (l/ \rho = K)\)
= 1.25 x 10^-3 Ω^-1 cm^-1 x 800Ω
= 1 cm^-1

11.

Charging anode:
PbSO_4(s) + 2e^- ➝ Pb_(s) + SO₄^2-_(aq)
Cathode:
PbSO_4(s) + 2H₂O_(I) ➝ PbO_2(s) + SO₄^2-_(aq) + 2e^-

12.

IF = 96500 C = charge of one mole of e^- charge of 6.022 x 10^-23 electron

Generic 2

5 x 2 = 10

13.

(i) The ion which has lower reduction I potential will be oxidised first at the anode.
(ii) Among (0.035V) and (0.44V), Fe²⁺ / Fe oxidation is easy because it has the lower reduction potential (-0.44V).

14.

\(\alpha =\frac { { \Lambda }_{ m } }{ { \Lambda }_{ m }^{ o } } \)

15.

The reciprocal of the resistance \((\frac{l}{R})\) gives the conductance of an electrolytic solution. The SI unit of conductance is Siemen (S).

16.

Given that
κ = 1.3 Sm^-1(for 0.1M KCl solution)
R = 190 Ω
\(\kappa = \frac{1}{R}(\frac{l}{A})\)
κ . R =\((\frac{l}{A})\) = (1.3 Sm^-1) (190Ω)
= 247 m^-1
\(\kappa_{(NaCl)} = \frac{1}{R_{(NaCl)}} (\frac{l}{A})\)
\(= \frac{1}{6.3 K\Omega}(247 m^{-1})\) (6.3KΩ = 6.3 x 10³Ω)
= 39.2 x 10^-3 Sm^-1
\(\Lambda_m = \frac{\kappa \times 10^{-3} mol^{-1} m^3}{M}\)
\(=\frac{39.2 \times 10 ^{-3}(Sm^{-1})10^-3 (mol^{-1} m^3)}{0.003}\)
\(\Lambda_m\) = 13.04 \(\times\) 10^-3Sm² mol^-1

17.

On dilution the concentration decreases. Conductivity decreases with decrease in concentration (or dilution) as the number of ions per unit volume that carry the current in a solution decrease on dilution.

Generic 3

4 x 3 = 12

18.

19.

Oxidation half cell : A metallic zinc strip that dips into an aqueous solution of zinc sulphate taken in a beaker.
Reduction half cell : A copper strip that dips into an aqueous solution of copper sulphate taken in a beaker.

20.

a) Equivalent conductance
\(\lambda_{\infty} \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} =\frac{1}{3} \lambda_{\infty} \mathrm{Al}^{3+}+\frac{1}{2} \lambda_{\infty} \mathrm{SO}_{4}^{2-} \)
\(=\left(\frac{1}{3} \times 189\right)+\frac{1}{2}(160) \)
\(=63+80=143 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~g} \mathrm{eq}^{-1} \)
b) Molar conduçtance
\(\mu_{\infty} \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} =2 \mu_{\infty} \mathrm{Al}^{3+}+3 \mu_{\infty} \mathrm{SO}_{4}^{2-} \)
=2(189) + 3(160)
=378 + 480
= \(858 \mathrm{~S} \mathrm{~cm}{ }^{2} \mathrm{~mol}^{-1}\)

21.

1. \(\kappa=\mathrm{C}\left(\frac{1}{\mathrm{~A}}\right)\)
2. \(\operatorname{let} \frac{l}{A}=x\)
i) \(0.01 \mathrm{M} \mathrm{KCl}: \kappa=0.01 x=10^{-2} x\)
ii) \(0.005 \mathrm{M} \mathrm{KCl}: \kappa=0.005 x=5 \times 10^{-3} x\)
iii) \(0.1 \mathrm{M} \mathrm{KCl} \quad: \kappa=0.1 x=10^{-1} x\)
iv) \(0.25 \mathrm{M} \mathrm{KCl}: \kappa=0.25 x=2.5 \times 10^{-1} x\)
v) \(0.5 \mathrm{M} \mathrm{KCl}: \mathrm{K}=0.5 x=5 \times 10^{-1} x\)
\(\therefore 5 \times 10^{-1} x>2.5 \times 10^{-1} x>10^{-1} x>10^{-2} x>5 \times 10^{-3} x\)
\((ie) 0.5 \mathrm{M} \mathrm{KCl}>0.25 \mathrm{M} \mathrm{KCl}>0.1 \mathrm{M} \mathrm{KCl}>0.01 \mathrm{M} \mathrm{KCl}>0.005 \mathrm{M} \mathrm{KCl}\)

Generic 5

2 x 5 = 10

22.

(i) Current: The flow of electrons through a wire or any conductor.
(ii) Electrode: A metallic rod / barf strip which conducts electrons into and out of a solution.
(iii) Anode: The electrode at which oxidation occurs. It sends electrons into the outer circuit. It has negative charge and is shown as (-) in cell diagrams.
(iv) Cathode: The electrode at which lect ons are received from the outer circuit. It has a positive charge and is known as (+) in the cell diagrams.
(v) Electrolyte is the salt solunon in a cell
(vi) Anode compartment: The compartment of cell in which oxidation half-reaction occurs. It contains the anode.
(vii) Cathode compartment: The compartment of the cell in which reduction half-reaction occurs. It contains the cathode.
(vii) Half-cell: Each half of an electrochemical cell, where oxidaton occurs, is called the half cell.

23.

Given:
Specific conductance = K = 5.55 x 10^-3 mho cm^-1
Concentration C = 1M
Volume = 1 litre.
Formula: Equivalent conductance = K x V
= \(\frac { k\times 1000 }{ C } \)
Solution:
\(\frac { { 5.55\times 10 }^{ -3 }\times 1000 }{ 1 } =5.55\times { 10 }^{ -3 }\times { 10 }^{ -3 }\)