Q. No.
Subtopic: Nernst Equation |
Level 4: Below 35%
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Find the pH, above which \(O_2\) will be evolved at anode:
\(\begin{aligned} & \mathrm{E}_{\mathrm{M}^{+2}(\mathrm{aq}) / \mathrm{M}(\mathrm{~s})}^{\circ}=0.997 \mathrm{~V}, \mathrm{E}_{\mathrm{O}_2(\mathrm{~g}) / \mathrm{H}_2 \mathrm{O}(\ell)}^{\circ}=+1.23 \mathrm{~V} \\ & \operatorname{Pt}(\mathrm{~s})\left|\mathrm{O}_2(\mathrm{~g})\right| \mathrm{H}^{+}(\mathrm{aq}) \| \mathrm{M}^{+2} \mid \mathrm{M} \end{aligned}\)
(Given that \(\left.2.303 \frac{\mathrm{RT}}{\mathrm{~F}}=0.059\right)\)
1. 4
2. 6
3. 10
4. 12
\(\begin{aligned} & \mathrm{E}_{\mathrm{M}^{+2}(\mathrm{aq}) / \mathrm{M}(\mathrm{~s})}^{\circ}=0.997 \mathrm{~V}, \mathrm{E}_{\mathrm{O}_2(\mathrm{~g}) / \mathrm{H}_2 \mathrm{O}(\ell)}^{\circ}=+1.23 \mathrm{~V} \\ & \operatorname{Pt}(\mathrm{~s})\left|\mathrm{O}_2(\mathrm{~g})\right| \mathrm{H}^{+}(\mathrm{aq}) \| \mathrm{M}^{+2} \mid \mathrm{M} \end{aligned}\)
(Given that \(\left.2.303 \frac{\mathrm{RT}}{\mathrm{~F}}=0.059\right)\)
1. 4
2. 6
3. 10
4. 12
Subtopic: Electrode & Electrode Potential | Nernst Equation | Relation between Emf, G, Kc & pH |
Level 3: 35%-60%
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Hints
For the electrochemical cell:
Pt | HSnO₂⁻, Sn(OH)₆²⁻, OH⁻ || Bi₂O₃, Bi | Pt
The reaction quotient (Q) is 10⁶.
Given:
E°(Sn(OH)₆²⁻ / HSnO₂⁻) = −0.90 V
E°(Bi₂O₃ / Bi) = −0.44 V
If the cell potential Ecell is expressed as x × 10⁻¹ V, find the value of x:
1. 22. 4
3. 6
4. 8
Subtopic: Nernst Equation |
73%
Level 2: 60%+
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Hints
A cell representation is given below:
\(Ag/ AgCl || FeCl_2 , FeCl_3 / Pt\)
Which of the following can increase the EMF of cell ?
(i) By increasing concentration of \(Fe^{2+}\)
(ii) By increasing concentration of \(Fe^{3+}\)
(iii) By decreasing concentration of \(Fe^{2+}\)
(iv) By decreasing concentration of \(Fe^{3+}\)
(v) By increasing concentration of \(Cl^-\)
1. i, iv, v
2. ii, iii, v
3. iii, iv, v
4. i, iii, v
\(Ag/ AgCl || FeCl_2 , FeCl_3 / Pt\)
Which of the following can increase the EMF of cell ?
(i) By increasing concentration of \(Fe^{2+}\)
(ii) By increasing concentration of \(Fe^{3+}\)
(iii) By decreasing concentration of \(Fe^{2+}\)
(iv) By decreasing concentration of \(Fe^{3+}\)
(v) By increasing concentration of \(Cl^-\)
1. i, iv, v
2. ii, iii, v
3. iii, iv, v
4. i, iii, v
Subtopic: Nernst Equation |
58%
Level 3: 35%-60%
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The EMF of a cell is given as 0.83 V. The cell can be represented as:
\(\mathrm{Ti}\left|\mathrm{Ti}^{+}(0.001 \mathrm{M}) \| \mathrm{Cu}^{2+}(0.01 \mathrm{M})\right| \mathrm{Cu}\)
The EMF of this cell could be increased by:
\(\mathrm{Ti}\left|\mathrm{Ti}^{+}(0.001 \mathrm{M}) \| \mathrm{Cu}^{2+}(0.01 \mathrm{M})\right| \mathrm{Cu}\)
The EMF of this cell could be increased by:
| 1. | increasing conc. of \(\mathrm{Cu}^{2+}\), keeping conc. of \(\mathrm{Ti}^{+}\) constant. |
| 2. | increasing conc. of \(\mathrm{Ti}^{+}\), keeping conc. of \(\mathrm{Cu}^{2+}\) constant. |
| 3. | increasing conc. of both \( \mathrm{Ti}^{+}\) and \( \mathrm{Cu}^{2+}\). |
| 4. | decreasing conc. of both \(\mathrm{Cu}^{2+}\) and \(\mathrm{Ti}^{+}\). |
Subtopic: Nernst Equation |
67%
Level 2: 60%+
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Which property or parameter of an electrolytic cell does not affect the electrical conductivity of the electrolyte solution?
1. Concentration of electrolyte
2. Nature of electrolyte added
3. Temperature
4. Nature of electrode
1. Concentration of electrolyte
2. Nature of electrolyte added
3. Temperature
4. Nature of electrode
Subtopic: Nernst Equation |
Level 3: 35%-60%
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Consider the following cell representation:
Pt/H2(1 atm)/H+(1 M) || Fe+3/Fe+2
Find the value of ratio of the concentration of Fe+2 to Fe+3:
[Given: Ecell = 0.712, E0cell = 0.771]
1. 12
2. 10
3. 16
4. 8
Pt/H2(1 atm)/H+(1 M) || Fe+3/Fe+2
Find the value of ratio of the concentration of Fe+2 to Fe+3:
[Given: Ecell = 0.712, E0cell = 0.771]
1. 12
2. 10
3. 16
4. 8
Subtopic: Nernst Equation |
81%
Level 1: 80%+
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Consider the following cell:
Pt|H2(1 bar)|H+(1 M) || M3+|M+
If the value of \(\frac{[M^{+}]}{[M^{3+}]} \) is 10x, then the value of ‘x’ is:
(Given: \(E_{M^{3+}/M^+}^o \)= 2V and \(E_{cell} \) = 1.1V)
1. 35
2. 40
3. 30
4. 50
Pt|H2(1 bar)|H+(1 M) || M3+|M+
If the value of \(\frac{[M^{+}]}{[M^{3+}]} \) is 10x, then the value of ‘x’ is:
(Given: \(E_{M^{3+}/M^+}^o \)= 2V and \(E_{cell} \) = 1.1V)
1. 35
2. 40
3. 30
4. 50
Subtopic: Nernst Equation |
71%
Level 2: 60%+
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Hints
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For a given cell at T K,
\(Pt/H_2 (g)(1 \ bar)/H^+(1 \ M) \ || \ Fe^{3+} /Fe^{2+}/Pt \)
E cell = 0.712 V
E0 cell = 0.770 V
If \([Fe^{2+}] \over [Fe^{3+}]\) is t, then the value of t in the expression \(({t \over 5})\) is:
(Given: \(({2.303 ~RT \over F} = 0.058) \))
1. 4
2. 6
3. 2
4. 1
\(Pt/H_2 (g)(1 \ bar)/H^+(1 \ M) \ || \ Fe^{3+} /Fe^{2+}/Pt \)
E cell = 0.712 V
E0 cell = 0.770 V
If \([Fe^{2+}] \over [Fe^{3+}]\) is t, then the value of t in the expression \(({t \over 5})\) is:
(Given: \(({2.303 ~RT \over F} = 0.058) \))
1. 4
2. 6
3. 2
4. 1
Subtopic: Nernst Equation |
74%
Level 2: 60%+
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Hints
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At \(298~ K\), the cell potential for the following electrochemical cell is measured to be \(0.31~\text{V}\):
\(\mathrm{Pt} \mid \mathrm{H}_2(\mathrm{~g}), 1 \text { bar }\left|\mathrm{H}^{+}(\mathrm{aq})\right|\left|\mathrm{Cu}^{2+}(\mathrm{aq})\right| \mathrm{Cu}(\mathrm{s})\)
If the pH of the acidic solution in the anodic half-cell is 3, and the concentration of \(\text{Cu}^{2+}\) in the cathodic half-cell is \(10^{-x}\text{ M}\), the value of x is:
(Given: \(\left.\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\ominus}=0.34 \mathrm{~V} \text { and } \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06 \mathrm{~V}\right)\)
\(\mathrm{Pt} \mid \mathrm{H}_2(\mathrm{~g}), 1 \text { bar }\left|\mathrm{H}^{+}(\mathrm{aq})\right|\left|\mathrm{Cu}^{2+}(\mathrm{aq})\right| \mathrm{Cu}(\mathrm{s})\)
If the pH of the acidic solution in the anodic half-cell is 3, and the concentration of \(\text{Cu}^{2+}\) in the cathodic half-cell is \(10^{-x}\text{ M}\), the value of x is:
(Given: \(\left.\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\ominus}=0.34 \mathrm{~V} \text { and } \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06 \mathrm{~V}\right)\)
| 1. | 7 | 2. | 3 |
| 3. | 11 | 4. | 6 |
Subtopic: Nernst Equation |
Level 3: 35%-60%
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