Q. No.Electrolysis of aqueous solution of \(CuSO_4\) is carried out, where 300 mg of copper is deposited (atomic mass of Cu = 63.54). After this 600 milli amp. current is further passed for 28 minutes. Calculate total volume of \(O_2\) released (in ml), given that 1 mole of a gas occupy 22.4 litres.
Given half reactions:
\(\mathrm{Cu}^{+2}(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(\mathrm{~s})\)
\(2 \mathrm{H}_2\mathrm{O}(l) \longrightarrow \mathrm{O}_2(\mathrm{~g})+4 \mathrm{H}^{+}(\mathrm{aq})+4 \mathrm{e}^{-}\)
1. 110
2. 111.15
3. 101.5
4. 80.5
Given half reactions:
\(\mathrm{Cu}^{+2}(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(\mathrm{~s})\)
\(2 \mathrm{H}_2\mathrm{O}(l) \longrightarrow \mathrm{O}_2(\mathrm{~g})+4 \mathrm{H}^{+}(\mathrm{aq})+4 \mathrm{e}^{-}\)
1. 110
2. 111.15
3. 101.5
4. 80.5
Subtopic: Electrode & Electrode Potential | Faraday’s Law of Electrolysis |
72%
Level 2: 60%+
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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 |
50%
Level 3: 35%-60%
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In the given reaction sequence, the standard electrode potentials (in volts) are provided for each step as follows:
\(\mathrm{FeO}_4^{2-} \xrightarrow{+2.0 \mathrm{~V}} \mathrm{Fe}^{3+} \xrightarrow{0.8 \mathrm{~V}} \mathrm{Fe}^{2+} \xrightarrow{-0.5 \mathrm{~V}} \mathrm{Fe}^0 \)
The value of \(\mathrm{E}_{\mathrm{FeO}_4^{2-} / \mathrm{Fe}^{2+}}^{\Theta}\) is:
1. 1.7 V
2. 1.2 V
3. 2.1 V
4. 1.4 V
\(\mathrm{FeO}_4^{2-} \xrightarrow{+2.0 \mathrm{~V}} \mathrm{Fe}^{3+} \xrightarrow{0.8 \mathrm{~V}} \mathrm{Fe}^{2+} \xrightarrow{-0.5 \mathrm{~V}} \mathrm{Fe}^0 \)
The value of \(\mathrm{E}_{\mathrm{FeO}_4^{2-} / \mathrm{Fe}^{2+}}^{\Theta}\) is:
1. 1.7 V
2. 1.2 V
3. 2.1 V
4. 1.4 V
Subtopic: Electrode & Electrode Potential |
57%
Level 3: 35%-60%
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If EMF of Hydrogen electrode at 25°C is zero in pure water then pressure of \(H_2\) in bar :
1. \(10^{-14}\)
2. \(10^{-7}\)
3. 1
4. 0.5
1. \(10^{-14}\)
2. \(10^{-7}\)
3. 1
4. 0.5
Subtopic: Electrode & Electrode Potential |
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Level 2: 60%+
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Consider the given cell reaction:
\(\begin{aligned} &\quad 2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_2 \\ & \mathrm{P}_{\mathrm{H}_2}=2 \mathrm{~atm} \\ & {\left[\mathrm{H}^{+}\right]=1 \mathrm{M}} \\ & \left(\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06\right) \end{aligned}\)
If \(\text{E}_{cell}\) for the reaction is given by \(-x \times 10^{-3} V\), the value of \(\text{x}\) is:
\(\begin{aligned} &\quad 2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_2 \\ & \mathrm{P}_{\mathrm{H}_2}=2 \mathrm{~atm} \\ & {\left[\mathrm{H}^{+}\right]=1 \mathrm{M}} \\ & \left(\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06\right) \end{aligned}\)
If \(\text{E}_{cell}\) for the reaction is given by \(-x \times 10^{-3} V\), the value of \(\text{x}\) is:
| 1. | 8 | 2. | 9 |
| 3. | 4 | 4. | 5 |
Subtopic: Electrode & Electrode Potential |
80%
Level 1: 80%+
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How many of the given species can be oxidised by \(\mathrm{MnO}_4^{-}\) in aqueous solution among the following?
[Given: \(\mathrm{E}_{\mathrm{MnO}_4^{-} / \mathrm{Mn}^{2+}}^{\circ}=1.33 \mathrm{~V}\) ]
(a) \(\mathrm{Cu}\left[\text { Given } \mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}=0.34 \mathrm{~V}\right. \text { ] }\)
(b) \(\mathrm{Cr} \text { [Given } \mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}}^{\circ}=-0.74 \mathrm{~V} \text { ] }\)
(c) \(\left.\mathrm{Ag} \text { [Given } \mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\circ}=0.80 \mathrm{~V}\right]\)
(d) \(\text { Au [Given } \mathrm{E}_{\mathrm{Au}^{+} / \mathrm{Au}}^{\circ}=1.68 \mathrm{~V} \text { ] }\)
1. (a), (b), (c), (d)
2. (a), (b), (c)
3. (b), (c), (d)
4. (d), (b), (c), (a)
[Given: \(\mathrm{E}_{\mathrm{MnO}_4^{-} / \mathrm{Mn}^{2+}}^{\circ}=1.33 \mathrm{~V}\) ]
(a) \(\mathrm{Cu}\left[\text { Given } \mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}=0.34 \mathrm{~V}\right. \text { ] }\)
(b) \(\mathrm{Cr} \text { [Given } \mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}}^{\circ}=-0.74 \mathrm{~V} \text { ] }\)
(c) \(\left.\mathrm{Ag} \text { [Given } \mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\circ}=0.80 \mathrm{~V}\right]\)
(d) \(\text { Au [Given } \mathrm{E}_{\mathrm{Au}^{+} / \mathrm{Au}}^{\circ}=1.68 \mathrm{~V} \text { ] }\)
1. (a), (b), (c), (d)
2. (a), (b), (c)
3. (b), (c), (d)
4. (d), (b), (c), (a)
Subtopic: Electrode & Electrode Potential |
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Level 1: 80%+
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Arrange following in increasing order of reduction potentials:
(a) \(Cl_2 /Cl^-\)
(b) \(I_2 /I^-\)
(c) \(Li^+ /Li \)
(d) \(Na^+ /Na \)
(e) \(Ag^+ /Ag \)
1. c < d < b < e < a
2. a < b < c < d < e
3. c < d < e < b < a
4. d < c < e < b < a
(a) \(Cl_2 /Cl^-\)
(b) \(I_2 /I^-\)
(c) \(Li^+ /Li \)
(d) \(Na^+ /Na \)
(e) \(Ag^+ /Ag \)
1. c < d < b < e < a
2. a < b < c < d < e
3. c < d < e < b < a
4. d < c < e < b < a
Subtopic: Electrode & Electrode Potential |
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Level 3: 35%-60%
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The correct order of reduction potentials of the following pairs is:
1. A > C > B > D > E
2. A > B > C > D > E
3. A > C > B > E > D
4. C > B > A > E > D
| A. | \(\text{Cl}_2/\text{Cl}^-\) |
| B. | \(\text{l}_2/\text{Cl}^-\) |
| C. | \(\text{Ag}^+/\text{Ag}\) |
| D. | \(\text{Na}^+/\text{Na}\) |
| E. | \(\text{Li}^+/\text{Li}\) |
1. A > C > B > D > E
2. A > B > C > D > E
3. A > C > B > E > D
4. C > B > A > E > D
Subtopic: Electrode & Electrode Potential |
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Consider the given half-cell reactions:
\(\begin{array}{ll} \mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}+\mathrm{e}^{-} & \mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^0=0.77 \mathrm{~V} \\ 2 \mathrm{I}^{-} \rightarrow \mathrm{I}_2+2 \mathrm{e}^{-} & \mathrm{E}_{\mathrm{I}_2 / \text{I}^-}^{\circ}=0.54 \mathrm{~V} \end{array}\)
The standard electrode potential for the spontaneous reaction in the cell is \(x\times 10^{-2}V~\text{at}~ 298~K.\)
The value of \(x\) is: (Nearest Integer)
1. 32
2. 23
3. 45
4. 17
\(\begin{array}{ll} \mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}+\mathrm{e}^{-} & \mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^0=0.77 \mathrm{~V} \\ 2 \mathrm{I}^{-} \rightarrow \mathrm{I}_2+2 \mathrm{e}^{-} & \mathrm{E}_{\mathrm{I}_2 / \text{I}^-}^{\circ}=0.54 \mathrm{~V} \end{array}\)
The standard electrode potential for the spontaneous reaction in the cell is \(x\times 10^{-2}V~\text{at}~ 298~K.\)
The value of \(x\) is: (Nearest Integer)
1. 32
2. 23
3. 45
4. 17
Subtopic: Electrode & Electrode Potential |
87%
Level 1: 80%+
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Which one of the following metal is best used for standard half electrode?
1. \(\Big(\frac{dE}{dT}\Big)_p=10^{-4} \)
2. \(\Big(\frac{dE}{dT}\Big)_p=2\times10^{-4} \)
3. \(\Big(\frac{dE}{dT}\Big)_p=0.1\times10^{-4} \)
4. \(\Big(\frac{dE}{dT}\Big)_p=0.2\times10^{-4} \)
1. \(\Big(\frac{dE}{dT}\Big)_p=10^{-4} \)
2. \(\Big(\frac{dE}{dT}\Big)_p=2\times10^{-4} \)
3. \(\Big(\frac{dE}{dT}\Big)_p=0.1\times10^{-4} \)
4. \(\Big(\frac{dE}{dT}\Big)_p=0.2\times10^{-4} \)
Subtopic: Electrode & Electrode Potential |
60%
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