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Q. No.Given the following reaction:
\(\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})\)
E°cell = 2 V at 298 K
The standard Gibbs energy for the given cell reaction in kJ mol–1 at 298 K is:
(Faraday’s constant, F = 96000 C mol–1)
1. –384
2. 384
3. 192
4. –192
\(\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})\)
E°cell = 2 V at 298 K
The standard Gibbs energy for the given cell reaction in kJ mol–1 at 298 K is:
(Faraday’s constant, F = 96000 C mol–1)
1. –384
2. 384
3. 192
4. –192
Subtopic: Nernst Equation | Relation between Emf, G, Kc & pH |
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= 1.1 V for Daniel cell. Which of the following expressions are correct descriptions of the state of equilibrium in this cell?
(a) 1.1 = KC
(b)
(c)
(d) log KC = 1.1
The correct choice among the given is -
1. (a, b)
2. (b, c)
3. (c, d)
4. (a, d)
Subtopic: Nernst Equation |
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Consider the following cell representation:
Pt/H2(1 atm)/H+(1 M) || Fe+3/Fe+2
The ratio of the concentration of Fe+2 to Fe+3 is-
[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
The ratio of the concentration of Fe+2 to Fe+3 is-
[Given Ecell = 0.712, E0cell = 0.771]
1. 12
2. 10
3. 16
4. 8
Subtopic: Nernst Equation |
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Level 1: 80%+
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For the cell, \({Mg}\left|{Mg}^{2+} \| {Cu}^{2+}\right| {Cu} \):
1. \(Mg\) is cathode
2. \(Cu\) is anode
3. The cell reaction is \({Mg}+{Cu}^{2+} \rightarrow {Mg}^{2+}+{Cu} \)
4. \(Cu \) is the oxidising agent
1. \(Mg\) is cathode
2. \(Cu\) is anode
3. The cell reaction is \({Mg}+{Cu}^{2+} \rightarrow {Mg}^{2+}+{Cu} \)
4. \(Cu \) is the oxidising agent
Subtopic: Nernst Equation |
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What is the electrode potential (in V) of the following electrode at 25 °C?
\(Ni^{2+} (0.1M)|Ni(s) \)
(Given: Standard reaction potential of \(Ni^{2+} |Ni \) is \(-0.25 \ V, {2.303 RT \over F} = 0.06)\)
1. -0.28 V
2. -0.34 V
3. -0.82 V
4. -0.12 V
\(Ni^{2+} (0.1M)|Ni(s) \)
(Given: Standard reaction potential of \(Ni^{2+} |Ni \) is \(-0.25 \ V, {2.303 RT \over F} = 0.06)\)
1. -0.28 V
2. -0.34 V
3. -0.82 V
4. -0.12 V
Subtopic: Nernst Equation |
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The correct cell representation for the reaction given below is:
\(\mathrm{H}_2+2 \mathrm{AgCl} \rightarrow 2 \mathrm{H}^{+}+2 \mathrm{Ag}+2 \mathrm{Cl}^{-} \)
1. \(\mathrm{Pt}\left|\mathrm{H}_2\right| \mathrm{HCl}|| \mathrm{AgCl} \mid \mathrm{Ag} \)
2. \(\mathrm{Pt}\left|\mathrm{H}_2\right| \mathrm{HCl}|| \mathrm{AgCl} \mid \mathrm{Pt} \)
3. \(\mathrm{Ag}|\mathrm{AgCl}| \mathrm{HCl}\left|\mathrm{H}_2\right| \mathrm{Pt} \)
4. \(\mathrm{Pt}|\mathrm{AgCl}| \mathrm{HC}|| \mathrm{H}_2 \mid \mathrm{Pt} \)
\(\mathrm{H}_2+2 \mathrm{AgCl} \rightarrow 2 \mathrm{H}^{+}+2 \mathrm{Ag}+2 \mathrm{Cl}^{-} \)
1. \(\mathrm{Pt}\left|\mathrm{H}_2\right| \mathrm{HCl}|| \mathrm{AgCl} \mid \mathrm{Ag} \)
2. \(\mathrm{Pt}\left|\mathrm{H}_2\right| \mathrm{HCl}|| \mathrm{AgCl} \mid \mathrm{Pt} \)
3. \(\mathrm{Ag}|\mathrm{AgCl}| \mathrm{HCl}\left|\mathrm{H}_2\right| \mathrm{Pt} \)
4. \(\mathrm{Pt}|\mathrm{AgCl}| \mathrm{HC}|| \mathrm{H}_2 \mid \mathrm{Pt} \)
Subtopic: Nernst Equation |
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Mark the correct Nernst equation representation for the following cell :
\(\mathrm{Fe}_{(s)}\left|\mathrm{Fe}^{2+}(0.001 \mathrm{M})\right|\left|\mathrm{H}^{+}(1 \mathrm{M})\right| \mathrm{H}_{2(g)}(1 \mathrm{bar}) \mid \mathrm{Pt}_{(s)}\)
\(\mathrm{Fe}_{(s)}\left|\mathrm{Fe}^{2+}(0.001 \mathrm{M})\right|\left|\mathrm{H}^{+}(1 \mathrm{M})\right| \mathrm{H}_{2(g)}(1 \mathrm{bar}) \mid \mathrm{Pt}_{(s)}\)
| 1. | \( E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.591}{2} \log \frac{\left[\mathrm{Fe}^{2+}\right]\left[\mathrm{H}^{+}\right]^2}{[\mathrm{Fe}]\left[\mathrm{H}_2\right]}\) |
| 2. | \(E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.591}{2} \log \frac{[\mathrm{Fe}]\left[\mathrm{H}^{+}\right]^2}{\left[\mathrm{Fe}^{2+}\right]\left[\mathrm{H}_2\right]} \) |
| 3. | \(E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.0591}{2} \log \frac{\left[\mathrm{Fe}^{2+}\right]\left[\mathrm{H}_2\right]}{[\mathrm{Fe}]\left[\mathrm{H}^{+}\right]^2} \) |
| 4. | \(E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.0591}{2} \log \frac{[\mathrm{Fe}]\left[\mathrm{H}_2\right]}{\left[\mathrm{Fe}^{2+}\right]\left[\mathrm{H}^{+}\right]^2}\) |
Subtopic: Nernst Equation |
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For the given electrochemical cell:
\( X\left|X^{2+}(0.001 \mathrm{~m}) \| Y^{2+}(0.01 \mathrm{~m})\right| Y \\ \text { at } 298 K \\ E_{\mathrm{X}^{2+} / X}^0=-0.76 \\ \mathrm{E}_{\mathrm{Y}^{2+} / Y}^0=+0.34 \\ \frac{2.303 R T}{F}=0.06 \)
If Ecell = t, then the value of 5t is:
(closest Integer)
\( X\left|X^{2+}(0.001 \mathrm{~m}) \| Y^{2+}(0.01 \mathrm{~m})\right| Y \\ \text { at } 298 K \\ E_{\mathrm{X}^{2+} / X}^0=-0.76 \\ \mathrm{E}_{\mathrm{Y}^{2+} / Y}^0=+0.34 \\ \frac{2.303 R T}{F}=0.06 \)
If Ecell = t, then the value of 5t is:
(closest Integer)
| 1. | 5.65 | 2. | 6.25 |
| 3. | 4.75 | 4. | 6.65 |
Subtopic: Nernst Equation |
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