Q. No.Calculate the standard electrode potential, E°(Cl⁻/AgCl/Ag), in millivolts, given:
E°(Ag⁺/Ag) = 0.79 V
Kₛₚ(AgCl) = 10⁻¹⁰
2.303 RT/F = 0.059 V
2. 200 mV
3. 300 mV
4. 400 mV
Subtopic: Relation between Emf, G, Kc & pH |
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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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Calculate the standard cell potential for the given reaction:
\(\mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Ag}(\mathrm{~s})\)
Given :
\(\begin{array}{ll} \mathrm{Ag}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{Ag} & \mathrm{E}^0=\mathrm{xV} \\ \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Fe} & \mathrm{E}^0=\mathrm{yV} \\ \mathrm{Fe}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} & \mathrm{E}^0=\mathrm{zV} \end{array}\)
1. x + y – z
2. x + 2y – 3z
3. y – 2x
4. x + 2y
\(\mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Ag}(\mathrm{~s})\)
Given :
\(\begin{array}{ll} \mathrm{Ag}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{Ag} & \mathrm{E}^0=\mathrm{xV} \\ \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Fe} & \mathrm{E}^0=\mathrm{yV} \\ \mathrm{Fe}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} & \mathrm{E}^0=\mathrm{zV} \end{array}\)
1. x + y – z
2. x + 2y – 3z
3. y – 2x
4. x + 2y
Subtopic: Relation between Emf, G, Kc & pH |
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Standard electrode potentials for a few half-cells are mentioned below:
\(\begin{aligned} & \mathrm{{E}_{{Cu}^{2+} / {Cu}}^{\circ}=0.34 {~V}, ~{E}_{{Zn}^{2+} / {Zn}}^o=-0.76 {~V}} \\ & \mathrm{{E}_{{Ag}^{+} / {Ag}}^{\circ}=0.80 {~V}, ~{E}_{{Mg}^{2+} / {Mg}}^{\circ}=-2.37 {~V}} \end{aligned}\)
Which one of the following cells gives the most negative value of ΔGº?
1. Zn|Zn2+ (1M) || Ag+ (1M)|Ag
2. Zn|Zn2+ (1M) || Mg2+(1M)|Mg
3. Ag|Ag+ (1M) || Mg2+(1M)|Mg
4. Cu|Cu2+(1M) || Ag+ (1M)|Ag
\(\begin{aligned} & \mathrm{{E}_{{Cu}^{2+} / {Cu}}^{\circ}=0.34 {~V}, ~{E}_{{Zn}^{2+} / {Zn}}^o=-0.76 {~V}} \\ & \mathrm{{E}_{{Ag}^{+} / {Ag}}^{\circ}=0.80 {~V}, ~{E}_{{Mg}^{2+} / {Mg}}^{\circ}=-2.37 {~V}} \end{aligned}\)
Which one of the following cells gives the most negative value of ΔGº?
1. Zn|Zn2+ (1M) || Ag+ (1M)|Ag
2. Zn|Zn2+ (1M) || Mg2+(1M)|Mg
3. Ag|Ag+ (1M) || Mg2+(1M)|Mg
4. Cu|Cu2+(1M) || Ag+ (1M)|Ag
Subtopic: Relation between Emf, G, Kc & pH |
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A hydrogen electrode is prepared by placing into a solution of \(\text{pH}=3.\) The magnitude of electrode potential at \(25^\circ\text{C}\) is given by \(\text{t}\times10^{-2}~\text{V}.\) Find out \(\text{t}\) : (Nearest integer)
1. 20
2. 18
3. 17
4. 19
1. 20
2. 18
3. 17
4. 19
Subtopic: Relation between Emf, G, Kc & pH |
76%
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For a given reaction in an acidic medium,
\(\small{5 e^{-}+8 H^{+} \ + \ Mn O_4^{-}(0.001 \ M) \rightarrow \mathrm{Mn}^{2+}(0.1 \ M) \ + \ 4 \mathrm{H}_2 \mathrm{O}} \)
(Given values are: \(E_{cell}^o \) = 1.54 V, \(E_{cell} \) = 1.2832 V)
The \(pH\) of the solution is:
1. 2.46
2. 3.52
3. 6.38
4. 1.02
\(\small{5 e^{-}+8 H^{+} \ + \ Mn O_4^{-}(0.001 \ M) \rightarrow \mathrm{Mn}^{2+}(0.1 \ M) \ + \ 4 \mathrm{H}_2 \mathrm{O}} \)
(Given values are: \(E_{cell}^o \) = 1.54 V, \(E_{cell} \) = 1.2832 V)
The \(pH\) of the solution is:
1. 2.46
2. 3.52
3. 6.38
4. 1.02
Subtopic: Relation between Emf, G, Kc & pH |
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For the following cell,
\(Pt(s)|H_2(g)|H^+(aq)||Ag^+|Ag(s);~~E^\circ_\text{cell}=0.302~V \)
Value of \(\Delta G^\circ \) is \([-X]\times10^3~J\), find the value of X:
[Report your answer to the nearest integer]
1. 29
2. 26
3. 32
4. 35
\(Pt(s)|H_2(g)|H^+(aq)||Ag^+|Ag(s);~~E^\circ_\text{cell}=0.302~V \)
Value of \(\Delta G^\circ \) is \([-X]\times10^3~J\), find the value of X:
[Report your answer to the nearest integer]
1. 29
2. 26
3. 32
4. 35
Subtopic: Relation between Emf, G, Kc & pH |
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Emf of the following cell is 0.576 V,
\(\mathrm{Pt}(\mathrm{s})\left|\mathrm{H}_{2}(\mathrm{g})(1 \mathrm{~atm})\right| \mathrm{H}^{+}(\mathrm{aq})\left|\mathrm{Cu}^{2+}(0.01 \mathrm{M})\right| \mathrm{Cu}(\mathrm{s})\)
What will be the pH of the anodic half-cell given ?
(Given: \(\mathrm{E}_{\mathrm{Cu}^{2+} | \mathrm{Cu}}^{0}=0.34 \mathrm{~V} \ \& \ \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06\))
[Report your answer to the nearest integer]
1. 6
2. 5
3. 7
4. 9
\(\mathrm{Pt}(\mathrm{s})\left|\mathrm{H}_{2}(\mathrm{g})(1 \mathrm{~atm})\right| \mathrm{H}^{+}(\mathrm{aq})\left|\mathrm{Cu}^{2+}(0.01 \mathrm{M})\right| \mathrm{Cu}(\mathrm{s})\)
What will be the pH of the anodic half-cell given ?
(Given: \(\mathrm{E}_{\mathrm{Cu}^{2+} | \mathrm{Cu}}^{0}=0.34 \mathrm{~V} \ \& \ \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06\))
[Report your answer to the nearest integer]
1. 6
2. 5
3. 7
4. 9
Subtopic: Relation between Emf, G, Kc & pH |
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What is the nearest integer value of \(x\) for the Gibbs free energy change at 298 K, expressed as \(x×10^{−1}~kJ mol^{−1}\) for the given reaction.
\(Cu(s)+Sn^{2+}(0.001 M)\rightarrow Cu^{2+}(0.01M)+Sn(s) \)
Given: \(\small{E^\ominus_{Cu^{2+}/Cu}=0.34~V;~E^\ominus_{Sn^{2+}/Sn}=-0.14~V;~F=96500~C~mol^{-1}}\)
1. 873
2. 983
3. 1002
4. 911
Subtopic: Relation between Emf, G, Kc & pH |
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If \(\Delta G \) of the following reaction is \([X]\times10^{-1}~\text{kJ} \), the value of \(X\) is:
\(Cu(s)+Sn^{2+}(0.01~\text M)\longrightarrow~Sn(s)+Cu^{2+}~(0.001~\text M)\\ \text{Given: }E^{0}_{Sn^{2+}/Sn}=-0.14~\text V,E^0_{Cu^{2+}/Cu}=0.34~\text V\)
1. 868.5
2. 879.8
3. 862.4
4. 873.1
\(Cu(s)+Sn^{2+}(0.01~\text M)\longrightarrow~Sn(s)+Cu^{2+}~(0.001~\text M)\\ \text{Given: }E^{0}_{Sn^{2+}/Sn}=-0.14~\text V,E^0_{Cu^{2+}/Cu}=0.34~\text V\)
1. 868.5
2. 879.8
3. 862.4
4. 873.1
Subtopic: Relation between Emf, G, Kc & pH |
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