QBManager

ELECTROCHEMISTRY

275879 ${Cu}^{+} (aq)$ is unstable in solution and undergoes simultaneous oxidation and reduction according to the reaction:
${Cu}^{+} (aq) ⇌ {Cu}^{2 +} (aq) + Cu (s)$
Choose correct $E^{0}$ for above reaction if
$E_{{Cu}^{2 +} / Cu}^{0} = 0.34 V$ and $E_{{Cu}^{2 +} / {Cu}^{+}}^{0} = 0.15 V$

1

- 0.38 V

2

+ 0.49 V

3

+ 0.38 V

4

- 0.19 V

Explanation:

Given data,
$E_{{cu}^{2 +} / cu}^{\circ} = 0.34 V, E_{{cu}^{2 +} / {cu}^{+}}^{\circ} = 0.15 V$
We know that, $Δ G^{o} = - {nE}^{o} F$
$2 {Cu}^{+} (aq) ⇌ {Cu}^{2 +} (aq) + Cu (s)$
$Cu (s) ⟶ {Cu}^{2 +} (aq) + 2 e^{-}, Δ G_{1}^{\circ} = - 2 \times (- 0.34) \times F$
${Cu}^{2 +} (aq) + e^{-} ⟶ {Cu}^{+} (aq), Δ G_{2}^{\circ} = - 1 \times (0.15) \times F \dots .$ . (ii)
On addition ${eq}^{n}$ (i) and ${eq}^{n}$ (ii)
$Cu (s) ⟶ {Cu}^{+} (aq) + e^{-}, Δ G_{3}^{\circ} = - 1 \times E_{3}^{\circ} \times F$
$Δ G_{3}^{\circ} = Δ G_{1}^{\circ} + Δ G_{2}^{\circ}$
$- 1 \times E_{3}^{\circ} \times F = (- 2 \times (- 0.34) \times F) + (- 1 \times 0.15 \times F)$
$- E_{3}^{\circ} \times F = 0.68 F - 0.15 F$
$E_{3}^{\circ} = - 0.53 V$
Reaction $2 {Cu}^{+} (aq) ⇌ {Cu}^{2 +} (aq) + Cu (s), E^{\circ} =$ ?
So, ${Cu}^{+} (aq) + e^{-} ⇌ Cu (s), E^{\circ} = 0.53 V$
${Cu}^{+} (aq) ⇌ {Cu}^{2 +} (aq) + e^{-}, E^{\circ} = - 0.15 V$
$2 {Cu}^{+} (aq) ⇌ {Cu}^{2 +} (aq) + Cu (s), E^{\circ} = + 0.38 V$

BITSAT - 2014

ELECTROCHEMISTRY

275880 Which of the following will give $H_{2 (g)}$ at cathode and $O_{2 (g)}$ at anode on electrolysis using platinum electrodes?

1 Molten

NaCl

2 Concentrated aq. Solution of

NaCl

3 Dilute aq. Solution of

NaCl

4 Solid

NaCl

Explanation:

$NaCl \to {Na}^{+} + {Cl}^{-}$
$H_{2} O \to H^{+} + {OH}^{-}$ At Anode.
$\begin{aligned} 2 {Cl}^{-} \to {Cl}_{2} + 2 e^{-}; E_{oxd}^{o} = 1.36 V \\ 4 {OH}^{-} \to O_{2} + 2 H_{2} O + 4 e^{-}; E_{oxd}^{o} = 1.23 V \end{aligned}$
At Cathode
$\begin{aligned} {Na}^{+} + e^{-} \to Na; E_{red}^{o} = - 2.71 V \\ 2 H^{+} + 2 e^{-} \to H_{2}; E_{red}^{o} = - 0.83 V \end{aligned}$

GUJCET-2014

ELECTROCHEMISTRY

275882 The standard emf of the cell $Zn | {Zn}^{2 +} | {Ag}^{+} | Ag$ is $1.56 V$ . If the standard reduction potential of $Ag$ is $0.8 V$ , the standard oxidation potential of $Zn$ is

1

- 0.76 V

2

+ 0.76 V

3

- 2.36 V

4

+ 2.36 V

Explanation:

$\begin{aligned} Given the cell reaction- \\ Zn | {Zn}^{2 +} | {Ag}^{+} | Ag \\ E_{cell}^{\circ} = 1.56 V \\ E_{anode}^{\circ} = ? \\ E_{cathode}^{\circ} = 0.8 V \\ E_{cell}^{o} = E_{cathode}^{\circ} - E_{anode}^{o} \\ 1.56 = 0.8 - (- E_{anode}^{\circ}) ∴ E_{anode}^{\circ} = - 0.8 + 1.56 = + 0.76 V \end{aligned}$

COMEDK 2014

ELECTROCHEMISTRY

275885 A hydrogen gas electrode is made by dipping platinum wire in a solution of $HCl$ of $pH = 10$ and by passing hydrogen gas around the platinum wire at one atm pressure. The oxidation potential of electrode would be

1

0.118 V

2

1.18 V

3

0.059 V

4

0.59 V

Explanation:

Given, $pH = 10$
$∵ pH = - \log [H^{+}]$
$∴ [H^{+}] = 10^{- pH} = 10^{- 10 M}$
For $2 H^{+} + 2 e^{-} ⟶ H_{2}$
$E_{Reduction} = E^{o} - \frac{0.0591}{2} \log \frac{P_{H_{2}}}{{[H^{+}]}^{2}} (∵ E_{SHE}^{o} = 0)$
$= 0 - \frac{0.0591}{2} \log \frac{1}{{(10^{- 10})}^{2}}$
$= 0 - \frac{0.0591}{2} \log 10^{20}$
$= \frac{0.0591}{2} \times 20 \log 10$
$E_{Reduction} = - 0.591 V$
$E_{Oxidation} = + 0.59 V$

(NEET- 2013)

ELECTROCHEMISTRY

275879 ${Cu}^{+} (aq)$ is unstable in solution and undergoes simultaneous oxidation and reduction according to the reaction:
${Cu}^{+} (aq) ⇌ {Cu}^{2 +} (aq) + Cu (s)$
Choose correct $E^{0}$ for above reaction if
$E_{{Cu}^{2 +} / Cu}^{0} = 0.34 V$ and $E_{{Cu}^{2 +} / {Cu}^{+}}^{0} = 0.15 V$

1

- 0.38 V

2

+ 0.49 V

3

+ 0.38 V

4

- 0.19 V

Explanation:

Given data,
$E_{{cu}^{2 +} / cu}^{\circ} = 0.34 V, E_{{cu}^{2 +} / {cu}^{+}}^{\circ} = 0.15 V$
We know that, $Δ G^{o} = - {nE}^{o} F$
$2 {Cu}^{+} (aq) ⇌ {Cu}^{2 +} (aq) + Cu (s)$
$Cu (s) ⟶ {Cu}^{2 +} (aq) + 2 e^{-}, Δ G_{1}^{\circ} = - 2 \times (- 0.34) \times F$
${Cu}^{2 +} (aq) + e^{-} ⟶ {Cu}^{+} (aq), Δ G_{2}^{\circ} = - 1 \times (0.15) \times F \dots .$ . (ii)
On addition ${eq}^{n}$ (i) and ${eq}^{n}$ (ii)
$Cu (s) ⟶ {Cu}^{+} (aq) + e^{-}, Δ G_{3}^{\circ} = - 1 \times E_{3}^{\circ} \times F$
$Δ G_{3}^{\circ} = Δ G_{1}^{\circ} + Δ G_{2}^{\circ}$
$- 1 \times E_{3}^{\circ} \times F = (- 2 \times (- 0.34) \times F) + (- 1 \times 0.15 \times F)$
$- E_{3}^{\circ} \times F = 0.68 F - 0.15 F$
$E_{3}^{\circ} = - 0.53 V$
Reaction $2 {Cu}^{+} (aq) ⇌ {Cu}^{2 +} (aq) + Cu (s), E^{\circ} =$ ?
So, ${Cu}^{+} (aq) + e^{-} ⇌ Cu (s), E^{\circ} = 0.53 V$
${Cu}^{+} (aq) ⇌ {Cu}^{2 +} (aq) + e^{-}, E^{\circ} = - 0.15 V$
$2 {Cu}^{+} (aq) ⇌ {Cu}^{2 +} (aq) + Cu (s), E^{\circ} = + 0.38 V$

BITSAT - 2014

ELECTROCHEMISTRY

275880 Which of the following will give $H_{2 (g)}$ at cathode and $O_{2 (g)}$ at anode on electrolysis using platinum electrodes?

1 Molten

NaCl

2 Concentrated aq. Solution of

NaCl

3 Dilute aq. Solution of

NaCl

4 Solid

NaCl

Explanation:

$NaCl \to {Na}^{+} + {Cl}^{-}$
$H_{2} O \to H^{+} + {OH}^{-}$ At Anode.
$\begin{aligned} 2 {Cl}^{-} \to {Cl}_{2} + 2 e^{-}; E_{oxd}^{o} = 1.36 V \\ 4 {OH}^{-} \to O_{2} + 2 H_{2} O + 4 e^{-}; E_{oxd}^{o} = 1.23 V \end{aligned}$
At Cathode
$\begin{aligned} {Na}^{+} + e^{-} \to Na; E_{red}^{o} = - 2.71 V \\ 2 H^{+} + 2 e^{-} \to H_{2}; E_{red}^{o} = - 0.83 V \end{aligned}$

GUJCET-2014

ELECTROCHEMISTRY

275882 The standard emf of the cell $Zn | {Zn}^{2 +} | {Ag}^{+} | Ag$ is $1.56 V$ . If the standard reduction potential of $Ag$ is $0.8 V$ , the standard oxidation potential of $Zn$ is

1

- 0.76 V

2

+ 0.76 V

3

- 2.36 V

4

+ 2.36 V

Explanation:

$\begin{aligned} Given the cell reaction- \\ Zn | {Zn}^{2 +} | {Ag}^{+} | Ag \\ E_{cell}^{\circ} = 1.56 V \\ E_{anode}^{\circ} = ? \\ E_{cathode}^{\circ} = 0.8 V \\ E_{cell}^{o} = E_{cathode}^{\circ} - E_{anode}^{o} \\ 1.56 = 0.8 - (- E_{anode}^{\circ}) ∴ E_{anode}^{\circ} = - 0.8 + 1.56 = + 0.76 V \end{aligned}$

COMEDK 2014

ELECTROCHEMISTRY

275885 A hydrogen gas electrode is made by dipping platinum wire in a solution of $HCl$ of $pH = 10$ and by passing hydrogen gas around the platinum wire at one atm pressure. The oxidation potential of electrode would be

1

0.118 V

2

1.18 V

3

0.059 V

4

0.59 V

Explanation:

Given, $pH = 10$
$∵ pH = - \log [H^{+}]$
$∴ [H^{+}] = 10^{- pH} = 10^{- 10 M}$
For $2 H^{+} + 2 e^{-} ⟶ H_{2}$
$E_{Reduction} = E^{o} - \frac{0.0591}{2} \log \frac{P_{H_{2}}}{{[H^{+}]}^{2}} (∵ E_{SHE}^{o} = 0)$
$= 0 - \frac{0.0591}{2} \log \frac{1}{{(10^{- 10})}^{2}}$
$= 0 - \frac{0.0591}{2} \log 10^{20}$
$= \frac{0.0591}{2} \times 20 \log 10$
$E_{Reduction} = - 0.591 V$
$E_{Oxidation} = + 0.59 V$

(NEET- 2013)

ELECTROCHEMISTRY

275879 ${Cu}^{+} (aq)$ is unstable in solution and undergoes simultaneous oxidation and reduction according to the reaction:
${Cu}^{+} (aq) ⇌ {Cu}^{2 +} (aq) + Cu (s)$
Choose correct $E^{0}$ for above reaction if
$E_{{Cu}^{2 +} / Cu}^{0} = 0.34 V$ and $E_{{Cu}^{2 +} / {Cu}^{+}}^{0} = 0.15 V$

1

- 0.38 V

2

+ 0.49 V

3

+ 0.38 V

4

- 0.19 V

Explanation:

Given data,
$E_{{cu}^{2 +} / cu}^{\circ} = 0.34 V, E_{{cu}^{2 +} / {cu}^{+}}^{\circ} = 0.15 V$
We know that, $Δ G^{o} = - {nE}^{o} F$
$2 {Cu}^{+} (aq) ⇌ {Cu}^{2 +} (aq) + Cu (s)$
$Cu (s) ⟶ {Cu}^{2 +} (aq) + 2 e^{-}, Δ G_{1}^{\circ} = - 2 \times (- 0.34) \times F$
${Cu}^{2 +} (aq) + e^{-} ⟶ {Cu}^{+} (aq), Δ G_{2}^{\circ} = - 1 \times (0.15) \times F \dots .$ . (ii)
On addition ${eq}^{n}$ (i) and ${eq}^{n}$ (ii)
$Cu (s) ⟶ {Cu}^{+} (aq) + e^{-}, Δ G_{3}^{\circ} = - 1 \times E_{3}^{\circ} \times F$
$Δ G_{3}^{\circ} = Δ G_{1}^{\circ} + Δ G_{2}^{\circ}$
$- 1 \times E_{3}^{\circ} \times F = (- 2 \times (- 0.34) \times F) + (- 1 \times 0.15 \times F)$
$- E_{3}^{\circ} \times F = 0.68 F - 0.15 F$
$E_{3}^{\circ} = - 0.53 V$
Reaction $2 {Cu}^{+} (aq) ⇌ {Cu}^{2 +} (aq) + Cu (s), E^{\circ} =$ ?
So, ${Cu}^{+} (aq) + e^{-} ⇌ Cu (s), E^{\circ} = 0.53 V$
${Cu}^{+} (aq) ⇌ {Cu}^{2 +} (aq) + e^{-}, E^{\circ} = - 0.15 V$
$2 {Cu}^{+} (aq) ⇌ {Cu}^{2 +} (aq) + Cu (s), E^{\circ} = + 0.38 V$

BITSAT - 2014

ELECTROCHEMISTRY

275880 Which of the following will give $H_{2 (g)}$ at cathode and $O_{2 (g)}$ at anode on electrolysis using platinum electrodes?

1 Molten

NaCl

2 Concentrated aq. Solution of

NaCl

3 Dilute aq. Solution of

NaCl

4 Solid

NaCl

Explanation:

$NaCl \to {Na}^{+} + {Cl}^{-}$
$H_{2} O \to H^{+} + {OH}^{-}$ At Anode.
$\begin{aligned} 2 {Cl}^{-} \to {Cl}_{2} + 2 e^{-}; E_{oxd}^{o} = 1.36 V \\ 4 {OH}^{-} \to O_{2} + 2 H_{2} O + 4 e^{-}; E_{oxd}^{o} = 1.23 V \end{aligned}$
At Cathode
$\begin{aligned} {Na}^{+} + e^{-} \to Na; E_{red}^{o} = - 2.71 V \\ 2 H^{+} + 2 e^{-} \to H_{2}; E_{red}^{o} = - 0.83 V \end{aligned}$

GUJCET-2014

ELECTROCHEMISTRY

275882 The standard emf of the cell $Zn | {Zn}^{2 +} | {Ag}^{+} | Ag$ is $1.56 V$ . If the standard reduction potential of $Ag$ is $0.8 V$ , the standard oxidation potential of $Zn$ is

1

- 0.76 V

2

+ 0.76 V

3

- 2.36 V

4

+ 2.36 V

Explanation:

$\begin{aligned} Given the cell reaction- \\ Zn | {Zn}^{2 +} | {Ag}^{+} | Ag \\ E_{cell}^{\circ} = 1.56 V \\ E_{anode}^{\circ} = ? \\ E_{cathode}^{\circ} = 0.8 V \\ E_{cell}^{o} = E_{cathode}^{\circ} - E_{anode}^{o} \\ 1.56 = 0.8 - (- E_{anode}^{\circ}) ∴ E_{anode}^{\circ} = - 0.8 + 1.56 = + 0.76 V \end{aligned}$

COMEDK 2014

ELECTROCHEMISTRY

275885 A hydrogen gas electrode is made by dipping platinum wire in a solution of $HCl$ of $pH = 10$ and by passing hydrogen gas around the platinum wire at one atm pressure. The oxidation potential of electrode would be

1

0.118 V

2

1.18 V

3

0.059 V

4

0.59 V

Explanation:

Given, $pH = 10$
$∵ pH = - \log [H^{+}]$
$∴ [H^{+}] = 10^{- pH} = 10^{- 10 M}$
For $2 H^{+} + 2 e^{-} ⟶ H_{2}$
$E_{Reduction} = E^{o} - \frac{0.0591}{2} \log \frac{P_{H_{2}}}{{[H^{+}]}^{2}} (∵ E_{SHE}^{o} = 0)$
$= 0 - \frac{0.0591}{2} \log \frac{1}{{(10^{- 10})}^{2}}$
$= 0 - \frac{0.0591}{2} \log 10^{20}$
$= \frac{0.0591}{2} \times 20 \log 10$
$E_{Reduction} = - 0.591 V$
$E_{Oxidation} = + 0.59 V$

(NEET- 2013)

ELECTROCHEMISTRY

275879 ${Cu}^{+} (aq)$ is unstable in solution and undergoes simultaneous oxidation and reduction according to the reaction:
${Cu}^{+} (aq) ⇌ {Cu}^{2 +} (aq) + Cu (s)$
Choose correct $E^{0}$ for above reaction if
$E_{{Cu}^{2 +} / Cu}^{0} = 0.34 V$ and $E_{{Cu}^{2 +} / {Cu}^{+}}^{0} = 0.15 V$

1

- 0.38 V

2

+ 0.49 V

3

+ 0.38 V

4

- 0.19 V

Explanation:

Given data,
$E_{{cu}^{2 +} / cu}^{\circ} = 0.34 V, E_{{cu}^{2 +} / {cu}^{+}}^{\circ} = 0.15 V$
We know that, $Δ G^{o} = - {nE}^{o} F$
$2 {Cu}^{+} (aq) ⇌ {Cu}^{2 +} (aq) + Cu (s)$
$Cu (s) ⟶ {Cu}^{2 +} (aq) + 2 e^{-}, Δ G_{1}^{\circ} = - 2 \times (- 0.34) \times F$
${Cu}^{2 +} (aq) + e^{-} ⟶ {Cu}^{+} (aq), Δ G_{2}^{\circ} = - 1 \times (0.15) \times F \dots .$ . (ii)
On addition ${eq}^{n}$ (i) and ${eq}^{n}$ (ii)
$Cu (s) ⟶ {Cu}^{+} (aq) + e^{-}, Δ G_{3}^{\circ} = - 1 \times E_{3}^{\circ} \times F$
$Δ G_{3}^{\circ} = Δ G_{1}^{\circ} + Δ G_{2}^{\circ}$
$- 1 \times E_{3}^{\circ} \times F = (- 2 \times (- 0.34) \times F) + (- 1 \times 0.15 \times F)$
$- E_{3}^{\circ} \times F = 0.68 F - 0.15 F$
$E_{3}^{\circ} = - 0.53 V$
Reaction $2 {Cu}^{+} (aq) ⇌ {Cu}^{2 +} (aq) + Cu (s), E^{\circ} =$ ?
So, ${Cu}^{+} (aq) + e^{-} ⇌ Cu (s), E^{\circ} = 0.53 V$
${Cu}^{+} (aq) ⇌ {Cu}^{2 +} (aq) + e^{-}, E^{\circ} = - 0.15 V$
$2 {Cu}^{+} (aq) ⇌ {Cu}^{2 +} (aq) + Cu (s), E^{\circ} = + 0.38 V$

BITSAT - 2014

ELECTROCHEMISTRY

275880 Which of the following will give $H_{2 (g)}$ at cathode and $O_{2 (g)}$ at anode on electrolysis using platinum electrodes?

1 Molten

NaCl

2 Concentrated aq. Solution of

NaCl

3 Dilute aq. Solution of

NaCl

4 Solid

NaCl

Explanation:

$NaCl \to {Na}^{+} + {Cl}^{-}$
$H_{2} O \to H^{+} + {OH}^{-}$ At Anode.
$\begin{aligned} 2 {Cl}^{-} \to {Cl}_{2} + 2 e^{-}; E_{oxd}^{o} = 1.36 V \\ 4 {OH}^{-} \to O_{2} + 2 H_{2} O + 4 e^{-}; E_{oxd}^{o} = 1.23 V \end{aligned}$
At Cathode
$\begin{aligned} {Na}^{+} + e^{-} \to Na; E_{red}^{o} = - 2.71 V \\ 2 H^{+} + 2 e^{-} \to H_{2}; E_{red}^{o} = - 0.83 V \end{aligned}$

GUJCET-2014

ELECTROCHEMISTRY

275882 The standard emf of the cell $Zn | {Zn}^{2 +} | {Ag}^{+} | Ag$ is $1.56 V$ . If the standard reduction potential of $Ag$ is $0.8 V$ , the standard oxidation potential of $Zn$ is

1

- 0.76 V

2

+ 0.76 V

3

- 2.36 V

4

+ 2.36 V

Explanation:

$\begin{aligned} Given the cell reaction- \\ Zn | {Zn}^{2 +} | {Ag}^{+} | Ag \\ E_{cell}^{\circ} = 1.56 V \\ E_{anode}^{\circ} = ? \\ E_{cathode}^{\circ} = 0.8 V \\ E_{cell}^{o} = E_{cathode}^{\circ} - E_{anode}^{o} \\ 1.56 = 0.8 - (- E_{anode}^{\circ}) ∴ E_{anode}^{\circ} = - 0.8 + 1.56 = + 0.76 V \end{aligned}$

COMEDK 2014

ELECTROCHEMISTRY

275885 A hydrogen gas electrode is made by dipping platinum wire in a solution of $HCl$ of $pH = 10$ and by passing hydrogen gas around the platinum wire at one atm pressure. The oxidation potential of electrode would be

1

0.118 V

2

1.18 V

3

0.059 V

4

0.59 V

Explanation:

Given, $pH = 10$
$∵ pH = - \log [H^{+}]$
$∴ [H^{+}] = 10^{- pH} = 10^{- 10 M}$
For $2 H^{+} + 2 e^{-} ⟶ H_{2}$
$E_{Reduction} = E^{o} - \frac{0.0591}{2} \log \frac{P_{H_{2}}}{{[H^{+}]}^{2}} (∵ E_{SHE}^{o} = 0)$
$= 0 - \frac{0.0591}{2} \log \frac{1}{{(10^{- 10})}^{2}}$
$= 0 - \frac{0.0591}{2} \log 10^{20}$
$= \frac{0.0591}{2} \times 20 \log 10$
$E_{Reduction} = - 0.591 V$
$E_{Oxidation} = + 0.59 V$

(NEET- 2013)

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