QBManager

ELECTROCHEMISTRY

275797 The standard electrode potential $(E^{0})$ values of ${Al}^{3 +} / Al, {Ag}^{+} / Ag, K^{+} / K$ and ${Cr}^{3 +} / Cr$ are $- 1.66 V$ , $0.88 V, - 2.93 V$ and $- 0.74 V$ , respectively. The correct decreasing order of reducing power of the metal is

1

Ag > Cr > Al > K

2

K > Al > Cr > Ag

3

K > Al > Ag > Cr

4

Al > K > Ag > Cr

Explanation:

More negative the value of standard reduction potential higher is the reduction power.
Reducing power $\propto \frac{1}{Standard reduction power}$
Thus, the correct decreasing order of reducing power of the metals is
$K (E_{K^{+} / K}^{\circ} = - 2.93 V) > Al (E_{{Al}^{3 +} / Al}^{\circ} = - 1.66 V)$
$> Cr (E_{{Cr}^{3 +} / Cr}^{\circ} = - 0.74 V) > Ag (E_{{Ag}^{+} / Ag}^{\circ} = + 0.80 V)$

(Odisha NEET 2019)

ELECTROCHEMISTRY

275800 At $298 K$ temperature, 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 potential of electrode would be?

1

0.59 V

2

0.118 V

3

1.18 V

4

0.059 V

Explanation:

For hydrogen electrode, oxidation half reaction is
If $pH = 10, [H^{+}] = 1 \times 10^{- pH} = 1 \times 10^{- 10} M$
From Nernst equation,
$E_{cell} = E_{cell}^{o} - \frac{0.0591}{2} \log \frac{{[H^{+}]}^{2}}{[H_{2}]}$
For hydrogen electrode, $E_{cell}^{o} = 0$
$E_{cell} = - \frac{0.0591}{2} \log \frac{{(10^{- 10})}^{2}}{1}$
$= 0.0591 \times \log 10^{10} = 0.0591 \times 10 = 0.591 V$

AIIMS 26 May 2019 (Morning)

ELECTROCHEMISTRY

275801 $E_{cell}$ of the following cell is
$Pt (s) ∣ H_{2} (g), 1$ bar $| H^{+} (1 M) | | H^{+} (0.1 M) |$
$H_{2} (g), 1$ bar $∣ Pt (s)$

1

\frac{- 2.303 RT}{F}

2

\frac{2.303 RT}{F}

3

\frac{- 2.303 RT}{2 F}

4

\frac{2.303 RT}{2 F}

5

\frac{RT}{2 F}

Explanation:

Given Cell
$Pt (s) ∣ H_{2} (g), 1$ bar $| H^{+} (1 M) | | H^{+} (0.1 M) | H_{2} (g), 1$ bar $Pt (s)$
For Nernst equation, $E_{cell} = E_{cell}^{o} - \frac{2.303 RT}{nF} \log [\frac{1}{H^{+}}]$
$\begin{aligned} E_{cell}^{o} & = 0 - \frac{2.303 RT}{F} \log [\frac{1}{0.1}] \\ = \frac{2.303 RT}{F} \log 10 \\ E_{cell} & = - \frac{2.303 RT}{F} \end{aligned}$

Kerala-CEE-2019

ELECTROCHEMISTRY

275802 Calculate E.M.F of following cell at $298 K$ $Zn (s) | {ZnSO}_{4} (0.01 M) | | {CuSO}_{4} (1.0 M) | Cu (s)$ if $E_{cell}^{o} = 2.0 V$

1

2.0 V

2

2.0592 V

3

2.0296 V

4

1.0508 V

Explanation:

Given, $E_{cell}^{0} = 2.0 V, E_{cell} =$ ?
At Anode :
$Zn \to {Zn}^{2 +} + 2 e^{-}$
At Cathode :
${Cu}^{2 +} + 2 e^{-} \to Cu$
Adding equation (i) and equation (ii)
$\begin{aligned} Zn + {Cu}^{2 +} \to {Zn}^{2 +} + Cu \\ E_{cell} = E_{cell}^{o} - \frac{0.0591}{n} \log \frac{[{Zn}^{2 +}]}{[{Cu}^{2 +}]} \\ = 2 - \frac{0.0591}{2} \log \frac{0.01}{1} \\ = 2.0592 V \end{aligned}$

MHT CET-03.05.2019

ELECTROCHEMISTRY

275797 The standard electrode potential $(E^{0})$ values of ${Al}^{3 +} / Al, {Ag}^{+} / Ag, K^{+} / K$ and ${Cr}^{3 +} / Cr$ are $- 1.66 V$ , $0.88 V, - 2.93 V$ and $- 0.74 V$ , respectively. The correct decreasing order of reducing power of the metal is

1

Ag > Cr > Al > K

2

K > Al > Cr > Ag

3

K > Al > Ag > Cr

4

Al > K > Ag > Cr

Explanation:

More negative the value of standard reduction potential higher is the reduction power.
Reducing power $\propto \frac{1}{Standard reduction power}$
Thus, the correct decreasing order of reducing power of the metals is
$K (E_{K^{+} / K}^{\circ} = - 2.93 V) > Al (E_{{Al}^{3 +} / Al}^{\circ} = - 1.66 V)$
$> Cr (E_{{Cr}^{3 +} / Cr}^{\circ} = - 0.74 V) > Ag (E_{{Ag}^{+} / Ag}^{\circ} = + 0.80 V)$

(Odisha NEET 2019)

ELECTROCHEMISTRY

275800 At $298 K$ temperature, 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 potential of electrode would be?

1

0.59 V

2

0.118 V

3

1.18 V

4

0.059 V

Explanation:

For hydrogen electrode, oxidation half reaction is
If $pH = 10, [H^{+}] = 1 \times 10^{- pH} = 1 \times 10^{- 10} M$
From Nernst equation,
$E_{cell} = E_{cell}^{o} - \frac{0.0591}{2} \log \frac{{[H^{+}]}^{2}}{[H_{2}]}$
For hydrogen electrode, $E_{cell}^{o} = 0$
$E_{cell} = - \frac{0.0591}{2} \log \frac{{(10^{- 10})}^{2}}{1}$
$= 0.0591 \times \log 10^{10} = 0.0591 \times 10 = 0.591 V$

AIIMS 26 May 2019 (Morning)

ELECTROCHEMISTRY

275801 $E_{cell}$ of the following cell is
$Pt (s) ∣ H_{2} (g), 1$ bar $| H^{+} (1 M) | | H^{+} (0.1 M) |$
$H_{2} (g), 1$ bar $∣ Pt (s)$

1

\frac{- 2.303 RT}{F}

2

\frac{2.303 RT}{F}

3

\frac{- 2.303 RT}{2 F}

4

\frac{2.303 RT}{2 F}

5

\frac{RT}{2 F}

Explanation:

Given Cell
$Pt (s) ∣ H_{2} (g), 1$ bar $| H^{+} (1 M) | | H^{+} (0.1 M) | H_{2} (g), 1$ bar $Pt (s)$
For Nernst equation, $E_{cell} = E_{cell}^{o} - \frac{2.303 RT}{nF} \log [\frac{1}{H^{+}}]$
$\begin{aligned} E_{cell}^{o} & = 0 - \frac{2.303 RT}{F} \log [\frac{1}{0.1}] \\ = \frac{2.303 RT}{F} \log 10 \\ E_{cell} & = - \frac{2.303 RT}{F} \end{aligned}$

Kerala-CEE-2019

ELECTROCHEMISTRY

275802 Calculate E.M.F of following cell at $298 K$ $Zn (s) | {ZnSO}_{4} (0.01 M) | | {CuSO}_{4} (1.0 M) | Cu (s)$ if $E_{cell}^{o} = 2.0 V$

1

2.0 V

2

2.0592 V

3

2.0296 V

4

1.0508 V

Explanation:

Given, $E_{cell}^{0} = 2.0 V, E_{cell} =$ ?
At Anode :
$Zn \to {Zn}^{2 +} + 2 e^{-}$
At Cathode :
${Cu}^{2 +} + 2 e^{-} \to Cu$
Adding equation (i) and equation (ii)
$\begin{aligned} Zn + {Cu}^{2 +} \to {Zn}^{2 +} + Cu \\ E_{cell} = E_{cell}^{o} - \frac{0.0591}{n} \log \frac{[{Zn}^{2 +}]}{[{Cu}^{2 +}]} \\ = 2 - \frac{0.0591}{2} \log \frac{0.01}{1} \\ = 2.0592 V \end{aligned}$

MHT CET-03.05.2019

ELECTROCHEMISTRY

275797 The standard electrode potential $(E^{0})$ values of ${Al}^{3 +} / Al, {Ag}^{+} / Ag, K^{+} / K$ and ${Cr}^{3 +} / Cr$ are $- 1.66 V$ , $0.88 V, - 2.93 V$ and $- 0.74 V$ , respectively. The correct decreasing order of reducing power of the metal is

1

Ag > Cr > Al > K

2

K > Al > Cr > Ag

3

K > Al > Ag > Cr

4

Al > K > Ag > Cr

Explanation:

More negative the value of standard reduction potential higher is the reduction power.
Reducing power $\propto \frac{1}{Standard reduction power}$
Thus, the correct decreasing order of reducing power of the metals is
$K (E_{K^{+} / K}^{\circ} = - 2.93 V) > Al (E_{{Al}^{3 +} / Al}^{\circ} = - 1.66 V)$
$> Cr (E_{{Cr}^{3 +} / Cr}^{\circ} = - 0.74 V) > Ag (E_{{Ag}^{+} / Ag}^{\circ} = + 0.80 V)$

(Odisha NEET 2019)

ELECTROCHEMISTRY

275800 At $298 K$ temperature, 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 potential of electrode would be?

1

0.59 V

2

0.118 V

3

1.18 V

4

0.059 V

Explanation:

For hydrogen electrode, oxidation half reaction is
If $pH = 10, [H^{+}] = 1 \times 10^{- pH} = 1 \times 10^{- 10} M$
From Nernst equation,
$E_{cell} = E_{cell}^{o} - \frac{0.0591}{2} \log \frac{{[H^{+}]}^{2}}{[H_{2}]}$
For hydrogen electrode, $E_{cell}^{o} = 0$
$E_{cell} = - \frac{0.0591}{2} \log \frac{{(10^{- 10})}^{2}}{1}$
$= 0.0591 \times \log 10^{10} = 0.0591 \times 10 = 0.591 V$

AIIMS 26 May 2019 (Morning)

ELECTROCHEMISTRY

275801 $E_{cell}$ of the following cell is
$Pt (s) ∣ H_{2} (g), 1$ bar $| H^{+} (1 M) | | H^{+} (0.1 M) |$
$H_{2} (g), 1$ bar $∣ Pt (s)$

1

\frac{- 2.303 RT}{F}

2

\frac{2.303 RT}{F}

3

\frac{- 2.303 RT}{2 F}

4

\frac{2.303 RT}{2 F}

5

\frac{RT}{2 F}

Explanation:

Given Cell
$Pt (s) ∣ H_{2} (g), 1$ bar $| H^{+} (1 M) | | H^{+} (0.1 M) | H_{2} (g), 1$ bar $Pt (s)$
For Nernst equation, $E_{cell} = E_{cell}^{o} - \frac{2.303 RT}{nF} \log [\frac{1}{H^{+}}]$
$\begin{aligned} E_{cell}^{o} & = 0 - \frac{2.303 RT}{F} \log [\frac{1}{0.1}] \\ = \frac{2.303 RT}{F} \log 10 \\ E_{cell} & = - \frac{2.303 RT}{F} \end{aligned}$

Kerala-CEE-2019

ELECTROCHEMISTRY

275802 Calculate E.M.F of following cell at $298 K$ $Zn (s) | {ZnSO}_{4} (0.01 M) | | {CuSO}_{4} (1.0 M) | Cu (s)$ if $E_{cell}^{o} = 2.0 V$

1

2.0 V

2

2.0592 V

3

2.0296 V

4

1.0508 V

Explanation:

Given, $E_{cell}^{0} = 2.0 V, E_{cell} =$ ?
At Anode :
$Zn \to {Zn}^{2 +} + 2 e^{-}$
At Cathode :
${Cu}^{2 +} + 2 e^{-} \to Cu$
Adding equation (i) and equation (ii)
$\begin{aligned} Zn + {Cu}^{2 +} \to {Zn}^{2 +} + Cu \\ E_{cell} = E_{cell}^{o} - \frac{0.0591}{n} \log \frac{[{Zn}^{2 +}]}{[{Cu}^{2 +}]} \\ = 2 - \frac{0.0591}{2} \log \frac{0.01}{1} \\ = 2.0592 V \end{aligned}$

MHT CET-03.05.2019

ELECTROCHEMISTRY

275797 The standard electrode potential $(E^{0})$ values of ${Al}^{3 +} / Al, {Ag}^{+} / Ag, K^{+} / K$ and ${Cr}^{3 +} / Cr$ are $- 1.66 V$ , $0.88 V, - 2.93 V$ and $- 0.74 V$ , respectively. The correct decreasing order of reducing power of the metal is

1

Ag > Cr > Al > K

2

K > Al > Cr > Ag

3

K > Al > Ag > Cr

4

Al > K > Ag > Cr

Explanation:

More negative the value of standard reduction potential higher is the reduction power.
Reducing power $\propto \frac{1}{Standard reduction power}$
Thus, the correct decreasing order of reducing power of the metals is
$K (E_{K^{+} / K}^{\circ} = - 2.93 V) > Al (E_{{Al}^{3 +} / Al}^{\circ} = - 1.66 V)$
$> Cr (E_{{Cr}^{3 +} / Cr}^{\circ} = - 0.74 V) > Ag (E_{{Ag}^{+} / Ag}^{\circ} = + 0.80 V)$

(Odisha NEET 2019)

ELECTROCHEMISTRY

275800 At $298 K$ temperature, 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 potential of electrode would be?

1

0.59 V

2

0.118 V

3

1.18 V

4

0.059 V

Explanation:

For hydrogen electrode, oxidation half reaction is
If $pH = 10, [H^{+}] = 1 \times 10^{- pH} = 1 \times 10^{- 10} M$
From Nernst equation,
$E_{cell} = E_{cell}^{o} - \frac{0.0591}{2} \log \frac{{[H^{+}]}^{2}}{[H_{2}]}$
For hydrogen electrode, $E_{cell}^{o} = 0$
$E_{cell} = - \frac{0.0591}{2} \log \frac{{(10^{- 10})}^{2}}{1}$
$= 0.0591 \times \log 10^{10} = 0.0591 \times 10 = 0.591 V$

AIIMS 26 May 2019 (Morning)

ELECTROCHEMISTRY

275801 $E_{cell}$ of the following cell is
$Pt (s) ∣ H_{2} (g), 1$ bar $| H^{+} (1 M) | | H^{+} (0.1 M) |$
$H_{2} (g), 1$ bar $∣ Pt (s)$

1

\frac{- 2.303 RT}{F}

2

\frac{2.303 RT}{F}

3

\frac{- 2.303 RT}{2 F}

4

\frac{2.303 RT}{2 F}

5

\frac{RT}{2 F}

Explanation:

Given Cell
$Pt (s) ∣ H_{2} (g), 1$ bar $| H^{+} (1 M) | | H^{+} (0.1 M) | H_{2} (g), 1$ bar $Pt (s)$
For Nernst equation, $E_{cell} = E_{cell}^{o} - \frac{2.303 RT}{nF} \log [\frac{1}{H^{+}}]$
$\begin{aligned} E_{cell}^{o} & = 0 - \frac{2.303 RT}{F} \log [\frac{1}{0.1}] \\ = \frac{2.303 RT}{F} \log 10 \\ E_{cell} & = - \frac{2.303 RT}{F} \end{aligned}$

Kerala-CEE-2019

ELECTROCHEMISTRY

275802 Calculate E.M.F of following cell at $298 K$ $Zn (s) | {ZnSO}_{4} (0.01 M) | | {CuSO}_{4} (1.0 M) | Cu (s)$ if $E_{cell}^{o} = 2.0 V$

1

2.0 V

2

2.0592 V

3

2.0296 V

4

1.0508 V

Explanation:

Given, $E_{cell}^{0} = 2.0 V, E_{cell} =$ ?
At Anode :
$Zn \to {Zn}^{2 +} + 2 e^{-}$
At Cathode :
${Cu}^{2 +} + 2 e^{-} \to Cu$
Adding equation (i) and equation (ii)
$\begin{aligned} Zn + {Cu}^{2 +} \to {Zn}^{2 +} + Cu \\ E_{cell} = E_{cell}^{o} - \frac{0.0591}{n} \log \frac{[{Zn}^{2 +}]}{[{Cu}^{2 +}]} \\ = 2 - \frac{0.0591}{2} \log \frac{0.01}{1} \\ = 2.0592 V \end{aligned}$

MHT CET-03.05.2019