QBManager

ELECTROCHEMISTRY

275785 The correct order of $E^{0}$ values of the given
elements is $(M^{2 +} + 2 e^{-} \to M (s))$

1

{Cu}^{2 +} > {Ni}^{2 +} > {Pb}^{2 +} > {Fe}^{2 +}

2

{Ni}^{2 +} > {Fe}^{2 +} > {Cu}^{2 +} > {Pb}^{2 +}

3

{Cu}^{2 +} > {Pb}^{2 +} > {Ni}^{2 +} > {Fe}^{2 +}

4

{Pb}^{2 +} > {Cu}^{2 +} > {Ni}^{2 +} > {Fe}^{2 +}

Explanation:

| Cathode Reduction Half Reaction | Standard Potential $E^{\circ}$ (Volts) |
| :--- | :--- |
| $M^{2 +} + 2 e^{-} \to M (s)$ | |
| ${Cu}^{2 +} + 2 e^{-} \to Cu (s)$ | -0.337 |
| ${Pb}^{2 +} + 2 e^{-} \to Pb (s)$ | -0.13 |
| ${Fe}^{2 +} + 2 e^{-} \to Fe (s)$ | -0.44 |

Shift-II

ELECTROCHEMISTRY

275809 For the cell,
$Z n (s) ∣ {Z n}^{2 +}$ (aq) \vert \vert $M^{x^{+}}$ (aq) $∣ M (s)$ , different half cells and their standard electrode potentials are given below.
| $M^{x}$ | | | | |
| :--- | :--- | :--- | :--- | :--- |
| $(aq) /$ | ${Au}^{3 +}$ | | | |
| $M (s) /$ | $\begin{array}{l} {Ag}^{+} \\ (Ag) / \end{array}$ | $\begin{array}{l} {Fe}^{3 +} (aq) / / {Fe}^{2 +} (aq) \end{array}$ | $\begin{array}{l} {Fe}^{2 +} \\ (aq) / \\ Fe (s) \end{array}$ | |
| $E^{0} M^{x} / M / V$ | 1.40 | 0.80 | 0.77 | -0.44 |
If $E_{{Zn}^{2 +} / Zn}^{\circ} = - 0.76 V$ , which cathode will give a maximum value of $E^{0}$ cell per electron transferred?

1

\frac{{Ag}^{+}}{Ag}

2

\frac{{Fe}^{2 +}}{Fe}

3

\frac{{Au}^{3 +}}{Au}

4

\frac{{Fe}^{3 +}}{{Fe}^{2 +}}

Explanation:

$E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}$
(i) For ${Au}^{3 +} / Au : E_{cell}^{\circ} = 1.40 - (- 0.76) = 2.16 V$ ;
(ii) For ${Ag}^{+} / Ag : E_{cell}^{\circ} = 0.80 - (- 0.76) = 1.56 V$
(iii) For ${Fe}^{3 +} / {Fe}^{2 +} : E_{cell}^{\circ} = 0.77 - (- 0.76) = 1.53 V$
(iv) For ${Fe}^{2 +} / Fe : E_{cell}^{\circ} = - 0.44 - (- 0.76) = 0.32 V$
$E_{cell}^{\circ}$ is maximum for $E_{{Au}^{3 +} (aq) / Au (s)}^{\circ}$ .
Note: The transfer of electron in the cell would not affect the $E_{cell}^{0}$ value.

JEE Main 2019

ELECTROCHEMISTRY

275856 The value $E_{rad}^{0}$ for metal, $A, B$ and $C$ are 0.34 volt $- 0.80$ and $- 0.46$ volt respectively state the correct order for their ability to act a reducing agent.

1

C > A > B

2

A > B > C

3

B > C > A

4

C > B > A

Explanation:

$B > C > A$
| Metal | $E_{cell}^{\circ}$ | Note |
| :--- | :--- | :--- |
| $A$ | $0.34 V$ | $\begin{array}{l} Highest E_{red}^{\circ} so highest \\ reducing agent. \end{array}$ |
| $B$ | $- 0.80 V$ | $\begin{array}{l} Least E_{red}^{\circ} no strong, \\ reducing agent \end{array}$ |
| $C$ | $- 0.46 V$ | $\begin{array}{l} Moderated E_{red}^{\circ} so \\ medium reducing agent \end{array}$ |
So, order reducing agent
$B > C > A$ .3

GUJCET-2016

ELECTROCHEMISTRY

275877 Match the following-
| Column I | | Column II | |
| :--- | :--- | :--- | :--- |
| A | Potential of hydrogen
electrode at $pH = 10$ | 1 | 0.76 V |
| B | ${Cu}^{2 +} ∣ Cu$ | 2 | 0.059 V |
| C | $Zn ∣ {Zn}^{2 +}$ | 3 | -0.591 V |
| D | $\frac{2.303 RT}{F}$ | 4 | 0.337 V |

1 A = 3, B = 1, C = 2, D = 5

2 A = 2, B = 5, C = 1, D = 4

3 A = 3, B = 4, C = 1, D = 2

4 A = 5, B = 1, C = 4, D = 2

Explanation:

Potential of Hydrogen electrode at $pH = 10$ is $- 0.591 V$
Reduction potential of ${Cu}^{2 +} ∣ Cu$ is $0.36 V$
Oxidation potential of $Zn ∣ {Zn}^{2 +}$ is $0.76 V$
$∴$ Value of $\frac{2.303 RT}{F} = \frac{2.303 \times 8.314 \times 298}{96500} = 0.059$

BCECE-2014

ELECTROCHEMISTRY

275785 The correct order of $E^{0}$ values of the given
elements is $(M^{2 +} + 2 e^{-} \to M (s))$

1

{Cu}^{2 +} > {Ni}^{2 +} > {Pb}^{2 +} > {Fe}^{2 +}

2

{Ni}^{2 +} > {Fe}^{2 +} > {Cu}^{2 +} > {Pb}^{2 +}

3

{Cu}^{2 +} > {Pb}^{2 +} > {Ni}^{2 +} > {Fe}^{2 +}

4

{Pb}^{2 +} > {Cu}^{2 +} > {Ni}^{2 +} > {Fe}^{2 +}

Explanation:

| Cathode Reduction Half Reaction | Standard Potential $E^{\circ}$ (Volts) |
| :--- | :--- |
| $M^{2 +} + 2 e^{-} \to M (s)$ | |
| ${Cu}^{2 +} + 2 e^{-} \to Cu (s)$ | -0.337 |
| ${Pb}^{2 +} + 2 e^{-} \to Pb (s)$ | -0.13 |
| ${Fe}^{2 +} + 2 e^{-} \to Fe (s)$ | -0.44 |

Shift-II

ELECTROCHEMISTRY

275809 For the cell,
$Z n (s) ∣ {Z n}^{2 +}$ (aq) \vert \vert $M^{x^{+}}$ (aq) $∣ M (s)$ , different half cells and their standard electrode potentials are given below.
| $M^{x}$ | | | | |
| :--- | :--- | :--- | :--- | :--- |
| $(aq) /$ | ${Au}^{3 +}$ | | | |
| $M (s) /$ | $\begin{array}{l} {Ag}^{+} \\ (Ag) / \end{array}$ | $\begin{array}{l} {Fe}^{3 +} (aq) / / {Fe}^{2 +} (aq) \end{array}$ | $\begin{array}{l} {Fe}^{2 +} \\ (aq) / \\ Fe (s) \end{array}$ | |
| $E^{0} M^{x} / M / V$ | 1.40 | 0.80 | 0.77 | -0.44 |
If $E_{{Zn}^{2 +} / Zn}^{\circ} = - 0.76 V$ , which cathode will give a maximum value of $E^{0}$ cell per electron transferred?

1

\frac{{Ag}^{+}}{Ag}

2

\frac{{Fe}^{2 +}}{Fe}

3

\frac{{Au}^{3 +}}{Au}

4

\frac{{Fe}^{3 +}}{{Fe}^{2 +}}

Explanation:

$E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}$
(i) For ${Au}^{3 +} / Au : E_{cell}^{\circ} = 1.40 - (- 0.76) = 2.16 V$ ;
(ii) For ${Ag}^{+} / Ag : E_{cell}^{\circ} = 0.80 - (- 0.76) = 1.56 V$
(iii) For ${Fe}^{3 +} / {Fe}^{2 +} : E_{cell}^{\circ} = 0.77 - (- 0.76) = 1.53 V$
(iv) For ${Fe}^{2 +} / Fe : E_{cell}^{\circ} = - 0.44 - (- 0.76) = 0.32 V$
$E_{cell}^{\circ}$ is maximum for $E_{{Au}^{3 +} (aq) / Au (s)}^{\circ}$ .
Note: The transfer of electron in the cell would not affect the $E_{cell}^{0}$ value.

JEE Main 2019

ELECTROCHEMISTRY

275856 The value $E_{rad}^{0}$ for metal, $A, B$ and $C$ are 0.34 volt $- 0.80$ and $- 0.46$ volt respectively state the correct order for their ability to act a reducing agent.

1

C > A > B

2

A > B > C

3

B > C > A

4

C > B > A

Explanation:

$B > C > A$
| Metal | $E_{cell}^{\circ}$ | Note |
| :--- | :--- | :--- |
| $A$ | $0.34 V$ | $\begin{array}{l} Highest E_{red}^{\circ} so highest \\ reducing agent. \end{array}$ |
| $B$ | $- 0.80 V$ | $\begin{array}{l} Least E_{red}^{\circ} no strong, \\ reducing agent \end{array}$ |
| $C$ | $- 0.46 V$ | $\begin{array}{l} Moderated E_{red}^{\circ} so \\ medium reducing agent \end{array}$ |
So, order reducing agent
$B > C > A$ .3

GUJCET-2016

ELECTROCHEMISTRY

275877 Match the following-
| Column I | | Column II | |
| :--- | :--- | :--- | :--- |
| A | Potential of hydrogen
electrode at $pH = 10$ | 1 | 0.76 V |
| B | ${Cu}^{2 +} ∣ Cu$ | 2 | 0.059 V |
| C | $Zn ∣ {Zn}^{2 +}$ | 3 | -0.591 V |
| D | $\frac{2.303 RT}{F}$ | 4 | 0.337 V |

1 A = 3, B = 1, C = 2, D = 5

2 A = 2, B = 5, C = 1, D = 4

3 A = 3, B = 4, C = 1, D = 2

4 A = 5, B = 1, C = 4, D = 2

Explanation:

Potential of Hydrogen electrode at $pH = 10$ is $- 0.591 V$
Reduction potential of ${Cu}^{2 +} ∣ Cu$ is $0.36 V$
Oxidation potential of $Zn ∣ {Zn}^{2 +}$ is $0.76 V$
$∴$ Value of $\frac{2.303 RT}{F} = \frac{2.303 \times 8.314 \times 298}{96500} = 0.059$

BCECE-2014

ELECTROCHEMISTRY

275785 The correct order of $E^{0}$ values of the given
elements is $(M^{2 +} + 2 e^{-} \to M (s))$

1

{Cu}^{2 +} > {Ni}^{2 +} > {Pb}^{2 +} > {Fe}^{2 +}

2

{Ni}^{2 +} > {Fe}^{2 +} > {Cu}^{2 +} > {Pb}^{2 +}

3

{Cu}^{2 +} > {Pb}^{2 +} > {Ni}^{2 +} > {Fe}^{2 +}

4

{Pb}^{2 +} > {Cu}^{2 +} > {Ni}^{2 +} > {Fe}^{2 +}

Explanation:

| Cathode Reduction Half Reaction | Standard Potential $E^{\circ}$ (Volts) |
| :--- | :--- |
| $M^{2 +} + 2 e^{-} \to M (s)$ | |
| ${Cu}^{2 +} + 2 e^{-} \to Cu (s)$ | -0.337 |
| ${Pb}^{2 +} + 2 e^{-} \to Pb (s)$ | -0.13 |
| ${Fe}^{2 +} + 2 e^{-} \to Fe (s)$ | -0.44 |

Shift-II

ELECTROCHEMISTRY

275809 For the cell,
$Z n (s) ∣ {Z n}^{2 +}$ (aq) \vert \vert $M^{x^{+}}$ (aq) $∣ M (s)$ , different half cells and their standard electrode potentials are given below.
| $M^{x}$ | | | | |
| :--- | :--- | :--- | :--- | :--- |
| $(aq) /$ | ${Au}^{3 +}$ | | | |
| $M (s) /$ | $\begin{array}{l} {Ag}^{+} \\ (Ag) / \end{array}$ | $\begin{array}{l} {Fe}^{3 +} (aq) / / {Fe}^{2 +} (aq) \end{array}$ | $\begin{array}{l} {Fe}^{2 +} \\ (aq) / \\ Fe (s) \end{array}$ | |
| $E^{0} M^{x} / M / V$ | 1.40 | 0.80 | 0.77 | -0.44 |
If $E_{{Zn}^{2 +} / Zn}^{\circ} = - 0.76 V$ , which cathode will give a maximum value of $E^{0}$ cell per electron transferred?

1

\frac{{Ag}^{+}}{Ag}

2

\frac{{Fe}^{2 +}}{Fe}

3

\frac{{Au}^{3 +}}{Au}

4

\frac{{Fe}^{3 +}}{{Fe}^{2 +}}

Explanation:

$E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}$
(i) For ${Au}^{3 +} / Au : E_{cell}^{\circ} = 1.40 - (- 0.76) = 2.16 V$ ;
(ii) For ${Ag}^{+} / Ag : E_{cell}^{\circ} = 0.80 - (- 0.76) = 1.56 V$
(iii) For ${Fe}^{3 +} / {Fe}^{2 +} : E_{cell}^{\circ} = 0.77 - (- 0.76) = 1.53 V$
(iv) For ${Fe}^{2 +} / Fe : E_{cell}^{\circ} = - 0.44 - (- 0.76) = 0.32 V$
$E_{cell}^{\circ}$ is maximum for $E_{{Au}^{3 +} (aq) / Au (s)}^{\circ}$ .
Note: The transfer of electron in the cell would not affect the $E_{cell}^{0}$ value.

JEE Main 2019

ELECTROCHEMISTRY

275856 The value $E_{rad}^{0}$ for metal, $A, B$ and $C$ are 0.34 volt $- 0.80$ and $- 0.46$ volt respectively state the correct order for their ability to act a reducing agent.

1

C > A > B

2

A > B > C

3

B > C > A

4

C > B > A

Explanation:

$B > C > A$
| Metal | $E_{cell}^{\circ}$ | Note |
| :--- | :--- | :--- |
| $A$ | $0.34 V$ | $\begin{array}{l} Highest E_{red}^{\circ} so highest \\ reducing agent. \end{array}$ |
| $B$ | $- 0.80 V$ | $\begin{array}{l} Least E_{red}^{\circ} no strong, \\ reducing agent \end{array}$ |
| $C$ | $- 0.46 V$ | $\begin{array}{l} Moderated E_{red}^{\circ} so \\ medium reducing agent \end{array}$ |
So, order reducing agent
$B > C > A$ .3

GUJCET-2016

ELECTROCHEMISTRY

275877 Match the following-
| Column I | | Column II | |
| :--- | :--- | :--- | :--- |
| A | Potential of hydrogen
electrode at $pH = 10$ | 1 | 0.76 V |
| B | ${Cu}^{2 +} ∣ Cu$ | 2 | 0.059 V |
| C | $Zn ∣ {Zn}^{2 +}$ | 3 | -0.591 V |
| D | $\frac{2.303 RT}{F}$ | 4 | 0.337 V |

1 A = 3, B = 1, C = 2, D = 5

2 A = 2, B = 5, C = 1, D = 4

3 A = 3, B = 4, C = 1, D = 2

4 A = 5, B = 1, C = 4, D = 2

Explanation:

Potential of Hydrogen electrode at $pH = 10$ is $- 0.591 V$
Reduction potential of ${Cu}^{2 +} ∣ Cu$ is $0.36 V$
Oxidation potential of $Zn ∣ {Zn}^{2 +}$ is $0.76 V$
$∴$ Value of $\frac{2.303 RT}{F} = \frac{2.303 \times 8.314 \times 298}{96500} = 0.059$

BCECE-2014

ELECTROCHEMISTRY

275785 The correct order of $E^{0}$ values of the given
elements is $(M^{2 +} + 2 e^{-} \to M (s))$

1

{Cu}^{2 +} > {Ni}^{2 +} > {Pb}^{2 +} > {Fe}^{2 +}

2

{Ni}^{2 +} > {Fe}^{2 +} > {Cu}^{2 +} > {Pb}^{2 +}

3

{Cu}^{2 +} > {Pb}^{2 +} > {Ni}^{2 +} > {Fe}^{2 +}

4

{Pb}^{2 +} > {Cu}^{2 +} > {Ni}^{2 +} > {Fe}^{2 +}

Explanation:

| Cathode Reduction Half Reaction | Standard Potential $E^{\circ}$ (Volts) |
| :--- | :--- |
| $M^{2 +} + 2 e^{-} \to M (s)$ | |
| ${Cu}^{2 +} + 2 e^{-} \to Cu (s)$ | -0.337 |
| ${Pb}^{2 +} + 2 e^{-} \to Pb (s)$ | -0.13 |
| ${Fe}^{2 +} + 2 e^{-} \to Fe (s)$ | -0.44 |

Shift-II

ELECTROCHEMISTRY

275809 For the cell,
$Z n (s) ∣ {Z n}^{2 +}$ (aq) \vert \vert $M^{x^{+}}$ (aq) $∣ M (s)$ , different half cells and their standard electrode potentials are given below.
| $M^{x}$ | | | | |
| :--- | :--- | :--- | :--- | :--- |
| $(aq) /$ | ${Au}^{3 +}$ | | | |
| $M (s) /$ | $\begin{array}{l} {Ag}^{+} \\ (Ag) / \end{array}$ | $\begin{array}{l} {Fe}^{3 +} (aq) / / {Fe}^{2 +} (aq) \end{array}$ | $\begin{array}{l} {Fe}^{2 +} \\ (aq) / \\ Fe (s) \end{array}$ | |
| $E^{0} M^{x} / M / V$ | 1.40 | 0.80 | 0.77 | -0.44 |
If $E_{{Zn}^{2 +} / Zn}^{\circ} = - 0.76 V$ , which cathode will give a maximum value of $E^{0}$ cell per electron transferred?

1

\frac{{Ag}^{+}}{Ag}

2

\frac{{Fe}^{2 +}}{Fe}

3

\frac{{Au}^{3 +}}{Au}

4

\frac{{Fe}^{3 +}}{{Fe}^{2 +}}

Explanation:

$E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}$
(i) For ${Au}^{3 +} / Au : E_{cell}^{\circ} = 1.40 - (- 0.76) = 2.16 V$ ;
(ii) For ${Ag}^{+} / Ag : E_{cell}^{\circ} = 0.80 - (- 0.76) = 1.56 V$
(iii) For ${Fe}^{3 +} / {Fe}^{2 +} : E_{cell}^{\circ} = 0.77 - (- 0.76) = 1.53 V$
(iv) For ${Fe}^{2 +} / Fe : E_{cell}^{\circ} = - 0.44 - (- 0.76) = 0.32 V$
$E_{cell}^{\circ}$ is maximum for $E_{{Au}^{3 +} (aq) / Au (s)}^{\circ}$ .
Note: The transfer of electron in the cell would not affect the $E_{cell}^{0}$ value.

JEE Main 2019

ELECTROCHEMISTRY

275856 The value $E_{rad}^{0}$ for metal, $A, B$ and $C$ are 0.34 volt $- 0.80$ and $- 0.46$ volt respectively state the correct order for their ability to act a reducing agent.

1

C > A > B

2

A > B > C

3

B > C > A

4

C > B > A

Explanation:

$B > C > A$
| Metal | $E_{cell}^{\circ}$ | Note |
| :--- | :--- | :--- |
| $A$ | $0.34 V$ | $\begin{array}{l} Highest E_{red}^{\circ} so highest \\ reducing agent. \end{array}$ |
| $B$ | $- 0.80 V$ | $\begin{array}{l} Least E_{red}^{\circ} no strong, \\ reducing agent \end{array}$ |
| $C$ | $- 0.46 V$ | $\begin{array}{l} Moderated E_{red}^{\circ} so \\ medium reducing agent \end{array}$ |
So, order reducing agent
$B > C > A$ .3

GUJCET-2016

ELECTROCHEMISTRY

275877 Match the following-
| Column I | | Column II | |
| :--- | :--- | :--- | :--- |
| A | Potential of hydrogen
electrode at $pH = 10$ | 1 | 0.76 V |
| B | ${Cu}^{2 +} ∣ Cu$ | 2 | 0.059 V |
| C | $Zn ∣ {Zn}^{2 +}$ | 3 | -0.591 V |
| D | $\frac{2.303 RT}{F}$ | 4 | 0.337 V |

1 A = 3, B = 1, C = 2, D = 5

2 A = 2, B = 5, C = 1, D = 4

3 A = 3, B = 4, C = 1, D = 2

4 A = 5, B = 1, C = 4, D = 2

Explanation:

Potential of Hydrogen electrode at $pH = 10$ is $- 0.591 V$
Reduction potential of ${Cu}^{2 +} ∣ Cu$ is $0.36 V$
Oxidation potential of $Zn ∣ {Zn}^{2 +}$ is $0.76 V$
$∴$ Value of $\frac{2.303 RT}{F} = \frac{2.303 \times 8.314 \times 298}{96500} = 0.059$

BCECE-2014

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