For the cell $\mathrm{Zn}(\mathrm{s})\left|\mathrm{Zn}^{2+}(\mathrm{aq})\right|\left|\mathrm{M}^{\mathrm{x}}+(\mathrm{aq})\right| \mathrm{M}(\mathrm{s})$, different half cells and their standard electrode potentials are given below: | $\mathrm{M}^{x+}(\mathrm{aq}) / \mathrm{M}(\mathrm{s})$ | $\mathrm{Au}^{3+}(\mathrm{aq}) / \mathrm{Au}(\mathrm{s})$ | $\mathrm{Ag}^{+}(\mathrm{aq}) / \mathrm{Ag}(\mathrm{s})$ | $\mathrm{Fe}^{3+}(\mathrm{aq}) / \mathrm{Fe}^{2+}(\mathrm{aq})$ | $\mathrm{Fe}^{2+}(\mathrm{aq}) / \mathrm{Fe}(\mathrm{s})$ | |---|---|---|---|---| | $\mathrm{E}_{\mathrm{M}^{x+}} / \mathrm{M}^{/(\mathrm{V})}$ | 1.40 | 0.80 | 0.77 | -0.44 | If $\mathrm{E}_{\mathrm{zn}^{2}+/ \mathrm{zn}}^{0}=-0.76 \mathrm{~V}$, which cathode will given a maximum value of $\mathrm{E}^{0}$ cell per electron transferred?

(A) $\mathrm{Ag}^{+} / \mathrm{Ag}$

(B) $\mathrm{Fe}^{3}+/ \mathrm{Fe}^{2+}$

(C) $\mathrm{Au}^{3}+/ \mathrm{Au}$

(D) $\mathrm{Fe}^{2+} / \mathrm{Fe}$

JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Redox Reactions and Electrochemistry

For the cell

Zn (s) Zn^{2 +} (aq) ∣ M^{x} + (aq) ∣ M (s)

, different half cells and their standard electrode potentials are given below:

$M^{x +} (aq) / M (s)$	$Au^{3 +} (aq) / Au (s)$	$Ag^{+} (aq) / Ag (s)$	$Fe^{3 +} (aq) / Fe^{2 +} (aq)$	$Fe^{2 +} (aq) / Fe (s)$
$E_{M^{x +}} / M^{/ (V)}$	1.40	0.80	0.77	-0.44

E_{zn^{2} + / zn}^{0} = - 0.76 V

, which cathode will given a maximum value of

E^{0}

cell per electron transferred?

Answer: (A)

The cell potential for each candidate cathode is

E_{ce l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{Zn X^{2 +} / Zn}^{\circ} = E_{c a t h o d e}^{\circ} + 0.76 V

, but the question asks for the value per electron transferred, so each

E_{ce l l}^{\circ}

must be divided by the number of electrons

n

moved in that reduction. For gold,

n = 3

and

E_{ce l l}^{\circ} = 1.40 + 0.76 = 2.16 V

, giving only

0.72 V

per electron. For silver,

n = 1

and

E_{ce l l}^{\circ} = 0.80 + 0.76 = 1.56 V

, i.e.

1.56 V

per electron. For the

Fe X^{3 +} / Fe X^{2 +}

couple,

n = 1

gives

0.77 + 0.76 = 1.53 V

per electron, and for

Fe X^{2 +} / Fe

n = 2

gives only

0.16 V

per electron. The largest value per electron transferred therefore belongs to the silver half-cell.

JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Single correct option