Consider the cell at \(25^{\circ} \mathrm{C}\) \(\mathrm{Zn}\left|\mathrm{Zn}^{2+}(\mathrm{aq}),(1 \mathrm{M})\right|\left|\mathrm{Fe}^{3+}(\mathrm{aq}), \mathrm{Fe}^{2+}(\mathrm{aq})\right| \mathrm{Pt}(\mathrm{s})\) The fraction of total iron present as \(\mathrm{Fe}^{3+}\) ion at the cell potential of 1.500 V is \(\mathrm{x} \times 10^{-2}\). The value of \(x\) is ______. (Nearest integer) (Given: \(\mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{0}=0.77 \mathrm{~V}, \mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{0}=-0.76 \mathrm{~V}\))

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Redox Reactions and Electrochemistry

Consider the cell at

2 5^{\circ} C

Zn Zn^{2 +} (aq), (1 M) Fe^{3 +} (aq), Fe^{2 +} (aq) Pt (s)

The fraction of total iron present as

Fe^{3 +}

ion at the cell potential of 1.500 V is

x \times 1 0^{- 2}

. The value of

x

is ______. (Nearest integer)
(Given:

E_{Fe^{3 +} / Fe^{2 +}}^{0} = 0.77 V, E_{Zn^{2 +} / Zn}^{0} = - 0.76 V

)

Answer: 24

Zinc is oxidised at the anode and iron(III) reduced at the inert cathode, so the cell reaction is

Zn + 2 F e^{3 +} \to Z n^{2 +} + 2 F e^{2 +}

with

n = 2

, and

E_{cell}^{\circ} = 0.77 - (- 0.76) = 1.53 V

The Nernst equation with

[Z n^{2 +}] = 1

M gives

1.500 = 1.53 - \frac{0.059}{2} lo g \frac{[ F e ^{2 +} ] ^{2}}{[ F e ^{3 +} ] ^{2}}

0.059 lo g \frac{[ F e ^{2 +} ]}{[ F e ^{3 +} ]} = 0.03

and

\frac{[ F e ^{2 +} ]}{[ F e ^{3 +} ]} = 1 0^{0.5085} = 3.23

. The fraction of the iron present as

F e^{3 +}

is then

\frac{1}{1 + 3.23} = 0.237 = 23.7 \times 1 0^{- 2}

. To the nearest integer the value asked for is

24

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Integer answer