For an electrochemical cell \(\mathrm{Sn}(\mathrm{s})\left|\mathrm{Sn}^{2+}(\mathrm{aq}, 1 \mathrm{M})\right|\left|\mathrm{Pb}^{2+}(\mathrm{aq}, 1 \mathrm{M})\right| \mathrm{Pb}(\mathrm{s})\) the ratio \(\frac{\left[\mathrm{Sn}^{2+}\right]}{\left[\mathrm{Pb}^{2+}\right]}\) when this cell attains equilibrium is \_\_\_\_. \[ \left(\text { Given : } \mathrm{E}_{\mathrm{Sn}^{2+} \mid \mathrm{Sn}}^{0}=-0.14 \mathrm{~V}, \mathrm{E}_{\mathrm{Pb}^{2+} \mid \mathrm{Pb}}^{0}=-0.13 \mathrm{~V}, \frac{2.303 \mathrm{RT}}{\mathrm{~F}}=0.06\right) \]

JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Redox Reactions and Electrochemistry

For an electrochemical cell

Sn (s) Sn^{2 +} (aq, 1 M) Pb^{2 +} (aq, 1 M) Pb (s)

the ratio

\frac{[ Sn ^{2 +} ]}{[ Pb ^{2 +} ]}

when this cell attains equilibrium is ____.

(Given : E_{Sn^{2 +} ∣ Sn}^{0} = - 0.14 V, E_{Pb^{2 +} ∣ Pb}^{0} = - 0.13 V, \frac{2.303 RT}{F} = 0.06)

Answer: 2.15

The spontaneous cell reaction is

Sn (s) + Pb X^{2 +} Sn X^{2 +} + Pb (s)

, with

E_{cell}^{\circ} = E_{Pb X^{2 +} / Pb}^{\circ} - E_{Sn X^{2 +} / Sn}^{\circ} = (- 0.13) - (- 0.14) = 0.01

V. At equilibrium

E_{cell} = 0

, and the Nernst equation with

n = 2

gives

0 = 0.01 - \frac{0.06}{2} lo g \frac{[ Sn X ^{2 +} ]}{[ Pb X ^{2 +} ]}

, so

lo g \frac{[ Sn X ^{2 +} ]}{[ Pb X ^{2 +} ]} = \frac{0.01}{0.03} = \frac{1}{3}

and the ratio is

1 0^{1/3} \approx 2.15

JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Numerical answer