The equilibrium constant for the reaction $\ce{Zn(s) + Sn^{2+}(aq) Zn^{2+}(aq) + Sn(s)}$ is $1 \times 10^{20}$ at 298 K. The magnitude of standard electrode potential of $\ce{Sn/Sn^{2+}}$ if $E_{\ce{Zn^{2+}/Zn}}^{\mathrm{o}}=-0.76~\mathrm{V}$ is $\_\_\_\_$ $\times 10^{-2}~\mathrm{V}$. (Nearest integer) Given: $\frac{2.303RT}{F}=0.059~\mathrm{V}$

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Redox Reactions and Electrochemistry

The equilibrium constant for the reaction

Zn (s) + Sn X^{2 +} (aq) Zn X^{2 +} (aq) + Sn (s)

1 \times 1 0^{20}

at 298 K. The magnitude of standard electrode potential of

Sn / Sn X^{2 +}

E_{Zn X^{2 +} / Zn}^{o} = - 0.76 V

____

\times 1 0^{- 2} V

. (Nearest integer)

Given:

\frac{2.303 R T}{F} = 0.059 V

Answer: 17

At equilibrium the standard cell potential is linked to the equilibrium constant by

E_{cell}^{\circ} = \frac{0.059}{n} lo g K

, and with two electrons transferred and

K = 1 \times 1 0^{20}

this gives

E_{cell}^{\circ} = \frac{0.059}{2} \times 20 = 0.59

V. In this cell zinc is oxidised and tin is reduced, so

E_{cell}^{\circ} = E_{Sn X^{2 +} / Sn}^{\circ} - E_{Zn X^{2 +} / Zn}^{\circ} = E_{Sn X^{2 +} / Sn}^{\circ} + 0.76

. Hence

E_{Sn X^{2 +} / Sn}^{\circ} = 0.59 - 0.76 = - 0.17

V, and the magnitude of the standard potential is

0.17

V, that is

17 \times 1 0^{- 2}

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Integer answer