For the given reactions $\mathrm{Sn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Sn}$ $\mathrm{Sn}^{4+}+4 \mathrm{e}^{-} \rightarrow \mathrm{Sn}$ The electrode potentials are; $\mathrm{E}_{\mathrm{Sn}^{2+} / \mathrm{Sn}}^{\mathrm{o}}=-0.140 \mathrm{~V}$ and $\mathrm{E}_{\mathrm{Sn}^{4+} / \mathrm{Sn}}^{\mathrm{o}}=0.010 \mathrm{~V}$. The magnitude of standard electrode potential for $\mathrm{Sn}^{4+} / \mathrm{Sn}^{2+}$ i.e. $\mathrm{E}_{\mathrm{Sn}^{4+} / \mathrm{Sn}^{2+}}^{\mathrm{o}}$ is ______ $\times 10^{-2} \mathrm{~V}$. (Nearest integer)

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Redox Reactions and Electrochemistry

For the given reactions

Sn^{2 +} + 2 e^{-} \to Sn

Sn^{4 +} + 4 e^{-} \to Sn

The electrode potentials are;

E_{Sn^{2 +} / Sn}^{o} = - 0.140 V

and

E_{Sn^{4 +} / Sn}^{o} = 0.010 V

. The magnitude of standard electrode potential for

Sn^{4 +} / Sn^{2 +}

i.e.

E_{Sn^{4 +} / Sn^{2 +}}^{o}

is ______

\times 1 0^{- 2} V

. (Nearest integer)

Answer: 16

Using the relation between Gibbs free energy and electrode potential:

For

Sn^{2 +} + 2 e^{-} \to Sn

Δ G_{1}^{0} = - 2 F (- 0.140) = + 0.280 F

For

Sn^{4 +} + 4 e^{-} \to Sn

Δ G_{2}^{0} = - 4 F (0.010) = - 0.040 F

For

Sn^{4 +} + 2 e^{-} \to Sn^{2 +}

Δ G_{3}^{0} = - 2 F E^{0}

Since

Δ G_{3}^{0} = Δ G_{2}^{0} - Δ G_{1}^{0}

- 2 F E^{0} = - 0.040 F - 0.280 F = - 0.320 F

E^{0} = 0.16

= 16 \times 1 0^{- 2}

V

Answer: 16

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Integer answer