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 ___________ $\times 10^{- 2} V$ . (Nearest integer)

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

answer is 16.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} {Sn}^{2 +} + 2 e^{-} \to Sn {ΔG}_{1}^{0} = + 2 \times 0 .140 \times F \\ {Sn}^{+ 4} + 4 e^{-} \to Sn {ΔG}_{2}^{0} = - 4 \times 0 .01 \times F \\ \bar{{Sn}^{+ 4} + 2 e^{-} \to {Sn}^{+ 2} {ΔG}_{3}^{0} = - 2 \times E^{0} {Sn}^{+ 4} / {Sn}^{+ 2} \times F} \\ {ΔG}_{3}^{0} = {ΔG}_{2}^{0} - {ΔG}_{1}^{0} \\ - 2 \times E^{0} \times F = - (0 .04 + 0 .28) \times F \\ E^{0} = 0 .16 volt = 16 \times 10^{- 2} V \end{array}$