Efficiency of the following cell is 84%A(s)+B2+(aq)→A2+(aq.)+B(s);ΔH=−285Then the standard electrode potential of the cell in Volt will be

Efficiency=ΔG0ΔH0=−nFE0ΔH0 Or 0.84=−2×E0×96500−285×1000E0=+1.24V

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Efficiency of the following cell is 84%
$A (s) + B^{2 +} (a q) \to A^{2 +} (a q .) + B (s); Δ H = - 285$
Then the standard electrode potential of the cell in Volt will be

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answer is 1.24.

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Efficiency $= \frac{Δ G^{0}}{Δ H^{0}}$ $= \frac{- n F E^{0}}{Δ H^{0}}$

$0.84 = \frac{- 2 \times E^{0} \times 96500}{- 285 \times 1000}$

$E^{0} = + 1.24 V$

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