${Sn}^{4 +} + \ {3e}^{-}\ \longrightarrow \ {Sn}^{2 +}$, $E^{{^\circ}}\ = \ 0.13\ V$ ${Br}_{2} + \ \ {2e}^{-}\ \ \longrightarrow \ {2Br}^{-},\ \ \ \ \ \$ $E^{{^\circ}}\ = \ 1.08\ V$ Calculate $K_{eq}$ for the cell reaction for the cell formed by two electrodes.

Question

Dr. Abubacker · Accepted Answer

(b)
Cell reaction will be
${Br}_{2} + \ \ {Sn}^{2 +}\  ightarrow \ {2Br}^{-}\  + \ \ {Sn}^{4 +},$ $E^{{^\circ}}$ = 0.95 V
$E_{cell}^{{^\circ}} = \ \frac{0.059}{2}\log K_{eq}$
$0.95\  = \ \frac{0.059}{2}\log K_{eq}\$
$\frac{0.95\  	imes 2}{0.059} = \ \log K_{eq}$
$K_{eq}\  \approx \ 10^{32}$

Q. ${Sn}^{4 +} + \ {3e}^{-}\ \longrightarrow \ {Sn}^{2 +}$, $E^{{^\circ}}\ = \ 0.13\ V$ ${Br}_{2} + \ \ {2e}^{-}\ \ \longrightarrow \ {2Br}^{-},\ \ \ \ \ \$ $E^{{^\circ}}\ = \ 1.08\ V$ Calculate $K_{eq}$ for the cell reaction for the cell formed by two electrodes.

Explanation

Reviewed by Dr. Abubacker

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