The cell, $Zn\ |\ {Zn}^{2 +}\ (1\ M)\ ||\ {Cu}^{2 +}\ (1\ M)\ |\ Cu$ ($E_{cell}^{{^\circ}} =$ 1.10 V), was allowed to be completely discharged at 298 K. The relative concentration of ${Zn}^{2 +}$ to ${Cu}^{2 +}\left( \frac{\lbrack{Zn}^{2 +}\rbrack}{\lbrack{Cu}^{2 +}\ \rbrack} \right)$ is

Question

Dr. Abubacker · Accepted Answer

(b)
Cell is completely discharged, it means equilibrium gets established,
$E_{cell}\  = \ 0\$
$Zn\ |\ {Zn}^{2 +}\ (1\ M)\ ||\ {Cu}^{2 +}\ (1\ M)\ |\ Cu$
Cell reaction : $Zn\  + \ {Cu}^{2 +}\  \rightleftharpoons \ {Zn}^{2 +}\  + \ Cu$
$K_{eq} = \ \ \frac{\lbrack{Zn}^{2 +}\rbrack}{\lbrack{Cu}^{2 +}\rbrack}$
We know,
$E_{cell}^{{^\circ}} = \ E_{cell}^{{^\circ}}\  - \ \frac{0.0591}{n}\log K_{eq}$
$E_{cell}^{{^\circ}} = \ \ \frac{0.0591}{n}\log K_{eq}$
Or $1.10\  = \ \frac{0.0591}{2}\log K_{eq}$
$K_{eq} = \ \frac{\lbrack{Zn}^{2 +}\rbrack}{\lbrack{Cu}^{2 +}\rbrack}\  = \ antilog\ \frac{2.20}{0.0591}$
= antilog 37.3

Q. The cell, $Zn\ |\ {Zn}^{2 +}\ (1\ M)\ ||\ {Cu}^{2 +}\ (1\ M)\ |\ Cu$ ($E_{cell}^{{^\circ}} =$ 1.10 V), was allowed to be completely discharged at 298 K. The relative concentration of ${Zn}^{2 +}$ to ${Cu}^{2 +}\left( \frac{\lbrack{Zn}^{2 +}\rbrack}{\lbrack{Cu}^{2 +}\ \rbrack} \right)$ is

Explanation

Reviewed by Dr. Abubacker

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