For the redox reaction Zn(s)+Cu2+(0.1M)→Zn2+(1M)+Cu(s) taking place in a cell Ecell ∘=1,10V the cell potential at 298 K will be

For the redox reaction $Zn (s) + {Cu}^{2 +} (0.1 M) \to {Zn}^{2 +} (1 M) + Cu (s)$ taking place in a cell $(E_{cell}^{\circ} = 1, 10 V)$ the cell potential at 298 $K$ will be

A
$0.82 V$
B
$2.14 V$
C
$1.80 V$
D
$1.07 V$

Solution:

$E_{cell} = E_{cell}^{\circ} - \frac{R T}{2 F} \ln \frac{[{Zn}^{2 +}]}{[{Cu}^{2 +}]} = (1.10 V) - (\frac{0.059 V}{2}) \log (\frac{1}{0.1}) = 1.07 V$

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