# Calculate emf of the following cell at 298 K: Mg(s) | Mg2+(0.1 M) || Cu2+ (0.01) | Cu(s) [Given Eocell = +2.71 V, 1 F = 96500 C mol–1] - Chemistry

Calculate emf of the following cell at 298 K:
Mg(s) | Mg2+(0.1 M) || Cu2+ (0.01) | Cu(s)
[Given Eocell = +2.71 V, 1 F = 96500 C mol–1]

#### Solution

The cell reaction can be represented as:
Mg(s) + Cu2+(aq.) → Mg2+(aq.) + Cu(s)

$Given,$
$E_{cell}^o = + 2 . 71 V$
$T = 298 K$
$\text{According to the Nernst equation:}$
$E = E_{cell}^o - \frac{0 . 0591}{2}\log\frac{[ {Mg}^{2 +} ]}{\left[ {Cu}^{2 +} \right]} = 2 . 71 - \frac{0 . 0591}{2}\log\frac{0 . 1}{0 . 01}$
$= 2 . 71 - 0 . 0295 \log 10 = 2 . 71 - 0 . 0295$
$= 2 . 6805 V$

Concept: Nernst Equation - Introduction
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