Question

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

Answer 1

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

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