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प्रश्न
Write the Nernst equation and emf of the following cell at 298 K:
Mg(s) | Mg2+ (0.001 M) || Cu2+ (0.0001 M) | Cu(s)
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उत्तर
The cell reaction is as follows:
\[\ce{Mg_{(s)} + Cu^{2+} (0.0001 M) -> Mg^{2+} (0.001 M) + Cu_{(s)}}\]
Hence, n = 2,
According to this, the Nernst equation will be as follows:
Ecell = \[\ce{E^{\circ}_{cell} - \frac{0.0591}{2} log_10 \frac{[Mg^{2+}]}{[Cu^{2+}]}}\]
∴ \[\ce{E_{cell} = (E^{\circ}_{Cu^{2+}/Cu} - E^{\circ}_{Mg^{2+}/Mg}) - \frac{0.0591}{2} log_10 \frac{0.001}{0.0001}}\]
= \[\ce{[+0.34 - (-2.37)] - \frac{0.0591}{2} log_10 10}\]
= 2.71 − 0.0295
= 2.68 V
संबंधित प्रश्न
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