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प्रश्न
Calculate emf of the following cell at 298 K:
\[\ce{Mg_{(s)} | Mg^{2+} (0.1 M) || Cu^{2+} (0.01) | Cu_{(s)}}\]
[Given \[\ce{E^{\circ}_{cell}}\] = +2.71 V, 1 F = 96500 C mol–1]
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उत्तर
Given: The given cell reaction is:
\[\ce{Mg_{(s)} | Mg^{2+} (0.1 M) || Cu^{2+} (0.01) | Cu_{(s)}}\]
Using the Nernst equation:
\[\ce{E_{cell} = E{^{\circ}_{cell}} - \frac{0.0591}{n} log \frac{[Mg^{2+}]}{[Cu^{2+}]}}\]
Where:
\[\ce{E^{\circ}_{cell}}\] = +2.71 V
n = 2 (number of electrons transferred)
Mg2+ = 0.1 M
Cu2+ = 0.01 M
\[\ce{E_{cell} = 2.71 - \frac{0.0591}{2} log \frac{0.1}{0.01}}\]
= 2.71 − 0.02955 log (10)
= 2.71 − 0.02955 × 1
= 2.71 − 0.02955
= 2.680 V
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