मराठी

Calculate emf of the following cell at 298 K: Mg⁢(s) | Mg⁢2+ (0.1M) |⁢| Cu⁢2+ (0.01) | Cu⁢(s) [Given E⁢𝐴∘cell = +2.71 V, 1 F = 96500 C mol–1] - Chemistry

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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]

संख्यात्मक
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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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2013-2014 (March) Delhi Set 1

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