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Question
Calculate the e.m.f. of the following cell.
\[\ce{Mg | Mg{^{++}_{(aq)}} || Ag{^+_{(aq)}} | Ag}\]
If \[\ce{E^\circ_{Mg}}\] = −2.37 Volt and \[\ce{E^\circ_{Ag}}\] = 0.8 Volt.
Numerical
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Solution
Given: \[\ce{E^\circ_{Mg}}\] = −2.37 Volt
\[\ce{E^\circ_{Ag}}\] = 0.8 Volt.
\[\ce{Mg | Mg{^{++}_{(aq)}} || Ag{^+_{(aq)}} | Ag}\]
Anode (Oxidation): \[\ce{Mg -> Mg++ + 2e-}\]
Cathode (Reduction): \[\ce{Ag+ + e- -> Ag}\]
The standard e.m.f. of the cell is calculated using the formula:
\[\ce{E{^{\circ}_{cell}} = E{^{\circ}_{cathode}} - E{^{\circ}_{anode}}}\]
= 0.80 − (−2.37)
= 0.80 + 2.37
= 3.17 V
∴ The e.m.f. of the given cell is 3.17 Volts.
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