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Question
Answer the following in brief.
Construct a galvanic cell from the electrodes Co3+ | Co and Mn2+/Mn. \[\ce{E{^{\circ}_{Co}}}\] = 1.82 V, \[\ce{E{^{\circ}_{Mn}}}\] = –1.18 V. Calculate \[\ce{E{^{\circ}_{cell}}}\]
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Solution
Given:
\[\ce{E{^{\circ}_{Co}}}\] = 1.82 V,
\[\ce{E{^{\circ}_{Mn}}}\] = –1.18 V.
To find: \[\ce{E{^{\circ}_{cell}}}\] and cell representation
Formulae: `"E"_"cell"^circ = "E"_"Cathode"^circ - "E"_"anode"^circ`
Calculation: Electrode reactions are
At anode: \[\ce{3(Mn_{(s)} -> Mn{^{2+}_{(aq)}} + 2e-}\]
At cathode: \[\ce{2(Co{^{3+}_{(aq)}} + 3e- -> Co_{(s)})}\]
The cell is composed of Mn (anode), Mn(s) \[\ce{Mn{^{2+}_{(aq)}}}\] and Co (cathode), \[\ce{Co{^{3+}_{(aq)}} | Co_{(s)}}\]
The cell is represented as:
\[\ce{Mn_{(s)} | Mn{^{2+}_{(aq)}} || Co{^{3+}_{(aq)}} | Co_{(s)}}\]
The standard electrode potential is given by
\[\ce{E{^{\circ}_{cell}} = E{^{\circ}_{cathode}} - E{^{\circ}_{anode}}}\]
= 1.82 V – (–1.18 V)
= 1.82 V + 1.18 V
= 3.00 V
The standard cell potential is 3.00 V.
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| Al | −1.66 |
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| Hg | +0.79 |
| Cu | +0.337 |
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