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Question
The electrode potentials for \[\ce{Cu{^{2+}_{(aq)}} + e- -> Cu{^+_{(aq)}}}\] and \[\ce{Cu{^{+}_{(aq)}} + e- -> Cu_{(s)}}\] are +0.15 V and +0.50 V respectively. The value of \[\ce{E^{\circ}_{{Cu^{2+}/{Cu}}}}\] will be:
Options
0.325 V
0.650 V
0.150 V
0.500 V
MCQ
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Solution
0.325 V
Explanation:
To find the overall standard electrode potential for:
\[\ce{Cu^2+ + 2e- −> Cu}\]
we combine the two half-reactions:
\[\ce{Cu^2+ + e- −> Cu+}\]; \[\ce{E^{\circ}_{1}}\] = +0.15 V
\[\ce{Cu+ + e- −> Cu}\]; \[\ce{E^{\circ}_{2}}\] = +0.50 V
Use the formula for combining two sequential redox steps:
\[\ce{E^{\circ}_{overall} = \frac{E^{\circ}_{1} + E^{\circ}_2}{2}}\]
= \[\ce{\frac{0.15 + 0.50}{2}}\]
= 0.325 V
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