ElectrochemistryHard
Question
The solution of CuSO4, in which copper rod is immersed, is diluted to 10 times. The reduction electrode potential
Options
A.increases by 0.0295 V
B.decreases by 0.0295 V
C.increases by 0.059 V
D.decreases by 0.059 V
Solution
$Cu^{2 +} + 2e^{-} \rightleftharpoons Cu$
$E = E^{o} - \frac{0.059}{2}.\log\frac{1}{\left\lbrack Cu^{+ 2} \right\rbrack} = E^{o} - \frac{0.059}{2}\log\frac{1}{0.1} $$$= E^{o} - 0.0295\text{ V}$$
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