Discussion
A copper electrode in 0.01 M Cu\( ^{2+} \), with \( E^\circ = 0.34 \) V, has a potential of about
Answer: A
Correct answer: 0.281 V
Applying the Nernst equation with \( n = 2 \),
\[ E = 0.34 + \frac{0.059}{2}\log(0.01) = 0.34 + 0.0295(-2) = 0.34 - 0.059 = 0.281 \ \mathrm{V} \]Diluting a hundredfold has lowered the potential by 59 millivolts, since two decades of concentration at 29.5 mV each. Adding rather than subtracting gives 0.399, which would mean dilution made the electrode more noble.
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