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A copper electrode is immersed in a 0.1 M solution of copper sulphate at 25 deg C. Determine the reduction electrode potential of this copper electrode. (Given: E°(Cu2+/Cu) = 0.34 V)
- 0.31 V
- 0.37 V
- 0.34 V
- 0.28 V
Correct answer: 0.31 V
Solution
For the reduction Cu2+ + 2e- -> Cu, the Nernst equation gives E = E° - (0.059/2)*log(1/[Cu2+]). With [Cu2+] = 0.1 M, log(1/0.1) = 1, so the potential drops slightly below the standard value, giving about 0.31 V.
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