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The reduction potential of a half-cell consisting of a Pt electrode in a solution of 1.5 M Fe²+ and 0.015 M Fe³+ at 25 deg C is (given E_std for Fe³+/Fe²+ = 0.770 V):
- 0.652 V
- 0.88 V
- 0.710 V
- 0.850 V
Correct answer: 0.710 V
Solution
Nernst equation for Fe³+ + e⁻ -> Fe²+: E = E_std - (0.0592/1)*log([Fe²+]/[Fe³+]). [Fe²+] = 1.5 M, [Fe³+] = 0.015 M. Ratio = 1.5/0.015 = 100. E = 0.770 - 0.0592*log(100) = 0.770 - 0.0592*2 = 0.770 - 0.1184 = 0.6516 V ≈ 0.652 V. Answer: 0.652 V.
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