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At 298 K, an electrode initially contains a 1 M K2Cr2O7 solution in an acidic buffer of pH 1.0. It is then supplied with 50% of the Sn required to completely convert all Cr2O7²− into Cr³+. If the pH stays unchanged, what is the electrode reduction potential? Given: E°(Cr2O7²−/Cr³+, H+) = 1.33 V, log 2 = 0.3, and 2.303RT/F = 0.06.

1. 1.285 V
2. 1.193 V
3. 1.187 V
4. None of these

Correct answer: 1.187 V

Solution

The correct option is derived from the Nernst equation, which accounts for the concentration of Cr2O7²− and the stoichiometry of the reaction. Since 50% of the required Sn is added, the concentration of Cr2O7²− is effectively halved, leading to a calculated reduction potential of 1.187 V.

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