STANDARD Chemistry Reduction Potential MCQ Question
What will be the reduction potential for the following half-cell reaction at 298 K? (Given : [Ag⁺] = 0.1 M and E°_cell = +0.80 V)
0.741 V
0.80 V
−0.80 V
−0.741 V
Correct Answer
Detailed Explanation
Using the Nernst equation, E = E° − (0.0591/n) log([Ag⁺]). Substituting the given values, E = 0.80 − 0.0591 log(1/0.1) = 0.741 V.
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