Cu 2+ + 2e − → Cu; log [ Cu 2+ ] vs . E red graph is of the type as shown in figure where OA = 0.34 V then electrode potential of the half cell of C u | C u 2 + ( 0.1 M ) will be

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Correct answer: A. - 0.34 + 0.0591 2 V

E Cu / Cu 2 + = E Cu / Cu 2 + 0 - 0 . 059 2 log Cu 2 + If log Cu 2 + = 0 i.e., Cu 2 + = 1 , then E Cu / Cu 2 + = E Cu / Cu 2 + o or E Cu / Cu 2 + o = - E Cu 2 + / Cu o = - 0 . 34 Now, E Cu / Cu 2 + = - 0 . 34 - 0 . 059 2 log 0 . 1 = - 0 . 34 + 0 . 059 2

Cu 2+ + 2e − → Cu; log [ Cu 2+ ] vs . E red graph is of the type as shown in figure where OA = 0.34 V then electrode potential of the half cell of C u | C u 2 + ( 0.1 M ) will be

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