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Emf of the cell  Zn /Zn+2//Cu+2/Cu  will be highest when

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answer is 2.

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Detailed Solution

Ecell = Ecello−0.0591nlog[Products]ReactantsAnode−Zn  → Zn+2+2e−Cathode   Cu+2 +2e-→Cu                  Zn + Cu+2 →  Zn+2 +CuE0cell is a constantEcell∝ −logZn+2Cu+2Lesser theZn+2Cu+2,higher will be the Ecell∴ (2) is correct
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