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$E_{{Cu}^{2 +} ∣ Cu}^{0} = + 0.337 V, E_{{Zn}^{2 +} | Zn}^{0} = - 0.762 V$
The EMF of the cell, $Zn |{Zn}^{2 +} (0 .1 M) ∥ {Cu}^{2 +} (0 .01 M)| Cu is$

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– 1.069 V

+1.099 V

+1.069 V

– 1.099 V

answer is A.

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Detailed Solution

$E_{cell}^{\circ} = E_{reduction} - E_{oxidation} = 0.337 - (- 0.762) = + 1.099 V$

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