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E.M.F of the cell reaction,
$2 {Ag}^{+} + Cu ⟶ 2 Ag + {Cu}^{2 +} is 0.46 V . If E_{{Cu}^{2 +} / Cu}^{0} is + 0.34 V, E_{{Ag}^{+} / Ag}^{0} is$

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0.80 V

0.12 V

0.40 V

1.60 V

answer is A.

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Detailed Solution

$E_{c e l l}^{o} = E_{c a t h o d e}^{o} - E_{a n o d e}^{o} E_{c e l l}^{o} = E_{A g^{+} / A g}^{o} - E_{C u^{2 +} / C u}^{o} 0.46 = E_{A g^{+} / A g}^{o} - (0.34) E_{A g^{+} / A g}^{o} = 0.46 + 0.34 = 0.80 V$

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