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What is the emf of the following cell (in volt)?
$C u_{(s)} | C u_{(a q, 2 M)}^{+ 2} | | Z n_{(a q, 2 M)}^{+ 2} | Z n_{(s)}$
Given,
$P t_{(s)} |H_{2 (g)} (1 b a r)| H_{(a q, 2 M)}^{+} |C u_{(a q, 1 M)}^{+ 2}| C u_{(s)} E = + 0.322 V P t_{(s)} | H_{2 (g)} (1 b a r) | | H_{a q, 1 M}^{+} | | Z n_{a q (1 M)}^{+ 2} | Z n_{(s)} E = - 0.760 V$
$\frac{R T}{F} = 0.06 \log 2 = 0.30$

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answer is -1.1.

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Detailed Solution

$E = E_{C u^{+ 2} / C u}^{0} - \frac{0.06}{2} \log {[H^{+}]}^{2} E_{C u^{+ 2} / C u}^{- 0} = 0.322 + \frac{0.06}{2} \times 2 \log^{2} = 0.322 + 0.018 = 0.34 V$

$∴$ For given cell $E = E_{C u / C u^{+ 2}}^{0} + E_{Z n^{+ 2} / Z n}^{0}$

$= - 0.34 - 0.760 = - 1.1 V$

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