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Standard electrode potential of cell $Pt, H_{2} / H^{+} ∥ {Ag}^{+} / Ag is (E_{{Ag}^{+} / Ag}^{0} = + 0.8 V)$

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- 1.2V

0.8 V

1.2 V

- 0.8V

answer is B.

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Detailed Solution

$\begin{array}{l} {Eg}_{Ag}^{0 +} / Ag = + 0.8 V; E_{H}^{0 +} / \frac{1}{2} H_{2} = 0 \\ ∴ E_{cell}^{0} = E_{cathode}^{0} - E_{anode}^{0} = 0.8 - 0 \\ E_{cell}^{0} = 0.8 V \end{array}$

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