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$E_{{Fe}^{2 +} / Fe}^{0}$ and $E_{{Ag}^{+} / Ag}^{\circ}$ are – 0.44 V and +0.8 V. Emf of the cell

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+0.36 V

-036 V

+1.24 V

-1.24 V

answer is C.

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Detailed Solution

$E_{cell}^{0} = E_{cathode}^{0} - E_{anode}^{0}$

$E_{cell}^{0} = + 0.8 - (- 0.44) = + 1.24 V$

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