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The standard oxidation potential E⁰ for the half cell reaction are
$\begin{array}{ll} Fe \to {Fe}^{2 +} + 2 e^{-}; E^{\circ} = + 0 .41 V \\ Cu \to {Cu}^{2 +} + 2 e^{-}; E^{\circ} = - 0 .34 V \end{array}$
EMF of the cell reaction is
$Fe (s) + {Cu}^{2 +} (aq) \to {Fe}^{2 +} (aq) + Cu (s)$

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0.75 V

-0.75 V

0.07 V

- 0.07 V

answer is A.

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Detailed Solution

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

= 0.34 – (– 0.41)
= 0.75 V

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