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$E_{{Sn}^{2 +} / S n}^{0} = - 0.14 v, E_{{Fe}^{+ 2} | Fe}^{0} = - 0.44 v$ ; The emf for the cell reaction Fe + Sn²⁺ → Fe²⁺ + Sn is

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+ 0.3 V

+ 0.58

– 0.3 V

– 0.58 V

answer is A.

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Detailed Solution

$E^{0} = E_{Cathode}^{0} - E_{Anode}^{0}$
= -0.1(-0.44) = +0.3V

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