If the half-cell reactions are given as
(i) ${Fe}^{2 +} + 2 e^{-} \to Fe (s); E^{o} = - 0.44 V$

(ii) $2 H^{+} (a q) + O_{2} (g) + 2 e \to H_{2} O (l); E^{o} = + 1.23 V$

The $E^{o}$ for the reaction
$F e (s) + 2 H^{+} + O_{2} (g) \to F e^{2 +} (a q) + H_{2} O (l)$

Question

TestHub · Accepted Answer

Subtracting equation (i) from (ii) we get equation (iii) as follows F e ( s ) + 2 H + + 1 2 O 2 → F e 2 + ( a q ) + H 2 O ( l ) …(iii) So, E o = E o 2 - E o 1 = + 1.23 - ( - 0.44 ) = + 1.67 V .

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