Question

If the half cell reactions are given as (i) $Fe ^{2+}(a q)+2 e^{-} \rightarrow Fe (s) ;\,\,E^{\circ}=-0.44\, V$ (ii) $2 H ^{+}(a q)+\frac{1}{2} O _{2}(g)+2 e^{-} \rightarrow H _{2} O (l)$ $E^{\circ}=+1.23\, V$ The $E^{\circ}$ for the reaction $Fe (s)+2 H ^{+}+\frac{1}{2} O _{2}(g) \rightarrow Fe ^{2+}(a q)+ H _{2} O (l)$ is

• A. + 1.67 V
• B. -1.67 V
• C. + 0.79 V
• D. -0.79 V

Answer 1

Answer: (A)

+ 1.67 V

Answer 2

$E_{\text {cell }}^{\circ} =E^{\circ}{}_{\text { cathode }}-E^{\circ}{ }_{\text {anode }}$
$=+1.23-(-0.44)$
$=+1.67\, V$