Question

If the half cell reactions are given as
(i) $F{e}^{2+}(aq)+2e^{-}\to Fe(s);E^{\circ }=-0.44V$
(ii) $2H^{+}(aq)+\frac{1}{2}{O}_2(g)+2e^{-}\to H_{2}O(l)$
$E^{\circ }=+1.23V$
The $E^{\circ }$ for the reaction
$Fe(s)+2H^{+}+\frac{1}{2}{O}_2(g)\to F{e}^{2+}(aq)+H_{2}O(l)$
is

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

Answer 1

Answer: (A)

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