Question

What is the cell potential (standard emf, $E^0$ ) for the reaction below?
$[E^0\left(F{e}^{2+}(aq)Fe\right)=0.44V$ and
$E^0\left(O_2(g)/H_{2}O/O{H}^{-}\right)=+0.4V]$
$2Fe(s)+O_2(g)+2H_{2}O(l)\leftrightharpoons 2F{e}^{2+}(\alpha q)+4O{H}^{-}(aq)$

• A. $E_{\text{cell }}^0=-0.48V$
• B. $E_{\text{ cell}}^0=-0.04V$
• C. $E_{\text{cell }}^0=+0.84V$
• D. $E_{\text{cell}}^0=+1.28V$

Answer 1

Answer: (C)

$E_{\text{cell }}^0=E_{\text{cathode }}^0-E_{\text{anode }}^0$
$=0.40-(-0.44)=0.84V$