Question

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

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

Answer 1

Answer: (C)

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