First slide

Nernst Equation

Question

Consider the following cell reaction
$\begin{array}{r} 2 Fe (s) + O_{2} (g) + 4 H^{+} (aq) ⟶ \\ 2 {Fe}^{2 +} (aq) + 2 H_{2} O (l), E^{\circ} = 1 .67 V \end{array}$
$At [{Fe}^{2 +}] = 10^{- 3} M, p_{O_{2}} = 0 .1 atm and pH = 3,$ the cell potential at 25^oC is

Easy

A

1.47 V
B

1.77 V
C

1.87 V
D

1.57 V

Solution

The half reactions are $[Fe (s) ⟶ {Fe}^{2 +} (aq) + 2 e^{-}] \times 2$
$\begin{array}{l} O_{2} (g) + 4 H^{+} + 4 e^{-} ⟶ 2 H_{2} O \\ 2 Fe (s) + O_{2} (g) + 4 H^{+} ⟶ 2 {Fe}^{2 +} (aq) + 2 H_{2} O (l) \end{array}$
$E = 1 .67 - \frac{0 .059}{4} \log \frac{{(10^{- 3})}^{2}}{{(10^{- 3})}^{4} (0 .1)} = 1 .57 V$

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