Question

At one time on Earth, iron was present mostly as iron(II). Later once plants had produced a significant quantity of oxygen in the atmosphere, the iron became oxidized to iron(III).

Show that Fe²⁺_(aq) can be spontaneously oxidized to Fe³⁺_(aq) by O_{2 (g)} at 25°C.

Assuming the following reasonable environmental conditions:

[Fe²⁺] = [Fe³⁺] = 1 x 10^-7 M
pH = 7.0
P_O2 = 160 mm Hg.

Answer 1

Answer 2

1. Standard Cell Potential

The two half-reactions:

• Oxidation (loses e⁻):
  ${\text{Fe}}^{2+}\to {\text{Fe}}^{3+}+e^{-}$
  $E_{\text{Fe}}^{\circ }=+0.77\text{V}$ (but flip the sign for oxidation: −0.77 V)

• Reduction (gains e⁻):
  $O_2+4H^{+}+4e^{-}\to 2H_{2}O$
  $E_{O_2}^{\circ }=+1.23\text{V}$

Total $E_{\text{cell}}^{\circ }$:
$1.23\text{V}-0.77\text{V}=+0.46\text{V}$ (spontaneous!)

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2. Adjust for Real Conditions (Nernst Equation)

Given:

• $[{\text{Fe}}^{2+}]=[{\text{Fe}}^{3+}]=10^{-7}\text{M}$

• $\text{pH}=7$ → $[H^{+}]=10^{-7}\text{M}$

• $P_{O_2}=160\text{mmHg}\approx 0.21\text{atm}$

Reaction Quotient ($Q$):

$$Q=\frac{[{\text{Fe}}^{3+}{]}^4}{[{\text{Fe}}^{2+}{]}^4[H^{+}{]}^4{P}_{O_2}}=\frac{(10^{-7}{)}^4}{(10^{-7}{)}^4(10^{-7}{)}^4\times 0.21}=\frac{1}{0.21\times 10^{-28}}\approx 4.76\times 10^{28}$$

Nernst Equation:

$$E_{\text{cell}}=0.46-\frac{0.0592}{4}\log (4.76\times 10^{28})$$

• $\log (4.76\times 10^{28})\approx 28.68$

• So: $0.46-(0.0148\times 28.68)\approx 0.46-0.42=+0.04\text{V}$