Consider the following cell reaction: 2Fe(s) + O2 (g) + 4H+ (aq) \rightarrow_{}^{} 2Fe2+ (aq) + 2H2O (ℓ) E° = 1.67 V At [Fe2+] = 10–3 M, P(O2) = 0.1 atm and pH = 3, the cell potential at 25° C is :

Engineering

Chemistry

Electrolytes and Types

Question

Consider the following cell reaction:

2Fe(s) + O₂ (g) + 4H⁺ (aq) $\to_{}^{}$ 2Fe²⁺ (aq) + 2H₂O (ℓ) E° = 1.67 V

At [Fe²⁺] = 10^–3 M, P(O₂) = 0.1 atm and pH = 3, the cell potential at 25° C is :

1.87 V

1.77 V

1.47 V

1.57 V

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$E_{cell} = E °_{cell} - \frac{. 0591}{4} \log \frac{[{Fe}^{+ 2}]}{[{Po}_{2}] {[H^{+}]}^{4}}$

$= 1.64 - \frac{.591}{4} \log \frac{{[10^{- 3}]}^{2}}{[.1] [10^{- 3}]}$

= 1.57 V

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