Consider the following cell reaction: \(2\mathrm{Fe}_{(\mathrm{s})}+\mathrm{O}_{2(\mathrm{~g})}+4\mathrm{H}^{+}_{(\mathrm{aq})} \rightarrow 2\mathrm{Fe}^{2+}_{(\mathrm{aq})}+2\mathrm{H}_{2}\mathrm{O}_{(\mathrm{l})} \quad \mathrm{E}^{\circ}=1.67\mathrm{~V}\) At \(\left[\mathrm{Fe}^{2+}\right]=10^{-3}\mathrm{M}\), \(\mathrm{P}\left(\mathrm{O}_{2}\right)=0.1\mathrm{~atm}\) and \(\mathrm{pH}=3\), the cell potential at \(25^{\circ}\mathrm{C}\) is

JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Redox Reactions and Electrochemistry

Consider the following cell reaction:

2 Fe_{(s)} + O_{2 (g)} + 4 H_{(aq)}^{+} \to 2 Fe_{(aq)}^{2 +} + 2 H_{2} O_{(l)} E^{\circ} = 1.67 V

[Fe^{2 +}] = 1 0^{- 3} M

P (O_{2}) = 0.1 atm

and

pH = 3

, the cell potential at

2 5^{\circ} C

Answer: (D)

Using the Nernst equation:

E_{cell} = E_{cell}^{\circ} - \frac{0.0591}{n} lo g Q

For the given reaction,

n = 4

electrons are transferred.

The reaction quotient:

Q = \frac{[ Fe ^{2 +} ] ^{2}}{P ( O _{2} ) [ H ^{+} ] ^{4}}

Given:

[Fe^{2 +}] = 1 0^{- 3} M

P (O_{2}) = 0.1 atm

pH = 3

[H^{+}] = 1 0^{- 3} M

Q = \frac{( 1 0 ^{- 3} ) ^{2}}{0.1 \times ( 1 0 ^{- 3} ) ^{4}} = \frac{1 0 ^{- 6}}{0.1 \times 1 0 ^{- 12}} = \frac{1 0 ^{- 6}}{1 0 ^{- 13}} = 1 0^{7}

E_{cell} = 1.67 - \frac{0.0591}{4} lo g (1 0^{7}) = 1.67 - \frac{0.0591 \times 7}{4} = 1.67 - 0.103 \approx 1.57 V

JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Single correct option