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Q.

Consider the following cell reaction,
$2 F e (s) + O_{2} (g) + 4 H^{+} (a q) \to 2 F e^{2 +} (a q) + 2 H_{2} O (l); E ° = 1.67 V$ $At [{Fe}^{2 +}] = 10^{- 3} M, p (O_{2}) = 0 .1 atm and pH = 3, the cell$ potential at 25°C is

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a

1.47V

b

1.57V

c

1.77V

d

1.87V

answer is D.

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Detailed Solution

$Here n = 4, and [H^{+}] = 10^{- 3} (as p H = 3)$
Applying Nernst equation
$E = E^{\circ} - \frac{0 .059}{n} l o g \frac{{[{F e}^{2 +}]}^{2}}{{[H^{+}]}^{4} (p_{O_{2}})} = 1 .67 - \frac{0 .059}{4} l o g \frac{{(10^{- 3})}^{2}}{{(10^{- 3})}^{4} \times 0 .1} = 1 .67 - \frac{0 .059}{4} l o g 10^{7} = 1 .67 - 0 .103 = 1 .567 V$

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Consider the following cell reaction, 2Fe(s)+O2(g)+4H+(aq)→2Fe2+(aq)+2H2O(l);E°=1.67 V At Fe2+=10−3M,pO2=0.1 atm and pH=3, the cell potential at 25°C is