Consider the following cell reaction, array cc & 2 Fe(s)+O_2(g)+4 H^ + (a q) & 2 Fe^ 2+ (a q)+2 H_2 O(l), E^ =1.67 V At & [Fe^ 2+ ]=10^ -3 M, P(O_2)=0.1 atm array and pH=3 , the cell potential at 25^ C is

Correct answer: D. 1.57 ~V The half reactions are Fe(s) Fe^ 2+ (aq)+2 e^ - × 2 array r O_ 2 (g)+4 H^ + +4 e^ - 2 H _ 2 O+2 Fe(s) +O _ 2 (g)+4 H^ + 2 Fe^ 2+ (a q)+2 H _2 O(l) E=E^ ^ -(0.059)/(4) (10^ -3 )^2 (10^ -3 )^4(0.1) =1.57 V array

JEE Advanced 2011Paper 2

Exam: JEE Advanced
Year: 2011
Paper: Paper 2
Subject: Chemistry
Chapter: Electrochemistry

JEE Advanced 2011 · Paper 2 · Chemistry · Electrochemistry

Consider the following cell reaction, array cc & 2 Fe(s)+O_2(g)+4...

Consider the following cell reaction,

\begin{array}{cc} & 2 \mathrm{Fe}(\mathrm{s})+\mathrm{O}_2(g)+4 \mathrm{H}^{+}(a q) \longrightarrow \\ & 2 \mathrm{Fe}^{2+}(a q)+2 \mathrm{H}_2 \mathrm{O}(l), E^{\circ}=1.67 \mathrm{ V} \\ \text{ At } & {\left[\mathrm{Fe}^{2+}\right]=10^{-3} \mathrm{M}, \mathrm{P}\left(\mathrm{O}_2\right)=0.1 \mathrm{ atm}} \end{array}

and

\mathrm{pH}=3

, the cell potential at

25^{\circ} \mathrm{C} is

Options

A $1.47 \mathrm{~V}$
B $1.77 \mathrm{~V}$
C $1.87 \mathrm{~V}$
D $1.57 \mathrm{~V}$

Correct answer

D. $1.57 \mathrm{~V}$

Step-by-step solution

The half reactions are

\mathrm{Fe}(s) \longrightarrow \mathrm{Fe}^{2+}(\mathrm{aq})+2 e^{-} \times 2

\begin{array}{r}\mathrm{O}_{2}(g)+4 \mathrm{H}^{+}+4 e^{-} \longrightarrow 2 \mathrm{H}}_{2} \mathrm{O}+2 \mathrm{Fe}(s) \\ +\mathrm{O}}_{2}(g)+4 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Fe}^{2+}(a q)+2 \mathrm{H}{ }_2 \mathrm{O}(l) \\ E=E^{{}^{\circ}}-\frac{0.059}{4} \log \frac{\left(10^{-3}\right)^2}{\left(10^{-3}\right)^4(0.1)}=1.57 \mathrm{ V} \end{array}

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