. Given, E_ Cr^ 3+ / Cr ^0=-0.72 ~V, E_ fe^ 2+ / Fe =-0.42 ~V . The potential for the cell Cr Cr^ 3+ (0.1 M) \| Fe^ 2+ (0.01 M) Fe is

Correct answer: C. 0.26 V aligned E_ cell ^0 & =E_ cathode -E_ anode & =-0.42-(-0.72)=0.3 aligned At anode [Cr Cr^ 3+ +3 e^ - ] × 2 At cathode [Fe^ 2+ +2 e^ - Fe] × 3 2 Cr+3 Fe^ 2+ arrow 2 Cr^ 3+ +3 Fe aligned & E=E_ cell ^0-(0.0591)/(6) [Cr^ 3+ ]^2 [Fe^ 2+ ]^3 & =0.3-(0.0591)/(6) ((0.1)^2)/((0.01)^3) & =0.26 V aligned

JIPMER 2019

Exam: JIPMER
Year: 2019
Subject: Chemistry
Chapter: Electrochemistry

JIPMER 2019 · Chemistry · Electrochemistry

. Given, E_ Cr^ 3+ / Cr ^0=-0.72 ~V, E_ fe^ 2+ / Fe =-0.42 ~V . The...

. Given,

\mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}}^0=-0.72 \mathrm{~V}, \mathrm{E}_{\mathrm{fe}^{2+} / \mathrm{Fe}}=-0.42 \mathrm{~V}

. The potential for the cell

\mathrm{Cr} \mid \mathrm{Cr}^{3+}(0.1 \mathrm{M}) \| \mathrm{Fe}^{2+}(0.01 \mathrm{M}) \mathrm{Fe}

Options

A-0.339 V
B-0.26 V
C0.26 V
D0.3 V

Correct answer

C. 0.26 V

Step-by-step solution

\begin{aligned} \mathrm{E}_{\text{cell }^0 & =\mathrm{E}_{\text{cathode }}-\mathrm{E}_{\text{anode }} \\ & =-0.42-(-0.72)=0.3 \end{aligned}

At anode

\left[\mathrm{Cr} \longrightarrow \mathrm{Cr}^{3+}+3 \mathrm{e}^{-}\right] \times 2

At cathode

\left[\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}\right] \times 3 2 \mathrm{Cr}+3 \mathrm{Fe}^{2+} \rightarrow 2 \mathrm{Cr}^{3+}+3 \mathrm{Fe}

\begin{aligned} \therefore \quad & \mathrm{E}=\mathrm{E}_{\text{cell }^0-\frac{0.0591}{6} \log \frac{\left[\mathrm{Cr}^{3+}\right]^2}{\left[\mathrm{Fe}^{2+}\right]^3} \\ & =0.3-\frac{0.0591}{6} \log\frac{(0.1)^2}{(0.01)^3} \\ & =0.26 \mathrm{ V}\end{aligned}

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