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JIPMER 2018
Exam
JIPMER
Year
2018
Subject
Chemistry
Chapter
Electrochemistry

JIPMER 2018 · Chemistry · Electrochemistry

If E^ (Zn^ 2+ , Zn)=-0.763 ~V and E^ (Fe^ 2+ , Fe)=-0.44 ~V , then...

If E∘(Zn2+,Zn)=−0.763 V\mathrm{E}^{\circ}\left(\mathrm{Zn}^{2+}, \mathrm{Zn}\right)=-0.763 \mathrm{~V} and E∘(Fe2+,Fe)=−0.44 V\mathrm{E}^{\circ}\left(\mathrm{Fe}^{2+}, \mathrm{Fe}\right)=-0.44 \mathrm{~V}, then the emf of the cell Zn∣Zn2+(a=0.00l)∥Fe2+(a=0.005)∣Fe\mathrm{Zn}\left|\mathrm{Zn}^{2+}(\mathrm{a}=0.00 \mathrm{l}) \| \mathrm{Fe}^{2+}(\mathrm{a}=0.005)\right| \mathrm{Fe} is

Options

  1. Aequal to 0.323 V
  2. Bless than 0.323 V
  3. Cgreater than 0.323 V
  4. Dequal to 1.103 V

Correct answer

C. greater than 0.323 V

Step-by-step solution

The cell reaction is Zn+Fe2+⟶Zn2++Fe\mathrm{Zn}+\mathrm{Fe}^{2+} \longrightarrow \mathrm{Zn}^{2+}+\mathrm{Fe} From Nernst equation Ecell =Ecell ∘−0.0591nlog⁡aZn2+aFe2+=(0.763−0.44)−0.05911log⁡0.0010.005=0.364 V\begin{aligned} & \mathrm{E}_{\text{cell }}=\mathrm{E}_{\text{cell }}^{\circ}-\frac{0.0591}{\mathrm{n}} \log \frac{\mathrm{a}_{\mathrm{Zn}^{2+}}}{\mathrm{a}_{\mathrm{Fe}^{2+}}} \\ & =(0.763-0.44)-\frac{0.0591}{1} \log \frac{0.001}{0.005} \\ & =0.364 \mathrm{~V} \\ & \end{aligned}
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