CBSE Class 12 Chemistry 2024: question 17

Hints

Open one hint at a time and try again before the next.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

1. Electrode reaction: $\mathrm{F}{\mathrm{e}}^{2+}(\mathrm{aq})+2{\mathrm{e}}^{-}\to \mathrm{Fe}(\mathrm{s})$, so $n=2$. Nernst equation: $E_{\mathrm{F}{\mathrm{e}}^{2+}/Fe}=E_{\mathrm{F}{\mathrm{e}}^{2+}/Fe}^{\circ }-\frac{0.059}{n}\log \frac{1}{[\mathrm{F}{\mathrm{e}}^{2+}]}$

   ½ mark

2. Substitution: $E=-0.45-\frac{0.059}{2}\log \frac{1}{0.01}=-0.45-0.0295\times \log 100=-0.45-0.0295\times 2$

   1 mark

3. $E=-0.45-0.059=-0.509\text{V}$

   ½ mark

Answer: $E_{\mathrm{F}{\mathrm{e}}^{2+}/Fe}=-0.509$ V

Revise the chapter

• Chapter: Electrochemistry (notes, formulas, where marks are lost and every past question)
• Topic: Nernst Equation, Gibbs Energy and Equilibrium Constant