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COMEDK · Chemistry · 18. Electrochemistry

What will be the emf of the following cell at \(25^{\circ} \mathrm{C}\) ?
\(\mathrm{Fe} / \mathrm{Fe}^{2+}(0.001 \mathrm{M})|| \mathrm{H}^{+}(0.01 \mathrm{M}) \mid \mathrm{H}_2(\mathrm{~g})\)
(1 Bar) \(\mid \mathrm{Pt}(s)\)
\(E_{\left(\mathrm{Fe}^{2+} / \mathrm{Fe}\right)}^{\circ}=-0.44 \mathrm{~V} ; E_{\left(\mathrm{H}^{+} / \mathrm{H}_2\right)}^{\circ}=0.00 \mathrm{~V}\)

  1. A \(0.44 \mathrm{~V}\)
  2. B \(-0.44 \mathrm{~V}\)
  3. C \(0.41 \mathrm{~V}\)
  4. D \(-0.41 \mathrm{~V}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(0.41 \mathrm{~V}\)

Step-by-step Solution

Detailed explanation

The cell reaction is given by: Anode: \(\text{Fe}(s) \rightarrow \text{Fe}^{2+}(aq) + 2e^{-}\) Cathode: \(2\text{H}^{+}(aq) + 2e^{-} \rightarrow \text{H}_{2}(g)\) Overall reaction: \(\text{Fe}(s) + 2\text{H}^{+}(aq) \rightarrow \text{Fe}^{2+}(aq) + \text{H}_{2}(g)\) The standard…