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JEE Mains · Chemistry · STD 12 - 2. Electrochemistry

An electrochemical cell is constructed using half cells in the direction of spontaneous change
\(Fe(OH)_2(s) + 2e^- \rightarrow Fe(s) + 2OH^-(aq) \quad E^\theta = -0.88\) V
and \(AgBr(s) + e^- \rightarrow Ag(s) + Br^-(aq) \quad E^\theta = +0.07\) V
Which of the following option is correct?

  1. A Overall reaction \(Fe(s) + 2OH^-(aq) + 2AgBr(s) \rightleftharpoons Fe(OH)_2(s) + 2Ag(s) + 2Br^-(aq)\)
  2. B \(E^\theta_{cell} = -0.95\) V
  3. C Fe is reduced in the electrochemical cell
  4. D \(E^\theta_{cell}\) is an extensive property
Verified Solution

Answer & Solution

Correct Answer

(A) Overall reaction \(Fe(s) + 2OH^-(aq) + 2AgBr(s) \rightleftharpoons Fe(OH)_2(s) + 2Ag(s) + 2Br^-(aq)\)

Step-by-step Solution

Detailed explanation

For a spontaneous reaction, the cell potential \(E^{\theta}_{cell}\) must be positive. The half-cell with the higher standard reduction potential acts as the cathode (reduction), and the one with the lower standard reduction potential acts as the anode (oxidation). Given…
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