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