An electrochemical cell is constructed using half cells in the direction of spontaneous change}
\[
Fe(OH)_2(s) + 2e^- \rightarrow Fe(s) + 2OH^- (aq) \qquad E^\circ = -0.88\,V
\]
\[
AgBr(s) + e^- \rightarrow Ag(s) + Br^- (aq) \qquad E^\circ = +0.07\,V
\]
Which of the following option is correct?
- Overall reaction \(Fe(s)+2OH^- (aq)+2AgBr(s) \rightarrow Fe(OH)_2(s)+2Ag(s)+2Br^- (aq)\)
- \(E^\circ_{cell} = -0.95\,V\)
- Fe is reduced in the electrochemical cell
- \(E^\circ_{cell}\) is an extensive property
The Correct Option is A
Solution and Explanation
- Reduction occurs at the cathode (higher reduction potential).
- Oxidation occurs at the anode.
- \(E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}\).
Step 1:Identify cathode and anode} Reduction potentials: \[ AgBr/Ag = +0.07\,V \] \[ Fe(OH)_2/Fe = -0.88\,V \] Thus \[ \text{Cathode: } AgBr + e^- \rightarrow Ag + Br^- \] \[ \text{Anode: } Fe + 2OH^- \rightarrow Fe(OH)_2 + 2e^- \]
Step 2:Balance electrons} Multiply silver reaction by \(2\): \[ 2AgBr + 2e^- \rightarrow 2Ag + 2Br^- \]
Step 3:Add the reactions} \[ Fe + 2OH^- + 2AgBr \rightarrow Fe(OH)_2 + 2Ag + 2Br^- \] Thus option (A) is correct.
Step 4:Cell potential} \[ E^\circ_{cell} = 0.07 - (-0.88) \] \[ = 0.95\,V \] Hence option (B) is incorrect.
Step 5:Other statements}
- Fe undergoes oxidation, not reduction.
- \(E^\circ_{cell}\) is an intensive property.
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