Question:medium
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?
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?
Updated On: Jun 6, 2026
- 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
Show Solution
The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
For a spontaneous change in an electrochemical cell, the cell potential \(E^0_{cell}\) must be positive.
The electrode with the higher reduction potential acts as the cathode (reduction), and the electrode with the lower reduction potential acts as the anode (oxidation).
Step 2: Key Formula or Approach:
1. Spontaneous cell: \(E^0_{cell} = E^0_{\text{cathode}} - E^0_{\text{anode}}>0\).
2. Overall reaction = Reduction half-reaction + Oxidation half-reaction.
Step 3: Detailed Explanation:
Identify the electrodes:
\(E^0_{\text{AgBr}/Ag} = +0.07 \text{ V}\) (Higher value, acts as Cathode).
\(E^0_{\text{Fe(OH)}_2/\text{Fe}} = -0.88 \text{ V}\) (Lower value, acts as Anode).
At Cathode (Reduction):
\[ [AgBr(s) + e^- \rightarrow Ag(s) + Br^-(aq)] \times 2 \]
\[ 2AgBr(s) + 2e^- \rightarrow 2Ag(s) + 2Br^-(aq) \]
At Anode (Oxidation):
\[ Fe(s) + 2OH^-(aq) \rightarrow Fe(OH)_2(s) + 2e^- \]
Adding the two half-reactions for the overall reaction:
\[ Fe(s) + 2OH^-(aq) + 2AgBr(s) \rightarrow Fe(OH)_2(s) + 2Ag(s) + 2Br^-(aq) \]
Calculating \(E^0_{cell}\):
\[ E^0_{cell} = E^0_{\text{cathode}} - E^0_{\text{anode}} = 0.07 - (-0.88) = +0.95 \text{ V} \].
Analysis of Options:
- Option (A) correctly lists the spontaneous overall reaction.
- Option (B) is incorrect as the calculated \(E^0_{cell}\) is positive (\(+0.95 \text{ V}\)).
- Option (C) is incorrect as Iron (Fe) is oxidized from \(0\) to \(+2\) oxidation state.
- Option (D) is incorrect because electrode potential is an intensive property (it does not depend on the amount of substance).
Step 4: Final Answer:
The correct option is (A).
For a spontaneous change in an electrochemical cell, the cell potential \(E^0_{cell}\) must be positive.
The electrode with the higher reduction potential acts as the cathode (reduction), and the electrode with the lower reduction potential acts as the anode (oxidation).
Step 2: Key Formula or Approach:
1. Spontaneous cell: \(E^0_{cell} = E^0_{\text{cathode}} - E^0_{\text{anode}}>0\).
2. Overall reaction = Reduction half-reaction + Oxidation half-reaction.
Step 3: Detailed Explanation:
Identify the electrodes:
\(E^0_{\text{AgBr}/Ag} = +0.07 \text{ V}\) (Higher value, acts as Cathode).
\(E^0_{\text{Fe(OH)}_2/\text{Fe}} = -0.88 \text{ V}\) (Lower value, acts as Anode).
At Cathode (Reduction):
\[ [AgBr(s) + e^- \rightarrow Ag(s) + Br^-(aq)] \times 2 \]
\[ 2AgBr(s) + 2e^- \rightarrow 2Ag(s) + 2Br^-(aq) \]
At Anode (Oxidation):
\[ Fe(s) + 2OH^-(aq) \rightarrow Fe(OH)_2(s) + 2e^- \]
Adding the two half-reactions for the overall reaction:
\[ Fe(s) + 2OH^-(aq) + 2AgBr(s) \rightarrow Fe(OH)_2(s) + 2Ag(s) + 2Br^-(aq) \]
Calculating \(E^0_{cell}\):
\[ E^0_{cell} = E^0_{\text{cathode}} - E^0_{\text{anode}} = 0.07 - (-0.88) = +0.95 \text{ V} \].
Analysis of Options:
- Option (A) correctly lists the spontaneous overall reaction.
- Option (B) is incorrect as the calculated \(E^0_{cell}\) is positive (\(+0.95 \text{ V}\)).
- Option (C) is incorrect as Iron (Fe) is oxidized from \(0\) to \(+2\) oxidation state.
- Option (D) is incorrect because electrode potential is an intensive property (it does not depend on the amount of substance).
Step 4: Final Answer:
The correct option is (A).
Was this answer helpful?
0
Top Questions on Electrochemical Cells and Gibbs Free Energy
Standard electrode potentials for a few half-cells are mentioned below:
- JEE Main - 2025
- Chemistry
- Electrochemical Cells and Gibbs Free Energy
Consider the following cell: $ \text{Pt}(s) \, \text{H}_2 (1 \, \text{atm}) | \text{H}^+ (1 \, \text{M}) | \text{Cr}_2\text{O}_7^{2-}, \, \text{Cr}^{3+} | \text{H}^+ (1 \, \text{M}) | \text{Pt}(s) $
Given: $ E^\circ_{\text{Cr}_2\text{O}_7^{2-}/\text{Cr}^{3+}} = 1.33 \, \text{V}, \quad \left[ \text{Cr}^{3+} \right]^2 / \left[ \text{Cr}_2\text{O}_7^{2-} \right] = 10^{-7} $
At equilibrium: $ \left[ \text{Cr}^{3+} \right]^2 / \left[ \text{Cr}_2\text{O}_7^{2-} \right] = 10^{-7} $
Objective: $ \text{Determine the pH at the cathode where } E_{\text{cell}} = 0. $- JEE Main - 2025
- Chemistry
- Electrochemical Cells and Gibbs Free Energy
- Consider the reaction \( aX \rightarrow bY \), for which the rate constant at 30°C is \( 1 \times 10^{-3} \, \text{mol}^{-1} \, \text{L} \, \text{s}^{-1} \). Which of the following statements are true?}
- (A) When concentration of \( X \) is increased to four times, the rate of reaction becomes 16 times.
- (B) The reaction is a second order reaction.
- (C) The half-life period is independent of the concentration of \( X \).
- (D) Decomposition of \( \text{N}_2\text{O}_5 \) is an example of the above reaction.
- (E)
- JEE Main - 2026
- Chemistry
- Electrochemical Cells and Gibbs Free Energy
- One half cell in a voltaic cell is constructed by dipping silver rod in $AgNO_3$ solution of unknown concentration, other half cell is Zn rod dipped in 1 molar solution of $ZnSO_4$.
A voltage of 1.60 V is measured at 298 K for this cell. What is the concentration of $Ag^+$ ions used in terms of $\log x$ ($x = [Ag^+]$)?
$E^\ominus_{Zn^{2+}/Zn} = -0.76 V$, $E^\ominus_{Ag^+/Ag} = +0.80 V$, $\frac{2.303 RT}{F} = 0.059 V$- JEE Main - 2026
- Chemistry
- Electrochemical Cells and Gibbs Free Energy
- Want to practice more? Try solving extra ecology questions todayView All Questions
