An electrochemical cell is constructed using half cells in the direction of spontaneous change Fe(OH)2(s) + 2e- -> Fe(s) + 2OH-(aq) E0 = -0.88 V and AgBr(s) + e- -> Ag(s) + Br-(aq) E0 = +0.07 V Which of the following option is correct ?

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Chemistry Q57 – JEE-MAIN 2026

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Home Chemistry ElectrochemistryQuestion #390

An electrochemical cell is constructed using half cells in the direction of spontaneous change

{Fe(OH)}_{2} (s) + 2 e^{-} \to Fe(s) + 2 {OH}^{-} (aq)

E^{0} = - 0.88 V

and

AgBr(s) + e^{-} \to Ag(s) + {Br}^{-} (aq)

E^{0} = + 0.07 V

Which of the following option is correct ?

Clue 1 of 3: Spontaneity and Electrode Potentials

For a spontaneous electrochemical cell, the standard cell potential ( $E_{cell}^{0}$ ) must be positive. This occurs when the species with the higher (more positive) standard reduction potential undergoes reduction, and the species with the lower (more negative) standard reduction potential undergoes oxidation.

🧠 Full Solution Path

Step 1: Determine Anode and Cathode✦ Active

Compare the given standard reduction potentials: $E^{0} (AgBr/Ag) = + 0.07 V$ and $E^{0} ({Fe(OH)}_{2} /Fe) = - 0.88 V$ . Since $+ 0.07 V$ is greater than $- 0.88 V$ , the $AgBr/Ag$ half-cell will act as the cathode (reduction), and the ${Fe(OH)}_{2} /Fe$ half-cell will act as the anode (oxidation).

\begin{array}{ll} Anode (Oxidation): & Fe(s) + 2 {OH}^{-} (aq) \to {Fe(OH)}_{2} (s) + 2 e^{-} \\ Cathode (Reduction): & AgBr(s) + e^{-} \to Ag(s) + {Br}^{-} (aq) \end{array}

Step 2: Calculate Standard Cell Potential (

E_{cell}^{0}

)○ Expand

The standard cell potential is calculated as the difference between the standard reduction potential of the cathode and the anode.

E_{cell}^{0} = E_{cathode}^{0} - E_{anode}^{0} = (+ 0.07 V) - (- 0.88 V) = + 0.07 V + 0.88 V = + 0.95 V

Since

E_{cell}^{0}

is positive, the reaction is spontaneous.

💡 Teacher's Secret Hint

Remember that $E_{cell}^{0}$ must be positive for a spontaneous reaction.

Step 3: Derive Overall Reaction and Evaluate Options○ Expand

To balance the electrons, multiply the cathode reaction by 2. Then, add the anode and balanced cathode reactions to get the overall reaction.

\begin{array}{l} Anode: Fe(s) + 2 {OH}^{-} (aq) \to {Fe(OH)}_{2} (s) + 2 e^{-} \\ Cathode: 2 AgBr(s) + 2 e^{-} \to 2 Ag(s) + 2 {Br}^{-} (aq) \\ Overall: Fe(s) + 2 {OH}^{-} (aq) + 2 AgBr(s) \to {Fe(OH)}_{2} (s) + 2 Ag(s) + 2 {Br}^{-} (aq) \end{array}

Now, let's evaluate the given options: * **Option 1:** The overall reaction matches our derived reaction. This is correct. * **Option 2:** States $E_{cell}^{0} = - 0.95 V$ , which is incorrect as we calculated $+ 0.95 V$ . * **Option 3:** States Fe is reduced. In the anode reaction, Fe goes from oxidation state 0 to +2, which is oxidation, not reduction. This is incorrect. * **Option 4:** States $E_{cell}^{0}$ is an extensive property. $E_{cell}^{0}$ is an intensive property, meaning it does not depend on the amount of substance. This is incorrect.

💡 Teacher's Secret Hint

Pay close attention to the direction of electron flow and changes in oxidation states to determine if a species is oxidized or reduced.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Chemistry Question 57 – JEE-MAIN 2026

Clue 1 of 3: Spontaneity and Electrode Potentials

🧠 Full Solution Path