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JEE Chemistry 2025 Question with Solution

For the given cell: Fe2+(aq)+Ag+(aq)→Fe3+(aq)+Ag(s)Fe^{2+}(aq) + Ag^+(aq) \to Fe^{3+}(aq) + Ag(s) The standard cell potential of the above reaction is given. The standard reduction potentials are given as: Ag++e−→AgE∘=x VAg^+ + e^- \to Ag \quad E^\circ = x \, V Fe2++2e−→FeE∘=y VFe^{2+} + 2e^- \to Fe \quad E^\circ = y \, V Fe3++3e−→FeE∘=z VFe^{3+} + 3e^- \to Fe \quad E^\circ = z \, V The correct answer is:

  • A

    x+y−zx + y - z

  • B

    x+2y−3zx + 2y - 3z

  • C

    y−2xy - 2x

  • D

    x+2yx + 2y

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The cell reaction is Fe2+(aq)+Ag+(aq)→Fe3+(aq)+Ag(s)Fe^{2+}(aq) + Ag^+(aq) \to Fe^{3+}(aq) + Ag(s). The given standard reduction potentials are Ag++e−→Ag, E∘=xAg^+ + e^- \to Ag,\ E^\circ = x, Fe2++2e−→Fe, E∘=yFe^{2+} + 2e^- \to Fe,\ E^\circ = y and Fe3++3e−→Fe, E∘=zFe^{3+} + 3e^- \to Fe,\ E^\circ = z.

Find: The expression for the standard cell potential.

At the cathode, silver ion is reduced:

Ag++e−→AgAg^+ + e^- \to Ag

So its reduction potential is xx.

To obtain the Fe3+/Fe2+Fe^{3+}/Fe^{2+} reduction potential, combine the given iron reduction reactions through Gibbs energy additivity, which gives:

E∘(Fe3++e−→Fe2+)=3z−2yE^\circ(Fe^{3+} + e^- \to Fe^{2+}) = 3z - 2y

But in the cell reaction, iron undergoes oxidation:

Fe2+→Fe3++e−Fe^{2+} \to Fe^{3+} + e^-

Therefore the corresponding oxidation potential is the negative of the above reduction potential.

Hence the standard cell potential is

Ecell∘=Ecathode∘−Eanode (reduction)∘=x−(3z−2y)=x+2y−3zE^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode\,(reduction)} = x - (3z - 2y) = x + 2y - 3z

Therefore, the correct option is B and the standard cell potential is x+2y−3zx + 2y - 3z.

The solution contains inconsistent intermediate statements, but its final conclusion and listed correct option both indicate x+2y−3zx + 2y - 3z.

Deriving the $$Fe^{3+}/Fe^{2+}$$ potential

Use the two given iron reduction half-reactions:

Fe3++3e−→FeE∘=zFe^{3+} + 3e^- \to Fe \qquad E^\circ = z Fe2++2e−→FeE∘=yFe^{2+} + 2e^- \to Fe \qquad E^\circ = y

Reverse the second reaction to write oxidation of iron metal to Fe2+Fe^{2+}:

Fe→Fe2++2e−Fe \to Fe^{2+} + 2e^-

Add it to the first reaction:

Fe3++3e−→FeFe→Fe2++2e−\begin{aligned} Fe^{3+} + 3e^- &\to Fe \\ Fe &\to Fe^{2+} + 2e^- \end{aligned}

This gives

Fe3++e−→Fe2+Fe^{3+} + e^- \to Fe^{2+}

Using ΔG∘=−nFE∘\Delta G^\circ = -nFE^\circ,

ΔG∘=−3Fz+2Fy=−F(3z−2y)\Delta G^\circ = -3Fz + 2Fy = -F(3z - 2y)

So,

E∘(Fe3+/Fe2+)=3z−2yE^\circ(Fe^{3+}/Fe^{2+}) = 3z - 2y

Now apply

Ecell∘=Ecathode∘−Eanode (reduction)∘E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode\,(reduction)}

with cathode potential xx and anode reduction potential 3z−2y3z - 2y:

Ecell∘=x−(3z−2y)=x+2y−3zE^\circ_{cell} = x - (3z - 2y) = x + 2y - 3z

Therefore, the correct option is B.

Common mistakes

  • Using E∘E^\circ values directly like ordinary algebraic quantities is incorrect when combining half-reactions. First convert through ΔG∘=−nFE∘\Delta G^\circ = -nFE^\circ, then recompute the net potential.

  • Treating Fe2+→Fe3++e−Fe^{2+} \to Fe^{3+} + e^- as if its potential were simply y−zy-z is wrong. The given iron half-reactions involve different electron counts, so the one-electron Fe3+/Fe2+Fe^{3+}/Fe^{2+} couple must be derived carefully.

  • Forgetting that the anode process in the cell is oxidation leads to sign errors. Use the reduction potential of the anode couple in Ecell∘=Ecathode∘−Eanode∘E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}, or reverse the sign consistently if writing oxidation potential.

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