Question 375 marksLong AnswerSection E

(a) Calculate emf and ΔG\Delta G for the following cell at 298 K:

Mg(s)/Mg2+(0.01 M)//Ag+(0.001 M)/Ag(s)Mg(s) / Mg^{2+}(0.01\text{ M}) // Ag^+(0.001\text{ M}) / Ag(s)

Given: E∘(Mg2+/Mg)=−2.37 VE^\circ(Mg^{2+}/Mg) = -2.37 \text{ V}, E∘(Ag+/Ag)=+0.80 VE^\circ(Ag^+/Ag) = +0.80 \text{ V}

[1F=96500 C mol−11F = 96500 \text{ C mol}^{-1}, log⁡10=1\log 10 = 1]

OR

(b) For the reaction:

2AgCl(s)+H2(g)(0.4 atm)→2Ag(s)+2H+(0.1 M)+2Cl−(0.2 M)2AgCl(s) + H_2(g) (0.4\text{ atm}) \to 2Ag(s) + 2H^+(0.1\text{ M}) + 2Cl^-(0.2\text{ M})

Calculate emf of the cell at 25 °C.

Given: ΔG∘=−43500 J mol−1\Delta G^\circ = -43500 \text{ J mol}^{-1}

[log⁡10=1\log 10 = 1, 1F=96500 C mol−11F = 96500 \text{ C mol}^{-1}]

Electrochemistry
Nernst equation: cell emf and Gibbs energy

Official answer

Primary answer — part (a)


Cell: Mg(s)∣Mg2+(0.01 M)∥Ag+(0.001 M)∣Ag(s)Mg(s) \mid Mg^{2+}(0.01\text{ M}) \| Ag^+(0.001\text{ M}) \mid Ag(s)


Step 1 — Standard cell potential


  • Ecell∘=Ecathode∘−Eanode∘=0.80−(−2.37)=E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.80 - (-2.37) = +3.17 V+3.17 \text{ V}

Step 2 — Cell reaction and n


  • Mg+2Ag+→Mg2++2AgMg + 2Ag^+ \to Mg^{2+} + 2Ag, so n=2n = 2.

Step 3 — Nernst equation (Q=[Mg2+]/[Ag+]2Q = [Mg^{2+}]/[Ag^+]^2)


  • Q=0.01/(0.001)2=0.01/10−6=Q = 0.01 / (0.001)^2 = 0.01 / 10^{-6} = 10410^4, so log⁡Q=4\log Q = 4.
  • Ecell=Ecell∘−(0.059/n)log⁡Q=3.17−(0.059/2)(4)E_{cell} = E^\circ_{cell} - (0.059/n) \log Q = 3.17 - (0.059/2)(4)
  • Ecell=3.17−0.118=E_{cell} = 3.17 - 0.118 = 3.052 V≈3.05 V3.052 \text{ V} \approx 3.05 \text{ V}

Step 4 — Gibbs energy


  • ΔG=−nF⋅Ecell=−(2)(96500)(3.052)\Delta G = -nF \cdot E_{cell} = -(2)(96500)(3.052)
  • ΔG=\Delta G = −5.89×105 J mol−1≈−589 kJ mol−1-5.89 \times 10^5 \text{ J mol}^{-1} \approx -589 \text{ kJ mol}^{-1}

OR — part (b) (brief)


2AgCl(s)+H2(g,0.4 atm)→2Ag(s)+2H+(0.1 M)+2Cl−(0.2 M)2AgCl(s) + H_2(g, 0.4\text{ atm}) \to 2Ag(s) + 2H^+(0.1\text{ M}) + 2Cl^-(0.2\text{ M}), n=2n = 2.


  • Ecell∘=−ΔG∘/(nF)=−(−43500)/(2×96500)=E^\circ_{cell} = -\Delta G^\circ/(nF) = -(-43500)/(2 \times 96500) = 0.225 V0.225 \text{ V}
  • Q=[H+]2[Cl−]2/p(H2)=(0.1)2(0.2)2/0.4=4×10−4/0.4=Q = [H^+]^2[Cl^-]^2 / p(H_2) = (0.1)^2(0.2)^2 / 0.4 = 4 \times 10^{-4} / 0.4 = 10−310^{-3}, log⁡Q=−3\log Q = -3.
  • Ecell=0.225−(0.059/2)(−3)=0.225+0.0885=E_{cell} = 0.225 - (0.059/2)(-3) = 0.225 + 0.0885 = 0.314 V≈0.31 V0.314 \text{ V} \approx 0.31 \text{ V}
Answer keywords:Nernst equationE°cell cathode minus anodeΔG = −nFEcellreaction quotient Qemf calculationMg Ag cellsilver chloride electroden = 2 electrons

Marking scheme

  1. 1.(a) 1 mark: Ecell∘=0.80−(−2.37)=3.17 VE^\circ_{cell} = 0.80 - (-2.37) = 3.17 \text{ V} with n=2n = 2 identified.
  2. 2.(a) 2 marks: correct Q=[Mg2+]/[Ag+]2=104Q = [Mg^{2+}]/[Ag^+]^2 = 10^4 and Nernst substitution giving Ecell≈3.05 VE_{cell} \approx 3.05 \text{ V}.
  3. 3.(a) 2 marks: ΔG=−nFEcell≈−589 kJ mol−1\Delta G = -nFE_{cell} \approx -589 \text{ kJ mol}^{-1} (≈−5.89×105 J\approx -5.89 \times 10^5 \text{ J}) with correct sign and units.
  4. 4.OR (b) 2 marks: Ecell∘=−ΔG∘/nF=0.225 VE^\circ_{cell} = -\Delta G^\circ/nF = 0.225 \text{ V}; 2 marks: Q=[H+]2[Cl−]2/p(H2)=10−3Q = [H^+]^2[Cl^-]^2/p(H_2) = 10^{-3}; 1 mark: Ecell≈0.31 VE_{cell} \approx 0.31 \text{ V}.

Hint

Ecell∘=Ecathode∘−Eanode∘E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}; then Ecell=Ecell∘−(0.059/n)log⁡QE_{cell} = E^\circ_{cell} - (0.059/n) \log Q with the correctly-exponentiated Q, and ΔG=−nF⋅Ecell\Delta G = -nF \cdot E_{cell}.

Quick oral answer

First Ecell∘=0.80−(−2.37)=3.17 VE^\circ_{cell} = 0.80 - (-2.37) = 3.17 \text{ V} with two electrons transferred; the Nernst equation with Q=[Mg2+]/[Ag+]2=104Q = [Mg^{2+}]/[Ag^+]^2 = 10^4 gives Ecell=3.17−0.0295×4=3.05 VE_{cell} = 3.17 - 0.0295 \times 4 = 3.05 \text{ V}, and ΔG=−nFEcell=−2×96500×3.05≈−589 kJ per mole\Delta G = -nFE_{cell} = -2 \times 96500 \times 3.05 \approx -589 \text{ kJ per mole}.

Common mistakes

  1. 1.Omitting the exponent on [Ag+][Ag^+] (writing Q=[Mg2+]/[Ag+]Q = [Mg^{2+}]/[Ag^+] instead of [Ag+]2[Ag^+]^2), which changes log Q and the emf.
  2. 2.Using Ecell∘E^\circ_{cell} instead of the actual EcellE_{cell} in ΔG=−nFEcell\Delta G = -nFE_{cell} when concentrations are non-standard.
  3. 3.Wrong n value: the Mg–Ag cell transfers 2 electrons (Ag+Ag^+ is a one-electron couple but two Ag+Ag^+ are reduced per MgMg).

Analysis and explanation

Concept — Nernst equation and ΔG\Delta G


The Nernst equation links the actual cell potential to concentrations:


  • Ecell=Ecell∘−(0.059/n)log⁡QE_{cell} = E^\circ_{cell} - (0.059/n) \log Q at 298 K,

and the free-energy change of the working cell is ΔG=−nF⋅Ecell\Delta G = -nF \cdot E_{cell} (use EcellE_{cell}, not Ecell∘E^\circ_{cell}, when non-standard concentrations are given).


Setting up Q correctly


  • For part (a), the balanced reaction Mg+2Ag+→Mg2++2AgMg + 2Ag^+ \to Mg^{2+} + 2Ag gives Q=[Mg2+]/[Ag+]2Q = [Mg^{2+}]/[Ag^+]^2; the square on [Ag+][Ag^+] is essential.
  • For part (b), solids (Ag, AgCl) are omitted and H₂ enters as its partial pressure: Q=[H+]2[Cl−]2/p(H2)Q = [H^+]^2[Cl^-]^2/p(H_2).

Exam trap


  • Forgetting the exponents in Q (the 2 in [Ag+]2[Ag^+]^2 or the squares in part b) is the most common scoring error.
  • Sign of ΔG\Delta G: it must come out negative here because E°cell and Ecell are positive (spontaneous cell).

Real-world


The silver–silver-chloride electrode of part (b) is a widely used, stable reference electrode in pH meters and biomedical sensors.

Interesting facts

The Nernst equation was formulated by Walther Nernst in 1889; he won the 1920 Nobel Prize in Chemistry for his work in thermochemistry.

The factor 0.059 V0.059 \text{ V} is 2.303RT/F2.303RT/F at 298 K298\text{ K} — it is temperature-dependent, so cell potentials shift measurably when the temperature changes.

Question details

Q37 of 37 questions

Board
CBSE
Class
Class 12
Subject
Chemistry
Year
2026
Section
Section E
Marks
5
Type
Long Answer
Chapter
Electrochemistry
Topic
Nernst equation: cell emf and Gibbs energy
Bloom's level
apply
Syllabus topic
Unit: Electrochemistry — Nernst Equation, Cell EMF, and Relation ΔG = −nFEcell
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

Why is n=2n = 2 for the Mg–Ag cell?

Magnesium loses 2 electrons (Mg→Mg2++2e−Mg \to Mg^{2+} + 2e^-) and two silver ions each gain one electron (2Ag++2e−→2Ag2Ag^+ + 2e^- \to 2Ag), so two electrons are transferred per overall reaction, giving n=2n = 2.

Do you use EcellE_{cell} or Ecell∘E^\circ_{cell} in ΔG\Delta G?

When concentrations are non-standard, use the actual EcellE_{cell} from the Nernst equation: ΔG=−nF⋅Ecell\Delta G = -nF \cdot E_{cell}. Ecell∘E^\circ_{cell} only gives ΔG∘\Delta G^\circ (standard free energy).

How is Q written for the AgCl–H₂ reaction?

Pure solids Ag and AgCl are excluded and hydrogen appears as its partial pressure, so Q=[H+]2[Cl−]2/p(H2)=(0.1)2(0.2)2/0.4=10−3Q = [H^+]^2[Cl^-]^2/p(H_2) = (0.1)^2(0.2)^2/0.4 = 10^{-3}.