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Fe-Ag cell reactions and EMF

Short answer

Fe anode, Ag cathode. Net reaction: Fe + 2Ag⁺ → Fe²⁺ + 2Ag. Cell: Fe | Fe²⁺ (0.5 M) || Ag⁺ (0.2 M) | Ag. Standard EMF = 0.80 - (-0.44) = 1.24 V. Using Nernst equation with Q=12.5, E = 1.24 - 0.0324 = 1.21 V; spontaneous.

From the given electrode couple E°(Fe/Fe²⁺) = 0.44 volt, [Fe²⁺] = 0.5 M and E°(Ag/Ag⁺) = −0.80 volt, [Ag⁺] = 0.2 M. Write the (i) electrode reaction (ii) net cell reaction (iii) cell notation (iv) EMF of Fe-Ag cell and spontaneity of the cell reaction.

Reading of data: the given values are oxidation potentials: Fe → Fe²⁺ + 2e⁻ (E° = +0.44 V) and Ag → Ag⁺ + e⁻ (E° = −0.80 V). So the reduction potential of Ag⁺/Ag is +0.80 V and that of Fe²⁺/Fe is −0.44 V. Fe has the higher oxidation potential, so it is the anode.

(i) Electrode reactions

  • Anode (oxidation): Fe → Fe²⁺ + 2e⁻
  • Cathode (reduction): 2Ag⁺ + 2e⁻ → 2Ag

(ii) Net cell reaction

Fe + 2Ag⁺ → Fe²⁺ + 2Ag (n = 2)

(iii) Cell notation

Fe | Fe²⁺ (0.5 M) || Ag⁺ (0.2 M) | Ag

(iv) EMF and spontaneity

E°cell = E°(cathode) − E°(anode), in reduction potentials = 0.80 − (−0.44) = 1.24 V

Nernst equation at 25°C:

E = E° − (0.0591/n) log Q
Q = [Fe²⁺]/[Ag⁺]² = 0.5/(0.2)² = 12.5
log Q = 1.097
E = 1.24 − (0.0591/2)(1.097)
E = 1.24 − 0.0324 = 1.208 V

Answer: EMF ≈ 1.21 V. Since the EMF is positive (ΔG = −nFE < 0), the cell reaction is spontaneous.

✓ Answer checked by Claude Opus 5.5 (high effort)

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