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Electrode potential; Cu/Ag cell EMF

Short answer

Electrode potential is the tendency at electrode–solution interface. Zn standard potential is measured by coupling with SHE; emf 0.76 V gives E° = -0.76 V. For the given cell, E° = 0.80 - 0.34 = 0.46 V; applying Nernst for 2e⁻ with Q = 0.2/(0.1)² = 20 gives E = 0.46 - 0.038 = 0.42 V.

Define electrode potential. How do you measure standard electrode potential of zinc electrode? Find the Emf of the following cell at 25°C. Cu/Cu²⁺(0.2M)//Ag⁺(0.1M)/Ag Given: E°Ag⁺/Ag = 0.80 V and E°Cu²⁺/Cu = −0.34 V

Electrode potential

The tendency of an electrode (metal) to lose or gain electrons when it is in contact with a solution of its own ions is called its electrode potential. It is the potential difference at the metal–solution interface. When ions are at 1 M, gases at 1 atm and temperature 25°C, it is called the standard electrode potential (E°).

Measuring standard electrode potential of zinc

A zinc rod dipped in 1 M ZnSO₄ is connected through a salt bridge to a standard hydrogen electrode (SHE, Pt, H₂ 1 atm | H⁺ 1 M, E° = 0).

Zn | Zn²⁺ (1 M) || H⁺ (1 M) | H₂ (1 atm), Pt

Zn is oxidised (anode), H⁺ is reduced (cathode). The measured emf of the cell is 0.76 V, and since E°SHE = 0: E°cell = 0 − E°(Zn²⁺/Zn) = 0.76 V, so E°(Zn²⁺/Zn) = −0.76 V.

EMF of Cu | Cu²⁺ (0.2 M) || Ag⁺ (0.1 M) | Ag

The given E°(Cu²⁺/Cu) = −0.34 V is taken as the oxidation potential of Cu (reduction potential of Cu²⁺/Cu is +0.34 V). Ag⁺/Ag = +0.80 V.

  • Anode: Cu → Cu²⁺ + 2e⁻
  • Cathode: 2Ag⁺ + 2e⁻ → 2Ag
  • Cell reaction: Cu + 2Ag⁺ → Cu²⁺ + 2Ag (n = 2)
E°cell = 0.80 − 0.34 = 0.46 V
Q = [Cu²⁺]/[Ag⁺]² = 0.2/(0.1)² = 20,  log Q = 1.301
E = E° − (0.0591/2) log Q
  = 0.46 − 0.02955 × 1.301
  = 0.46 − 0.0384 = 0.4216 V

Answer: EMF ≈ 0.42 V.

✓ Answer checked by Claude Opus 5.5 (high effort)

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