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Normal hydrogen electrode; Ni/Cu cell emf

Short answer

Normal hydrogen electrode consists of Pt foil coated with Pt black, immersed in 1 M H⁺ with H₂ at 1 atm and 25°C; potential taken as zero. For Ni|Ni²⁺(0.1 M)||Cu²⁺(0.2 M)|Cu, Ni anode, Cu cathode: E°=0.59 V, Q=0.5, EMF=0.60 V using Nernst equation.

What is normal hydrogen electrode? A galvanic cell consists of a metallic nickel plate immersed in 0.1M Ni(NO₃)₂ solution and a metallic plate of copper in 0.2M CuSO₄ solution. Calculate the emf of this cell. E°Ni²⁺/Ni = −0.25 V, E°Cu²⁺/Cu = +0.34 V

Normal hydrogen electrode

A reference electrode made of platinum foil coated with platinum black, dipped in 1 N (1 M) H⁺ solution with pure H₂ gas at 1 atm bubbled around it at 25°C. Its potential is taken as zero. Notation: Pt, H₂ (1 atm) | H⁺ (1 M).

EMF of the Ni–Cu cell

Ni | Ni²⁺ (0.1 M) || Cu²⁺ (0.2 M) | Cu

Anode   : Ni → Ni²⁺ + 2e⁻
Cathode : Cu²⁺ + 2e⁻ → Cu
Cell    : Ni + Cu²⁺ → Ni²⁺ + Cu         (n = 2)

E°cell = 0.34 − (−0.25) = 0.59 V
Q = [Ni²⁺]/[Cu²⁺] = 0.1/0.2 = 0.5 ,  log Q = −0.301
E = 0.59 − 0.02955 × (−0.301)
  = 0.59 + 0.0089 = 0.599 V

Answer: EMF ≈ 0.60 V.

✓ Answer checked by Claude Opus 5.5 (high effort)

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