Skip to main content

Standard hydrogen electrode; Mg/Ag cell emf

Short answer

The standard hydrogen electrode is the primary reference (E°=0) with Pt in 1 M H⁺ and H₂ at 1 atm. In Mg|Mg²⁺(0.1M)||Ag⁺(1M)|Ag, Mg is anode, Ag cathode; E°cell = 3.17 V. Using Nernst equation with Q=0.1, EMF = 3.20 V.

What is meant by standard hydrogen electrode? Explain briefly with diagram. Calculate the emf of the following cell at 25°C Mg/Mg²⁺ (0.1M)//Ag⁺(1M)/Ag. Given E°Mg²⁺/Mg = −2.37 V, E°Ag⁺/Ag = +0.80 V

Standard hydrogen electrode (SHE)

The SHE is the primary reference electrode, whose potential is arbitrarily taken as zero. It is a platinum foil coated with platinum black, dipped in 1 M H⁺ (1 M HCl), with pure H₂ gas at 1 atm bubbled over it at 25°C.

   H₂ gas (1 atm)
        |
   ┌────┴────┐
   │ glass   │
   │ tube    │
   │  Pt foil│  (coated with
   │  (black)│   Pt black)
 ──┴─────────┴──
   1 M HCl (H⁺ = 1 M)
  • Notation: Pt, H₂ (1 atm) | H⁺ (1 M)
  • Reaction: H₂ ⇌ 2H⁺ + 2e⁻ (as anode) or 2H⁺ + 2e⁻ → H₂ (as cathode)
  • Pt black adsorbs H₂ and acts as an inert conductor; E°(SHE) = 0.00 V. It is used to find the potential of other electrodes by coupling.

EMF of Mg | Mg²⁺ (0.1 M) || Ag⁺ (1 M) | Ag

Anode  : Mg → Mg²⁺ + 2e⁻
Cathode: 2Ag⁺ + 2e⁻ → 2Ag
Cell   : Mg + 2Ag⁺ → Mg²⁺ + 2Ag     (n = 2)

E°cell = 0.80 − (−2.37) = 3.17 V
Q = [Mg²⁺]/[Ag⁺]² = 0.1/1² = 0.1,  log Q = −1
E = 3.17 − (0.0591/2)(−1)
  = 3.17 + 0.0296 = 3.1996 V

Answer: EMF ≈ 3.20 V.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Engineering Chemistry chapter 1 questions →