Chemistry · Electrochemistry · NEET
For a single reduction half-reaction like Mⁿ⁺ + n e⁻ → M, at 298 K you write: E = E° − (0.059/n) log(1/[Mⁿ⁺]) The number n is how many electrons the half-reaction uses. The reduced solid or gas (like solid metal M) is not put in Q; only the ion in solution goes inside the log. So for a single electrode, Q = 1/[ion]. This is the same idea as the full-cell Nernst equation, but applied to only one electrode.
The hydrogen electrode is 2H⁺ + 2e⁻ → H₂(g). Its potential is: E = 0 − (0.059/2) log(P_H₂ / [H⁺]²) More H⁺ means a bigger [H⁺]², so log becomes negative, and E goes UP. More H⁺ means a lower pH. So low pH (acidic) gives a higher (more positive) potential, and high pH (basic) gives a lower (more negative) potential. Rule of thumb: as pH rises by 1 unit at fixed 1 atm H₂, E drops by about 0.059 V.
E° of the hydrogen electrode is 0 V by definition, but that is only at standard conditions ([H⁺] = 1 M, P_H₂ = 1 atm). In pure water [H⁺] = 10⁻⁷ M, which is not standard, so E is not automatically zero. E only becomes zero again if you also change the H₂ pressure so that P_H₂/[H⁺]² = 1. This is exactly what NEET 2016 tested.
Use the number of electrons in that half-reaction as written. Cu²⁺ + 2e⁻ → Cu uses n = 2. Ag⁺ + e⁻ → Ag uses n = 1. 2H⁺ + 2e⁻ → H₂ uses n = 2. Getting n wrong is the most common NEET mistake, because it changes the 0.059/n factor.
Never put a pure solid inside Q. For a metal electrode only the ion counts, so Q = 1/[Mⁿ⁺]. For a gas electrode like H₂ you DO include the gas as its pressure. So for the H₂ electrode, Q = [H⁺]²/P_H₂ (oxidised over reduced) or the inverse depending on how you write the reaction; keep the reduced form in the denominator when the reaction is written as reduction.
The pressure of H₂ required to make the potential of the H₂ electrode zero in pure water at 298 K is:
Calculate the emf of the half cell: Pt(s) | H₂(g, 2 atm) | HCl(aq, 0.02 M). Given E°(H⁺/H₂) = 0 V, 2.303RT/F = 0.059, log 2 = 0.3010.
Find the emf of the cell reaction Ni(s) + 2Ag⁺(0.001 M) → Ni²⁺(0.001 M) + 2Ag(s). Given E°cell = 1.05 V, 2.303RT/F = 0.059 at 298 K.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
No, it is the same equation. For one electrode you only include that electrode's ion in Q; for a full cell you include ions from both sides. Both use E = E° − (0.059/n) log Q at 298 K.
At fixed 1 atm H₂ and 298 K, the potential changes by about 0.059 V for every 1 unit change in pH. Higher pH lowers the potential; lower pH raises it.
It equals 2.303RT/F at 298 K. It combines the gas constant R, temperature T (298 K), Faraday constant F, and the factor 2.303 that changes natural log to base-10 log. NEET usually gives it to you.
They are equal: log(1/[ion]) = −log[ion]. Use whichever is faster. For a metal electrode Mⁿ⁺ + n e⁻ → M, E = E° + (0.059/n) log[Mⁿ⁺], which is the same as E = E° − (0.059/n) log(1/[Mⁿ⁺]).
