Chemistry · Electrochemistry · NEET
It depends on WHICH ion. Look at Q = [products]/[reactants]. If you increase a REACTANT ion (like Cu²⁺ at the cathode of a Zn-Cu cell), Q gets smaller, log Q is more negative, so E = E° − (0.059/n)logQ goes UP. If you increase a PRODUCT ion (like Zn²⁺ at the anode), Q gets bigger, log Q is positive, so E goes DOWN. So there is no single 'always up' rule — you must see if the ion is a reactant or a product.
E° (standard EMF) IS fixed — it is measured only when every ion is exactly 1 M. Real cells rarely have 1 M everywhere. The Nernst equation, E = E° − (0.059/n) log Q, is a correction term that adjusts E° for the actual concentrations. The E° part never moves; the log Q part slides the real EMF up or down as ions change. That is why the same cell can give different voltages.
For Zn | Zn²⁺ || Cu²⁺ | Cu the cell reaction is Zn + Cu²⁺ → Zn²⁺ + Cu. Solids (Zn, Cu) are left out. So Q = [Zn²⁺]/[Cu²⁺] and n = 2. E = E° − (0.059/2) log([Zn²⁺]/[Cu²⁺]). Increasing Cu²⁺ makes the fraction smaller (EMF up); increasing Zn²⁺ makes it bigger (EMF down).
It FALLS. Both changes push Q = [Zn²⁺]/[Cu²⁺] higher. A bigger Q gives a positive log Q, and E = E° − (0.059/2)logQ subtracts more, so the EMF drops. This is exactly the NEET 2017 question: swapping 0.01 M and 1.0 M made E₂ smaller than E₁ (E₁ > E₂).
When [Zn²⁺] = [Cu²⁺], the ratio Q = 1, and log 1 = 0. The whole correction term vanishes, so E = E°. This is the easiest checkpoint: equal ion concentrations means the cell reads its standard EMF. As reaction proceeds and Q keeps rising toward K, EMF keeps falling until it reaches 0 (dead battery, equilibrium).
For Pt | H₂ | H⁺, higher [H⁺] (lower pH, more acidic) makes the electrode potential more positive; lower [H⁺] (higher pH) makes it more negative. Using the NEET key convention E = E° − (0.059/2) log([H⁺]²/P(H₂)), more H⁺ ions raise E. This is why the H₂ electrode potential is only zero at [H⁺] = 1 M and 1 atm.
In the Daniell cell Zn | ZnSO₄(0.01 M) || CuSO₄(1.0 M) | Cu the emf is E₁. When [ZnSO₄] is changed to 1.0 M and [CuSO₄] to 0.01 M, the emf becomes E₂. Which relationship is correct? (Given RT/F = 0.059)
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)
Calculate the emf of the half cell Pt(s) | H₂(g, 2 atm) | HCl(aq, 0.02 M). E°(H⁺/H₂) = 0 V. (Given 2.303RT/F = 0.059, log 2 = 0.3010)
Try the real previous-year questions from this chapter — each with the answer and a full solution.
More reactant ions (cathode side, like Cu²⁺) push EMF up. More product ions (anode side, like Zn²⁺) pull EMF down. This comes from E = E° − (0.059/n) log([products]/[reactants]).
When all ion concentrations are 1 M, or when the reactant-to-product ratio makes Q = 1 (so log Q = 0). Then the correction term is zero and E = E°.
At 298 K, the term 2.303RT/F equals about 0.059 V. Dividing by n (electrons transferred) gives the slope for each unit change in log Q. NEET almost always gives you this value in the question.
Yes. As the reaction runs, product ions build up and reactant ions fall, so Q keeps rising. When Q reaches the equilibrium constant K, E becomes 0 — the cell is 'dead' and no current flows.
A concentration cell has the SAME electrode and same ion on both sides but at different concentrations. Its E° = 0, so its entire EMF comes only from the concentration difference through the Nernst term. Current flows until both sides become equal.
