Chemistry · Electrochemistry · NEET
You use it whenever the ions in a cell are NOT at 1 mol/L (standard) concentration. E°cell is the voltage only at 1 M and 298 K. The Nernst equation corrects this for real concentrations. At 298 K: Ecell = E°cell − (0.059/n) log Q. If every ion is exactly 1 M, then Q = 1, log 1 = 0, and Ecell = E°cell. So E°cell is just the special case of the Nernst equation.
n is the number of electrons transferred in the balanced cell reaction. For the Daniell cell Zn + Cu2+ → Zn2+ + Cu, two electrons move, so n = 2. Q is the reaction quotient: products over reactants, using concentrations of ions (and pressures for gases). Pure solids and pure liquids are NOT included (their value is 1). For Zn + Cu2+ → Zn2+ + Cu, Q = [Zn2+]/[Cu2+].
The full equation has RT/F and a ln term. NCERT converts ln to 2.303 log and plugs in R = 8.314, T = 298 K, F = 96500. This gives 2.303RT/F = 0.059 V (some books round to 0.0591). So 0.059 only works at 298 K (25 °C). If the exam gives a different temperature, you cannot use 0.059 — you must recompute. NEET usually gives you '2.303RT/F = 0.059' in the question.
It depends on WHERE the ion sits in Q. More PRODUCT ions make Q bigger, so log Q is more positive, so you subtract more and Ecell DROPS. More REACTANT ions make Q smaller, log Q negative, so Ecell RISES. Quick rule for a Daniell cell: increasing [Cu2+] (the reactant ion at the cathode) raises EMF; increasing [Zn2+] lowers it.
For a single electrode Mn+ + ne− → M, NCERT writes E = E° − (0.059/n) log(1/[Mn+]). For the whole cell you use E°cell = E°cathode − E°anode and Q made from all ions. Same idea, two levels. Single-electrode form is what you use for the hydrogen electrode and pH problems (see the next concept).
In the electrochemical (Daniell) cell Zn | ZnSO4 (0.01 M) || CuSO4 (1.0 M) | Cu, the emf is E1. When the concentration of ZnSO4 is changed to 1.0 M and that of CuSO4 to 0.01 M, the emf changes to E2. Which is the correct relationship between E1 and E2? (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)
For a cell involving one electron E°cell = 0.59 V at 298 K. The equilibrium constant for the cell reaction is: (Given 2.303RT/F = 0.059 V)
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Ecell = E°cell − (0.059/n) log Q at 298 K, where n is electrons transferred and Q is the reaction quotient. It gives the cell voltage at any concentration.
As the reaction proceeds, products build up and Q rises, so the driving force falls. Subtracting (0.059/n) log Q lowers the voltage until it reaches zero at equilibrium (that is why Ecell = 0 gives you Kc).
No. Pure solids (like Zn or Cu metal) and pure liquids have activity 1, so they are left out. Only ion concentrations and gas pressures appear in Q. This is a common NEET slip.
No. 0.059 V is only for 298 K (25 °C) because it comes from 2.303RT/F at that temperature. At another temperature you must recalculate. NEET normally states the value for you.
Q uses the ACTUAL concentrations at any moment; Kc is the special value of Q at equilibrium, when Ecell = 0. Setting Ecell = 0 in the Nernst equation lets you find Kc from E°cell.
