Chemistry · Electrochemistry · NEET
At 298 K (25°C), use log K = (n × E°cell) / 0.059. Here n is the number of electrons transferred in the balanced cell reaction and E°cell is the standard cell potential in volts. After you get log K, take antilog (10 to that power) to get K. Example: if n = 1 and E°cell = 0.59 V, then log K = 0.59/0.059 = 10, so K = 10^10.
It comes from joining two facts. First, ΔG° = -nFE°cell. Second, ΔG° = -RT ln K. Setting them equal gives nFE°cell = RT ln K. Rearranging and switching ln to log (2.303 log) gives E°cell = (2.303RT/nF) log K. At 298 K, 2.303RT/F = 0.059 V, so E°cell = (0.059/n) log K. You do NOT need to derive this in the exam — just remember the final form.
n is the total number of electrons that move from anode to cathode in the balanced overall reaction. Balance the two half-reactions so the electrons cancel; the number that cancels is n. For 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂, two electrons move, so n = 2. Getting n wrong is the most common mistake, so always balance first.
0.059 V (sometimes written 0.0591 V) is the value of 2.303RT/F at T = 298 K. It is a constant only at that temperature. NEET usually gives you '2.303RT/F = 0.059 V' in the question, so use exactly what they give. If they give 0.0591, use 0.0591.
Yes. Positive E°cell means log K is positive, so K > 1, and the reaction favours products (spontaneous). Negative E°cell means log K is negative, so K < 1, and the reaction favours reactants (non-spontaneous). If E°cell = 0, then K = 1 and the cell is at equilibrium.
Both exist, so read carefully. The ΔG° form uses natural log: ΔG° = -RT ln K. The E°cell form uses base-10 log: E°cell = (0.059/n) log K. The 0.059 already includes the 2.303 factor that converts ln to log, so when you see 0.059, always use log (base 10).
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 at T = 298 K)
If the E°cell for a given reaction has a negative value, which of the following gives the correct relationships for the values of ΔG° and K_eq?
For the cell reaction 2Fe³⁺(aq) + 2I⁻(aq) → 2Fe²⁺(aq) + I₂(aq), E°cell = 0.24 V at 298 K. The standard Gibbs energy (ΔrG°) of the cell reaction is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
A large K (much greater than 1) means the reaction goes almost fully to products before stopping. It matches a positive E°cell and a negative ΔG°, so the cell reaction is spontaneous and useful for making electricity.
No. The value 0.059 V is only correct at 298 K (25°C). At other temperatures use the full form E°cell = (2.303RT/nF) log K. NEET questions almost always use 298 K, so 0.059 is fine there.
It is the same equilibrium constant idea. The K you get from E°cell is the equilibrium constant of the overall cell (redox) reaction, written in terms of ion concentrations. It behaves like any K: large K means products favoured.
For NEET you can treat K as a number without units. The formula log K = nE°cell/0.059 gives a pure number, and options are given as plain powers of 10.
