Calculate the emf of the cell in which the following spontaneous reaction takes place at 298 K : 2Fe³⁺(aq) + 2I^-(aq) 2Fe²⁺(aq) + I₂(s) Given: E^ _ Fe³⁺/Fe²⁺ = 0.77 V E^ _ I₂/I^- = 0.54 V [Fe³⁺] = 0.01 M [Fe²⁺] = 0.1 M [I^-] = 0.1 M ( Take 2.303 RT F = 0.059 at 298 K )

Step-by-step solution

The standard cell potential is calculated from the standard reduction potentials: E^ _ cell = E^ _ cathode - E^ _ anode = 0.77 V - 0.54 V = 0.23 V The balanced redox reaction involves the transfer of n = 2 electrons. The reaction quotient Q is given by: Q = [Fe²⁺]^2 [Fe³⁺]^2 [I^-]^2 (since solid I₂ has an active mass of 1) Substituting the given concentrations: Q = (0.1)^2 (0.01)^2 (0.1)^2 = 10⁻² 10⁻⁴ 10⁻² = 10⁻² 10⁻⁶ = 10^4 Applying the Nernst equation: E_ cell = E^ _ cell - 0.059 n Q E_ cell = 0.23 - 0.059 2 (10^