Q1(l). Iron redox-electrode potential

Answer in brief

E=E°-RT/Fln (a_(Fe²⁺))/(a_(Fe³⁺)).

Key points

• Write the reduction half-reaction Fe³⁺ + e⁻ → Fe²⁺, so n=1. Platinum is an inert electron conductor and does not enter the reaction quotient.
• Combining ΔᵣG=ΔᵣG°+RTln Q with ΔᵣG=-nFE gives E=E°-(RT/nF)ln Q.
• Here Q=a_(Fe²⁺)/a_(Fe³⁺). At 298 K, E=E°-0.05916log₁₀ Q volts. Activities can be approximated by concentration ratios only in an appropriate dilute medium.