Q1(i). Nernst equation for the redox electrode Pt | Cr⁶⁺, Cr³⁺

Answer in brief

E = E° + (RT/3F) ln[a(Cr⁶⁺)/a(Cr³⁺)]; at 25 °C, E = E° + 0.0197 log[a(Cr⁶⁺)/a(Cr³⁺)] V.

Key points

• Electrode reaction (written as reduction): Cr⁶⁺ + 3e⁻ ⇌ Cr³⁺; the inert Pt only carries electrons.
• Reaction isotherm: ΔG = ΔG° + RT ln[a(Cr³⁺)/a(Cr⁶⁺)].
• Electrical work: ΔG = −nFE and ΔG° = −nFE° with n = 3; dividing by −3F gives E = E° + (RT/3F) ln[a(Cr⁶⁺)/a(Cr³⁺)].
• At 25 °C, 2.3026RT/F = 0.05916 V, so E = E° + 0.0197 log[a(Cr⁶⁺)/a(Cr³⁺)].
• Real form: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ ⇌ 2Cr³⁺ + 7H₂O, so E = E° + (0.05916/6) log([Cr₂O₇²⁻][H⁺]¹⁴/[Cr³⁺]²); E depends strongly on pH.