We have an oxidation–reduction system: [Fe(CN)6]^3– + e^– [Fe(CN)6]^4–; E° = +0.36 V. The ratio of concentrations of oxidized and reduced from at which the potential of the system becomes 0.24 V,

Question

We have an oxidation–reduction system: [Fe(CN)6]^3– + e^– <=> [Fe(CN)6]^4–; E° = +0.36 V. The ratio of concentrations of oxidized and reduced from at which the potential of the system becomes 0.24 V, is [Given: 2.303RT/F = 0.06]

Prashant Kotian · Accepted Answer

Correct answer: D. Using the Nernst equation: E = E° - (0.06/1) * log([Reduced]/[Oxidized]). Substituting values: 0.24 = 0.36 - 0.06 * log([Fe(CN)6^4-]/[Fe(CN)6^3-]). -0.12 = -0.06 * log([Red]/[Ox]). log([Red]/[Ox]) = 2, meaning [Red]/[Ox] = 100/1. The question asks for the ratio of oxidized to reduced form, which is 1:100. Therefore, correct answer is D.

💡 Solution & Explanation

Practice 22,000+ questions like this