A constant current flowed for 2 h through a potassium iodide solution, oxidizing the iodide ion to iodine. At the end of the experiment, the iodine was titrated with 72 ml of 1.0 M–Na2S2O3 solution. W

Question

A constant current flowed for 2 h through a potassium iodide solution, oxidizing the iodide ion to iodine. At the end of the experiment, the iodine was titrated with 72 ml of 1.0 M–Na2S2O3 solution. What was the average rate of current flow, in ampere?

Prashant Kotian · Accepted Answer

Correct answer: A. Titration reaction: I2 + 2S2O3^2– -> 2I– + S4O6^2–. Moles of Na2S2O3 = 72 mL × 1.0 M = 72 mmol. Equivalents of I2 produced = equivalents of S2O3^2– = 72 × 10^–3 eq. Total charge Q = equivalents × 96500 = 72 × 10^–3 × 96500 = 6948 C. Current I = Q / t = 6948 / (2 × 3600 s) = 6948 / 7200 = 0.965 A. Therefore, correct answer is A.

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