ElectrochemistryHard
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?
Options
A.0.965 A
B.1.93 A
C.0.483 A
D.0.0965 A
Solution
$\frac{Q}{F} = n_{eq}\text{ of }I^{-} = n_{eq}Na_{2}S_{2}O_{3}$
Or, $\frac{i \times 2 \times 3600}{96500} = \frac{72 \times 1.0}{1000} \times 1 \Rightarrow i = 0.965\text{ A}$
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