80 mL sample solution of KI was electrolyzed for 3.0 minute, using a constant current. The produced iodine required 37.2 mL of 0.25 M sodium thiosulphate solution and unreacted KI required 36.5 mL of

Question

80 mL sample solution of KI was electrolyzed for 3.0 minute, using a constant current. The produced iodine required 37.2 mL of 0.25 M sodium thiosulphate solution and unreacted KI required 36.5 mL of 0.02 M acidic solution of KMnO 4 ​ . Determine the current (in ampere) passed during the electrolysis.

TestHub · Accepted Answer

No. of equivalents of KI electrolysed=No.of equivalents of $I_{2}$ produced=No.of equivalents of $Na_{2} S_{2} O_{3}$ .

No. of equivalents of $Na_{2} S_{2} O_{3} = 0.25 \times 37.2 \times 1 0^{- 3} = 9.3 \times 1 0^{- 3}$

No. of equivalents $= \frac{i.t}{96500} = \frac{i \times 3 \times 60}{96500} \Rightarrow i = 5 A$

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