ElectrochemistryHard

Question

In an analytical determination of arsenic, a solution containing arsenious acid, H3AsO3, KI and a small amount of starch is electrolysed. The electrolysis produces free I2 from I− ions and the I2 immediately oxidizes the arsenious acid to hydrogen arsenate ion, HAsO42−.

I2 + H3AsO3 + H2O → 2I− + HAsO42− + 4H+

When the oxidation of arsenic is complete, the free iodine combines with the starch to give a deep blue colour. If during a particular run, it takes 96.5 s for a current of 1.68 mA to give an end point (indicated by the blue colour), how many g of arsenic are present in the solution. (As = 75)

Options

A.6.3 × 10−4 g
B.1.26 × 10−4 g
C.3.15 × 10−5 g
D.6.3 × 10−5 g

Solution

$n_{As} = n_{H_{3}AsO_{3}} = n\text{ and }n_{eq}H_{3}AsO_{3} = n_{eq}I_{2} = \frac{Q}{F}$

Or, $\frac{w}{75} \times 2 = \frac{1.68 \times 10^{- 3} \times 96.5}{96500} \Rightarrow w = 6.3 \times 10^{- 5}\text{ gm}$

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