ElectrochemistryHard
Question
Copper sulphate solution (250 ml) was electrolysed using a platinum anode and a copper cathode. A constant current of 2 mA was passed for 19.3 min. It was found that after electrolysis the absorbance of the solution was reduced to 50% of its original value. Calculate the concentration of copper sulphate in the solution to begin with.
Options
A.9.6 × 10−5 M
B.4.8 × 10−5 M
C.2.4 × 10−5 M
D.1.2 × 10−5 M
Solution
$n_{eq}Cu^{2 +}\text{ reduced = }\frac{Q}{F} \Rightarrow n \times 2 = \frac{2 \times 10^{- 3} \times 19.3 \times 60}{96500}$
$\therefore n = 1.2 \times 10^{- 5} $$$\therefore\left\lbrack CuSO_{4} \right\rbrack_{0} = \frac{\left( 1.2 \times 10^{- 5} \times 2 \right)}{250} \times 1000 = 9.6 \times 10^{- 5}\text{ M}$$
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