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2L of 1M CuSO₄ aqueous solution is electrolysed using inert electrodes. If 1F of charge is used up, the concentration of cupric ions left out : [ assume that vol. of the solution remain same ]

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1.5 M

0.75 M

0.5 M

1.25 M

answer is B.

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Detailed Solution

$M = \frac{n}{v}$

$1 = \frac{n}{2}$ $∴$ $n_{C u^{2 +}}$ Initially present are 2

1F can remove 0.5 mole Cu⁺² ions

$∴ n_{{Cu}^{2 +}}$ left are = 2 – 0.5

= 1.5

$∴ | {Cu}^{2 +} | = \frac{1.5}{2} molar$

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