Question

A current of 1.70 amp is passed through 500 mL of 0.16 M solution of ZnSO₄ for 230 seconds with a current efficiency of 90%. Find the molarity of Zn²⁺ after the deposition of zinc. Assume that the volume of the solution remains constant during electrolysis.

Answer 1

Number of moles of ZnSO₄ present in 500 mL of 0.16 M solution

= `("Molarity" xx "Volume")/1000`

= `(0.16 xx 500)/1000`

= 0.80

In solution, ZnSO₄ ionises as:

\[\ce{\underset{1 mole}{ZnSO4_{(aq)}} <=> \underset{1 mole}{Zn{^{2+}_{(aq)}}} + SO^{2-}_{4(aq)}}\]

∴ Number of moles of Zn²⁺ in the given solution (before electrolysis) = 0.08

Electric current passed in the solution

= `1.70 xx 90/100 xx 230`

= 351.9 C  ...(∵ Efficiency of current is 90%.)

The process involved in the deposition of Zn²⁺ as zinc is:

\[\ce{\underset{1 mole}{Zn^{2+}} + \underset{2 moles}{2 e-} -> Zn}\]

Electricity required to deposit 1 mole of Zn²⁺ ions = 2F

= 2 × 96500 C

= 193000 C

∴ Number of moles of Zn²⁺ deposited by 351.9 C electricity

= `1/193000 xx 351.9`

= 1.82 × 10⁻³

Number of moles of Zn²⁺ left in the solution after electrolysis

= 0.08 − 1.82 × 10⁻³

= 0.0782

∴ `"Molarity of solution" = "No. of moles"/"Volume in litres"`

= `0.0782/0.5`  ...(∵ Volume of solution = 500 mL = 0.5 L)

= 0.1564 M