Question

A current of 1.70 A is passed through 300.0 mL of 0.160 M solution of $ZnS{O}_4$ for 230 s with a current efficiencty of 90 %. The molarity (M) of $Z{n}^{2+}$ after the deposition of Zn is $154\times {10}^{-x}$, then what is the value of x?
Assume the volume of the solution to remain constant during the electrolysis.

Answer 1

$I=\frac{1.70\times 90}{100}A$

$\therefore$ Equivalent of $Z{n}^{2+}$ lost = $\frac{It}{96500}$

$=\frac{1.70\times 90\times 230}{100\times 96500}=3.646\times {10}^{-3}$

Milliequivalent of $Z{n}^{2+}lost=3.646$

Initially milliequivalent of $Z{n}^{2+}$ = 300 0.160 2 = 96

($\therefore M\times =NforZ{n}^{2+}\Rightarrow mEq=N\times V_{\left(mL\right)}$)

$\therefore MilliequivalentofZ{n}^{2+}$ left in solution = 96 3.646 = 92.354

$\therefore \left[ZnS{O}_4\right]=\frac{92.254}{2\times 300}=0.154M$