Question

On passing electric current of one ampere for $16$ min and $5 \,s$ through one litre solution of $ CuCl_{2} $ , all copper solution was deposited at cathode. The strength of $ CuCl_{2} $ solution was (molar mass of $Cu = 63.5$, faraday constant $= 96500\, C/mol)$

• A. 0.2 N
• B. 0.01 N
• C. 0.001 N
• D. 0.02 N

Answer 1

Answer: (B)

Key Idea According to Faraday's first law $W=Z I t$
where, $W=$ mass of substance deposited (gram)
$I=$ current (amperes)
$t=$ time $($ second $)$
$Z=$ electrochemical equivalent
and $ Z=\frac{E}{96500}$
where, $E=$ equivalent weight
$E=\frac{\text { atomic weight }}{\text { valency }}$
Here, $ I=1 A$,
Molar mass of copper $=63.5$
$t=16 \min$ and $5 \,s$
$=16 \times 60+5=965 \,s$
Using formula
$m =\frac{E}{96500} \times I \times t $
$=\frac{63.5 / 2}{96500} \times 1 \times 965$
$=\frac{63.5}{200}=0.3175 \,g$
Volume of $CuCl _{2}$ solution $=1 \,L$
$\because$ Normality
$=\frac{\text { weight of solute/eq. wt. of solute }}{\text { volume of solution (in litre) }}$
$\therefore $ Normality of $CuCl _{2}$ solution $=\frac{0.3175}{31.75 \times 1}$
$=0.01 \,N$