Question

On passing $5$ ampere current through an aqueous solution of an unknown salt of $Pd$ for $2.15$ hour, $10.64 \,g$ of $Pd ^{n+}$ get deposited at cathode. The value of $n$ is (At.wt. of $P d=106.4)$

• A. 2
• B. 3
• C. 3.5
• D. 4

Answer 1

Answer: (D)

$P d^{n+}+n e^{-} \rightarrow P d$

Applying, $w=Z \times I \times t$

$w=10.64 \,g , Z =\frac{\text { Eq. wt. }}{96500}$

$I=5$ ampere, $t=2.15$ hour $=2.15 \times 60 \times 60 \,s$

$\therefore 10.64=\frac{106.4}{n} \times \frac{1}{96500} \times 5 \times 2.15 \times 60 \times 60$

$\Rightarrow n=\frac{106.4}{10.64} \times \frac{5 \times 2.15 \times 3600}{96500} \approx 4$

Alternatively, $10.64 g$ of $Pd$ gets deposited by charge

$Q=I \times t=5 \times 2.15 \times 60 \times 60 \,C$

$\therefore 1$ mole i.e. $106.4 \,g$ of $Pd$ will be deposited by charge

$=\frac{5 \times 2.15 \times 3600}{10.64} \times 106.4=387000\, C $

$387000 \,C =\frac{387000}{96500} F \approx 4\, F$

We know, if a reaction involves $n$ electrons then $n$ Faradays deposit 1 mole of the substance. Thus, $n=4$