Question

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

• A. 2
• B. 3
• C. 3.5
• D. 4

Answer 1

Answer: (D)

$P{d}^{n+}+n{e}^{-}\to Pd$

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

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

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

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

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

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

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

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

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

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

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