Question

How many kgs of silver would be deposited at cathode if $0.0342d{m}^3$ of oxygen measured at NTP is liberated at the anode when silver nitrate solution is electrolyzed between platinum electrodes?(Given : Atomic masses of Ag = 108, O =16)

• A. 0.22 gm
• B. 0.44 gm
• C. 0.66 gm
• D. None of the above

Answer 1

The correct option is C 0.66 gm

$AgN{O}_3\to A{g}^{+}+N{O}_3^{-}$

$2H_{2}O\to A{H}^{+}+O_2+A{e}^{-}$

For 1 mole $O_2\Rightarrow 4F$ change released

At NTP

$1$ mole $O_2=22.4L$

$1L=\frac{1}{22.4}$ mole $O_2$

$0.0342L=\frac{0.0342}{22.4}=0.00153$ mole $O_2$

For $0.00153$ mole $\Rightarrow 4\times 0.00153F$

$\Rightarrow 0.00611F$

$\Rightarrow 0.00611\times 96500$

$\Rightarrow 589.34C$

$96500C$ or $1F$ give $\theta$ $\Rightarrow 1$ mole $Ag=108gAg$

$589.34C$ give$\theta$ $\Rightarrow \frac{589.34\times 108}{96500}$

$\Rightarrow 0.66gm$