Question

An electric current is passed through silver and water voltmeters connected in series. The cathode of the silver voltmeter weighed $0.05$ g more at the end of electrolysis. The volume of $O_2$ evolved at the anode of water voltmeter is:

• A. $5.6$ $c{m}^3$
• B. $11.2$ $c{m}^3$
• C. $22.4$ $c{m}^3$
• D. $2.8$ $c{m}^3$

Answer 1

The correct option is D $2.8$ $c{m}^3$

An electric current is passed through silver and water voltmeters connected in series.

The cathode of the silver voltmeter weighed 0.05g more at the end of electrolysis.

The volume of $O_2$ evolved at the anode of water voltmeter is $2.8c{m}^3$

According to faradays second law :

$\frac{W_{Ag}}{W_{O_2}}=\frac{E_{Ag}}{E_{O_2}}$

$\frac{0.05g}{W_{O_2}}=\frac{108}{8}$

$W_{O_2}=\frac{0.054\times 8}{108}=4\times {10}^{-3}$g

8 g of oxygen at STP occupies a volume of $5600c{m}^3$

$\therefore 4\times {10}^{-3}$ g of oxygen will occupy a volume of $=\frac{4\times {10}^{-3}\times 5600}{8}=2.8c{m}^3$

Hence the correct option is D.