Question

An electric current is passed through silver voltmeter connected to a water voltmeter. The cathode of the silver voltmeter weighed 0.108 g more at the end of the electrolysis. The volume of $O_2$ evolved at STP:
A. $56c{m^3}$
B. $550c{m^3}$
C. $5.6c{m^3}$
D. $11.2c{m^3}$

Answer 1

Hint: We will write the individual reactions going on anode and cathode electrodes and find out the amount of the electric charge passed through the electrodes, then find out the volume of the oxygen liberated by that much amount of charge using faraday’s law.

Complete step by step answer:
Cathode is the electrode where the reduction takes place, and the cathode is made out of silver, so the reaction at cathode will be
a. $Ag + {e^ - }\xrightarrow{{red}}A{g^ - }$
b. We see that one faraday charge deposit one mole of silver (108 g)
c. (one mole of electrons=1 Faraday)

Oxidation of water will take place at the anode to form oxygen gas, and the balanced reaction at anode will be
a. ${H_2}O\xrightarrow{{oxd}}{O_2} + 4{H^ + } + 4{e^ - }$
b. We see that four Faraday of charge is responsible for the liberation of one mole (32 g) of oxygen gas.
c. It’s given that 0.108 g of extra silver is weighed at the end, that is 0.108 g of silver has deposited.

From the reaction at cathode, we know that one Faraday electric charge is responsible for the deposition of one mole (180g) silver. Using unitary method

$180 g \to 1 F $
$1 g \to \dfrac{1}{{180}}F $
$0.108 g \to \dfrac{1}{{180}} \times 0.108 F $
$= {10^{ - 3}}F $

Now, we will calculate how much oxygen is liberated by the same amount of charge. We know from the reaction at anode that 4Faraday charge is responsible for the liberation of one mole (32g) oxygen

$4F \to 32 g $
$1F \to \dfrac{{32}}{4} g = 8 g$
Now, ${10^{ - 3}}F \to \dfrac{8}{{{{10}^{ - 3}}}} g $
$= 0.008 g $

At STP one mole (32 g) oxygen occupies 22.4 liters of volume
$32 g \to 22.4 l $
$0.008 g \to \dfrac{{22.4}}{{32}} \times 0.008 l$
$= 5.6 l $
$32 g \to 22.4 l $
$0.008 g \to \dfrac{{22.4}}{{32}} \times 0.008 l $
$= 5.6 l = 5.6c{m^3} $

The correct option is C.

Note:
Oxidation can also be defined as the loss of electrons, while reduction is also defined as the gain of electrons.