Question

Q coloumbs of charge deposits 'X'g of Mg. Now,
to deposit same quantity of Ag, charge required
in coloumbs is
​

Answer 1

Given info : Q Coulombs of charge deposits X g of Mg.

To find : to deposit same amount of Ag, charge required in Coulombs is .....

Solution : half reduction reaction of Mg is ...

Mg²⁺ + 2e¯ ⇒Mg

here it is clear that, 2 mole of electron is required to deposit 1 mole of Mg.

so, 2mole of electron deposits 24 g of Mg

⇒2/24 × X mole of electron deposits X g of Mg.

i.e., Q = X/12 mole of electron .....(1)

Now half reduction reaction of Ag,

Ag⁺ + e¯⇒Ag

i.e., 1 mole of electron deposits 1 mole of Ag.

⇒1/108 × x mole of electron deposits x g of Ag.

charge = x/108 = x/(12 × 9) = (x/12) × 1/9

from eq (1).

Charge = Q/9

Therefore charge required to deposit X g of Ag is Q/9.