Question

In an electrolysis of AgNO₃ solution, 0.7 g of Ag is deposited after a certain period of time. Calculate the quantity of electricity required in coulomb. (Molar mass of Ag is 107.9 g mol^-1.)

Answer 1

Given : 

Mass of Ag deposited = 0.7 g 

Molar mass of Ag = 107.9 g mol^-1 

Quantity of electricity = Q = ? 

Reduction half reaction is,

Ag⁺ + e^- → Ag

1 mole of Ag = 107.9 g Ag requires 1 mole of electrons

∴ 0.7 g Ag will require,

\(\frac{0.7}{107.9}\) = 6.49 × 10^-3 mole of electrons

∵ 1 mole of electrons carry 96500 C charge

∴ 6.49 × 10^-3 mole of electrons will carry,

96500 × 6.49 × 10^-3 = 626 C

∴ Quantity of electricity required = 626 C,