In an electrolysis of AgNO3 solution, 0.7 g of Ag is deposited after a

1.	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» 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,

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»

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,

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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.)

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