Question

. A certain quantity of electricity is passed through an aqueous

solution of AgNO3 and cupric salt solution connected in series. The

amount of silver deposited is 1.08 g. What is the amount of copper

deposited ? (Atomic wt. of Cu=63.54, Ag=108 g/mole).​

Answer 1

0.32 g

Step-by-step explanation:

First, we find the number of moles n of silver deposited.

n = m/M where m = mass of silver deposited = 1.08 g and M = Atomic weight of silver = 108 g/mol

n = m/M = 1.08 g/108 g/mol = 0.01 mol

Now, the quantity of electricity Q = nF where n = number of moles of silver = 0.01 mol and F = Faraday's constant = 96500 C/mol

Q= nF = 0.01 mol × 96500 C/mol = 965 C

By the chemical equation for copper,

Cu²⁺ + 2e ⇒ Cu

2 moles of electrons deposits 1 mol of copper, that is

2 × 96500 C deposits 1 mol of copper, then 965 C will deposit x mol

x = 1 mol × 965 C/2 × 96500 C = 1/200 mol = 0.005 mol of copper

So, we find the mass of 0.005 mol of copper using

n = m/M where m = mass of copper deposited = unknown and M = Atomic weight of copper = 63.54 g/mol and n = number of moles of copper = 0.005 mol

m = nM

= 0.005 mol × 63.54 g/mol

= 0.32 g