Answer:
0.32 g
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
