Question

. A certain quantity of electricity is passed through an aqueous

Answer 1

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