Question

A current of 4 amperes is passed through a molten solution for 45 minutes. 2.977 g of metal is deposited. Calculate the charge carried by the metal cation if its atomic mass is 106.4 g mol⁻¹.

Answer 1

Given: Current, (I) = 4 A

Time, (t) = 45 min = 45 × 60 = 2700 s

Mass of metal deposited, (m) = 2.977 g

Atomic mass (Molar mass), (M) = 106.4 g mol⁻¹

Faraday’s constant, (F) = 96500 C mol⁻¹

Total charge (Q) = I × t

= 4 × 2700

= 10800 C

Using Faraday’s first law of electrolysis:

Equivalent mass = `"Molar mass"/n`

= `106.4/n`

Mass deposited = `(Q xx "Equivalent mass")/F`

2.977 = `(10800 xx (106.4 // n))/96500`

2.977 = `(10800 xx 106.4)/(n xx 96500)`

n = `(10800 xx 106.4)/(2.977 xx 96500)`

= `1149120/287280.5`

= 3.999

≈ 4

∴ The charge carried by the metal cation is +4, i.e., the metal ion is M⁴⁺.