Question

A constant current was passed through a solution of \[\ce{AuCl^-_4}\] between gold electrodes. After a period of 14 minutes, the cathode recorded an increase in mass of 1.808 g. Calculate the quantity of current passed in solution. What was the magnitude of current? (Atomic mass of gold = 197.0)

Answer 1

For \[\ce{AuCl^-_4}\]​, the reduction at the cathode is:

\[\ce{AuCl^-_4 + 3e- -> Au + 4Cl-}\]

So, 3 moles of electrons deposit 1 mole of Au.

Moles of Au = `1.808/197`

= 0.00917 mol

Moles of e⁻ = 0.00917 × 3

= 0.02751 mol

Q = n × F

= 0.02751 × 96500

= 2654.7 C

Time t = 14 minutes

= 14 × 60

= 840 seconds

`I = Q/t`

= `2654.7/840`

= 3.16 A