Question

Approximately how many moles of Al3+ are reduced when 0.1 faraday of charge passes through a cell during the production of Al? (Note: Assume there is excess Al3+ available and that Al3+ is reduced to Al metal only.)

A. 0.033 molB. 0.050 molC. 0.067 molD. 0.10 mol

Answer 1

0.033 mol

Step-by-step explanation:

1 faraday is require to 1 mol of electron

0.1 faraday is required for 0.1 mol of electron

3 mol of electron is required for 1 mole Al3+ to Al

0.1 mole of electron will therefore need ( 0.1 mol × 1 mol ) / 3 mol = 0.033 mol