Answer to Question #111536 in General Chemistry for Abdallah

Question

A 2.500 gram sheet of copper undergoes electrolysis at 0.5 amps for 45 minutes. The final mass of Cu(65.35 g/ol) foil was grams
How many moles of electrons were passed ?
How many moles of copper were oxidized ?
What was the oxidation number of the copper ions created by the electrolysis ?

Answer 1

We have the following electrolysis equation:

Cu⁰ - ne^- = Cuⁿ⁺

As the weight of the plate decreases 2.5 - 2.04 = 0.46 g, the number of moles of copper oxidized is:

0.46/65.35 = 0.00704 mol

Find the oxidation number of the copper ions:

m = M*I*t/nF (Faraday's laws of electrolysis), where m - how many grams the weight of the plate decreased and n - oxidation number of the copper ions. From this equation we express n:

n = M*I*t/mF = (65.35*0.5*2700)/(0.46*96500) = 1.99 ~ 2

Correct electrolysis equation is:

Cu⁰ - 2e^- = Cu²⁺

Based on this equation 2*0.00704 = 0.0141 mol of electrons was passed.