Question

How many grams of Cu (63.55 g/mol) may be deposited from a Cu2+ solution during electrolysis by a current of 3.00 A for 5 hours? Please report an integer. F = 96485 C/mol

Answer 1

Quantity of electricity $(Q)$ $=$

Current $×$ time

$=3×5×3600$

$=54000$ coulombs

$54000$ coulombs $=\frac{54000}{96485}Faraday$

$=0.5597Faraday$

2 Faradays are needed to discharge one mole of $Cu^{2+}$ ions

$\therefore0.5597$ Faraday will discharge

$\frac{0.5597}{2}×1=0.2799$ moles of Cu_(s)

Mass of Cu_(s) discharged $= 0.2799×63.55$

$=17.7845g$

$\approx18g$