Question

An acidic solution containing copper ions is electrolyzed, producing gaseous oxygen (from water) at the anode and copper at the cathode. For every 16.0 g of oxygen generated 63.5 g of copper plates out. What is the oxidation state of the copper in the solution?

Short answer

Copper has an oxidation state of +2.

Step-by-step solution

Definition of electrolysis

Electrolysis is a technique that uses a direct electric current to promote a chemical reaction that would otherwise be non-spontaneous.

Finding oxidation reaction

16.0 g of oxygen gas is generated or every 63.5 g of metallic copper,

$\begin{array}{c}\text{n}\left({\text{O}}_{\text{2}}\right)\text{=}\frac{\text{16.0g}}{{\text{31.998gmol}}^{\text{-1}}}\\ \text{=0.500mol\hspace{0.33em}\hspace{0.33em}}\end{array}$

$\begin{array}{c}\text{n}\left({\text{O}}_{\text{2}}\right){\text{=-}}_{\text{molar}}{\text{mass}}^{\text{-}}\text{n}\\ \text{n(Cu)=}\frac{\text{63.5g}}{{\text{63.5gmol}}^{\text{-1}}}\text{n}\\ \text{=1.00mol}\\ \text{nMr}\left({\text{O}}_{\text{2}}\right){\text{=31.998gmol}}^{\text{-1}}\text{n}\\ \text{n(Cu)=}\frac{\text{mass}}{\text{molarmass}}\text{n}\\ \text{m(Cu)=63.5g}\\ {\text{Mr(Cu)=63.5gmol}}^{\text{-1}}\end{array}$

Finding the moles of electrons

The number of moles of electrons created by$\text{0.500m\hspace{1em}}$

$\text{0.500mol×}\frac{{\text{4mole}}^{\text{-}}}{\text{1mol}}{\text{=2mole}}^{\text{-}}$

Four moles of electrons are produced per oxygen gas

Finding the reduction half reaction of copper

The reduction half reaction of copper is

${\text{Cu}}^{\text{2+}}{\text{(aq)+2e}}^{\text{-}}\to \text{Cu(s)\hspace{0.33em}\hspace{0.33em}\hspace{0.22em}}$

$\text{1.00mol}$of copper gain$\text{2mol}$ of electrons

Therefore Copper has an oxidation state of +2.