Question

3. (14%)
The metal copper is oxidized in nitric acid according to the equation:
3 Cu + 8 HNO3 → 3 Cu (NO3) 2 + 2 NO + 4 H2O

a) Show how you set the oxidation-reduction reaction and get the equation here
above (show as many steps in the process)

b) You receive 0.3855 g of copper-containing metal together with other dissolved substances
easily set up. You find that to dissolve the copper and form
copper nitrate Cu (NO3) 2 requires 22.5 ml of 0.575 M nitric acid. What is the ratio
copper in the sample?

Show how you calculate thx :*

Answer 1

Ans 3

Part a

The unbalanced reaction

Cu + HNO₃ = Cu(NO₃)₂ + NO + H₂O

Assign oxidation numbers for each atom within brackets

Cu(0) + H(+1)N(+5)O(-2)₃ = Cu(+2)[N(+5)O(-2)₃​​​​​​]₂ + N(+2)O(-2) + H(+1)₂O(-2)

The two half cell reactions are

Oxidation reaction

O

Cu(0) = Cu(+2)[N(+5)O(-2)₃​​​​​​]₂

Cu is being oxidized from 0 to +2

Reduction reaction

R:

H(+1)N(+5)O(-2)₃ = N(+2)O(-2)

N is being reduced from 5 to 2

Balance the atoms except hydrogen and oxygen.

Oxidation reaction

Cu + 2HNO₃ = Cu(NO₃)₂

Reduction reaction

HNO₃ = NO

Balance the oxygen atoms by adding H2O

O:

Cu + 2HNO₃ = Cu(NO₃)₂

R:

HNO₃ = NO + 2H₂O

Balance the hydrogen atoms by adding H⁺

O:

Cu + 2HNO₃ = Cu(NO₃)₂ + 2H⁺

R:

HNO₃ + 3H⁺ = NO + 2H₂O

Balance the charge.

O:

Cu + 2HNO₃ = Cu(NO₃)₂ + 2H⁺ + 2e^-

R:

HNO₃ + 3H⁺ + 3e^- = NO + 2H₂O

Make electron balance

O:

Cu + 2HNO₃ = Cu(NO₃)₂ + 2H⁺ + 2e^-

Multiply by 3

3Cu + 6HNO₃ = 3Cu(NO₃)₂ + 6H⁺+ 6e^-

R:

HNO₃ + 3H⁺ + 3e^- = NO + 2H₂O

Multiply by 2

2HNO₃ + 6H⁺ + 6e^- = 2NO + 4H₂O

Add half reactions

3Cu + 8HNO₃ + 6H⁺ + 6e^- = 3Cu(NO₃)₂ + 2NO + 6H⁺ + 4H₂O + 6e^-

The balanced reaction

3Cu + 8HNO₃ = 3Cu(NO₃)₂ + 2NO + 4H₂O

Part b

Mass of Cu = 0.3855 g

Moles of Cu = mass/molecular weight

= (0.3855g) / (63.546g/mol)

= 0.006066 mol

Volume of HNO3 = 22.5 mL x 1L/1000 mL = 0.0225 L

Molarity of HNO3 = 0.575 mol/L

Moles of HNO3 = molarity x volume

= 0.575 mol/L x 0.0225 L

= 0.0129375 mol

Mass of HNO3 = moles x molecular weight

= 0.0129375 mol x 63.01 g/mol

= 0.8152 g

Total mass of reactants = 0.3858 + 0.8152 = 1.201 g

Mass ratio of copper in the sample = mass of Cu/total mass

= 0.3858/1.201

= 0.3212

= 32.12%