Writing Full Equations from Ionic Half-Equations

Question 1

for the half equations Zn(s) = Zn2+(aq) and Cu2+(aq) + 2e- = Cu(s). how would you construct the full ionic equation for this

Answer 1

• add the two half-equations together so the electrons cancel out
• Zn(s) + Cu2+ = Zn2+(aq) + Cu(s)

Question 2

for the half equations Fe2+(aq) = Fe3+(aq) + e- and Cl2(g) + 2e- = 2Cl-(aq), what is the full ionic equation

Answer 2

• you need to balance out the equations so the electrons in each one are equal
• 2Fe2+ = 2Fe3+ + 2e- and Cl2 + 2e- = 2Cl-
• giving 2Fe2+ + Cl2 = 2Fe3+ 2Cl-

Question 3

for the half equations Fe2+ = Fe3+ + e- and MnO4- + 8H+ + 5e- = Mn2+ + 4H2O, what is the full ionic equation

Answer 3

• multiply the first one by 5 to equate electrons
• 5Fe2+ = 5Fe3+ + 5e-
• 5Fe2+ + MnO4- + 8H+ = 5Fe3+ + Mn2+ + 4H2O

Question 4

for the half equations MnO4- + 8H+ + 5e- = Mn2+ + 4H2O and H2O2 = 2H+ + O2 + 2e-, what is the full ionic equation

Answer 4

• lowest multiple of 2 and 5 is 10
• so you multiply first one by 2 and second one by 5
• 2MnO4- + 16H+ + 10e- = 2Mn2+ + 8H2O
• 5H2O2 = 10H+ + 5O2 + 10e-
• giving: 2MnO4- + 16H+ 5H2O2 = 2Mn2+ + 8H2O + 10H+ + 5O2

Question 5

what would you further do to complete the ionic equation

Answer 5

• cancel out common ions on either side

- in this case it would be the H+ ions

Question 6

what would the full ionic equation therefore be

Answer 6

• 16 - 10 = 6, so you have 6H+ on the left and none on the right
• 2MnO4- + 6H+ + 5H2O2 = 2Mn2+ + 8H2O + 5O2