Writing full equations from ionic half-equations

Question 1

combine the following two half ionic equations into one full equation:
Zn(s) = Zn2+(aq) + 2e-
Cu2+(aq) + 2e- = Cu(s)

Answer 1

Zn(s) + Cu2+(aq) = Zn2+(aq) + Cu(s)

Question 2

combine the two half ionic equations into one full equation:
Fe2+ = Fe3+ + e-
Cl2 + 2e- = 2Cl-

Answer 2

the equation containing the Fe2+ must be multiplied by 2 to get the same number of electrons as the chlorine equation.
2Fe2+ + Cl2 = 2Fe3+ + 2Cl-

Question 3

use these pairs ionic half equations to construct a full ionic equation:
FeO4 2- + 8H+ + 3e- = Fe3+ + 4H2O
C2O4 2- = 2CO2 + 2e-

Answer 3

FeO4 2- + 8H+ + 3e- = Fe3+ + 4H2O needs to be multiplies by 2 and C2O4 2- = 2CO2 + 2e- by 3 to get balanced numbers of electrons
2FeO4 2- + 16H+ + 6e- = 2Fe3+ + 8H2O
3C2O4 2- = 6CO2 + 6e-
so the full ionic equation is:
2FeO4 2- + 16H+ 3C2O4 2- = 2Fe3+ + 8H2O +6CO2

Question 4

this is not a complete full ionic equation because:

2MnO4 - + 16H+ + 5H2O2= 2Mn2+ + 8H2O + 10H+ + 5O2

Answer 4

the H+ ions on each side are not balanced,

2MnO4 - + 6H+ + 5H2O2 = 2Mn2+ + 8H2O + 5O2