Balancing and half-equations for redox reactions??

Question

Write half-equations for the following redox reactions and use them to balance the full equations.

i) IO3^-1 + I^-1 + H^+1 => I2 + H2O

ii) V^+3 + H2O2 => VO^+2 + H^+1

iii) SO2 + H2O + I2 => H^+1 + SO4^-2 + I^-1

I just never had this kinda question b4, so i dont know how to answer it.

So if you can, then that would be great! thanks

Answer 1

i)  IO3^-1 + I^-1 + H^+1 => I2 + H2O

First, break the reaction into two half reactions, one for IO31- forming I2, the other for I1- forming I2.

IO31- ----->  I2

I1- -----> I2

Next, balance the iodines in each half reaction.

2IO31- ---->  I2

2I1- ---->  I2

Now, balance the oxygens by adding water.

2IO31- ---->  I2 + 3H2O

2I1- ---->  I2

Next, balance the hydrogens by adding H+.

6H+ +  2IO31- ---->  I2 + 3H2O

2I1- ---->  I2

Then, balance the charge by adding e- to the more positive side of each half-reaction.

6H+ +  2IO31- + 4e-  ---->  I2 + 3H2O

2I1- ---->  I2 + 2e-

Make the electrons lost and gained equal. Multiply the first equation by 1 and the second equation by 2, then add the two equations together.

6H+ +  2IO31- + 4e-  ---->  I2 + 3H2O

4I1- ---->  2I2 + 4e-

6H+ + 4I1- + 2IO31- ----> 2I2 + I2 + 3H2O

Simplify the I2

6H+ + 4I1- + 2IO31- ----> 3I2 + 6H2O

The equation is now balanced for an acidic solution (H+/H2O).

ii)  V^+3 + H2O2 => VO^+2 + H^+1

Following the same steps as shown in i):

V3+  ---->  VO2+

H2O2 ----> H1+

2H2O + V3+  ---->  VO2+

H2O2  ---->  H1+ + 2H2O

2H2O + V3+  ---->  VO2+ + 4H+

H2O2 + 3H+ ---->  H1+ + 2H2O

2H2O + V3+  ---->  VO2+ + 4H+ + 3e-

H2O2 + 3H+ + 2e-  ---->  H1+ + 2H2O

Multiply the first equation by 2 and the second by 3 to equalize the electrons gained and lost. Then, add the two half reactions, simplifying the waters and H+.

4H2O + 2V3+  ---->  2VO2+ + 8H+ + 6e-

3H2O2 + 9H+ + 6e-  ---->  3H1+ + 6H2O

3H2O2 + 2V3+ ---->  2VO2+ + 2H+ + 2H2O

iii) I'll leave the third equation for you to balance. Just follow the same steps as explained in i). Good luck.