# Balance following oxidation reduction reactions using oxidation states method?

a) Cl2(g) + Al(s) --> Al3+(aq) + Cl-

b.) O2(g) + H2O(l) + Pb(s) --> Pb(OH)2(s)

c) H+(aq) + MnO4-(aq) + Fe2+(aq) --> Mn2+(aq) + Fe3+(aq) + H2O(l)

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1. Cl2(g) + Al(s) --> Al3+(aq) + Cl-

Cl2 + 2e --> 2Cl^-
Al ===> Al^3+ + 3e

Multiply equation 1 by 3 and equation 21 by 2 and add. Cancel the electrons.

Here is MnO4^- from part c done this way in detail
MnO4^- ==> Mn^2+

1. The oxidation state Mn on the left is +7; on the right is +2. Add electrons to the appropriate side to balance the change in electrons.

2. MnO4^- + 5e ==> Mn^2+

3. Count the charge on the left and right, then add H^+ (since this is an acid solution but add OH^- if a basic solution) to the appropriate side to balance the charge..
charge on left is -6; on the right is +2 so I add 8H^+ to left.

4. MnO4^- + 5e + 8H^+ ==> Mn^2+

5. Then add H2O (makes no difference in the previous step if you added H^+ or OH^-) to the appropriate side to balance.
MnO4^- + 5e + 8H^+ ==> Mn^2+ + 4H2O

6. Check it.
a. Mn goes from 7 to 2; add 5e to 7 to make 2.
b. count charge on left and right to see if it balances. 2+ on left; 2+ on right
d. Check that atoms balance.
1Mn on left and right; 8H on left and right; 4 O on left and right.
Success.

Here is an example of one (same MnO4^-) using OH but I won't go through the explanation. You can read that step by step from above.

MnO4^- --> MnO2
Mn is 7 on left and 4 on right. Add 3e
MnO4^- + 3e ==> MnO2

Charge on left is -4; on right is 0. Add 4OH^- to right.

MnO4^- + 3e ==> MnO2 + 4OH^-

Then add H2O to the left.]
MnO4^- + 3e + 2H2O ==> MnO2 + 4OH^-

I'll let you check it but I know it's right because I checked it too. ALWAYS check it; that way you will know if you have it right or wrong. If wrong you can redo it and get it right. There is NEVER and excuse for THINKING you have it right and getting it wrong or THINKING it wrong and getting it right. You CAN know it's wrong but not know how to fix it to right. ;-)

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