Chem--PLEASE HELP 911
posted by JO .
Balance the following redox reaction and identify the oxidizing and reducing agents
A- Fe2+(aq) + Cr2072-(aq) -->Fe3+(aq) + Cr3+(aq) (acidic)
B- N2H4(g) + ClO3-(aq) -->NO2(g) + Cl- (basic)
Here is a site that shows you how to balance redox equations. Repost with specific questions. I can help you through it but you must learn how to do this.
A- 14H + 6Fe^2+ + Cr2O7^2- --> 6Fe^3+ + 2Cr ^3+ + 7H2O
B- 3N2H4 +4Cl03- --> 6NO2+ 12Cl- + 6H2O
This was all I was able to come up with, am I at lease headed in the right direction?
A is ok. Good work.
B is not right. For example, N and H are OK but I see 4 Cl on the left and 12 on the right. I see 12 O on the left and 18 on the right.
ClO3^- ==> Cl^-
Cl is +5 on the left and -1 on the right, so
ClO3^- + 6e ==> Cl^-
I see -7 charge on the left; -1 on the right so I add 6 OH^- to the right.
ClO3^- + 6e ==> Cl^- + 6OH^-
Now add H2O the left to balance.
ClO3^- + 6e + 3H2O ==> Cl^- + 6OH^-
N2H4 ==> NO2
First we balance the N so we are comparing apples and oranges when we look at electron change.
N2H4 ==> 2NO2
Both N on the left = -4; on the right both N are +8 which is a change of -12e so
N2H4 ==> 2NO2 + 12e
The change on the left is zero and -12 on the right, we add OH to the left
12OH^- + N2H4 ==> 2NO2 + 12e
Add to H2O on the other side to balanced
12OH^- + N2H4 ==> 2NO2 + 12e + 8H2O
Multiply equantion 1 by 2 and equation 2 by 1 and add. Cancel common ions/molecules/charges that appear on both sides. I get this final equation.
2ClO3^- + N2H4 ==> 2Cl^- + 2NO2 + 2H2O