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Posted by on Thursday, April 25, 2013 at 12:22pm.

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)

  • Chem--PLEASE HELP 911 - , Thursday, April 25, 2013 at 2:07pm

    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.
    http://www.chemteam.info/Redox/Redox.html

  • Chem--PLEASE HELP 911 - , Thursday, April 25, 2013 at 4:00pm

    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?

  • Chem--PLEASE HELP 911 - , Thursday, April 25, 2013 at 9:27pm

    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

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