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balance each of the following redox reactions occurring in acidic aqueous solution.

Balance each of the following redox reactions occu
Balance each of the following redox reactions occurring in acidic solution
a) SO32- (s) +MNO4- —> SO42- +Mn2+
b) S2O32-+ Cl2 —> SO42-+Cl-

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asked by Sunshine
Aug 5, 2009
You need to know and understand how to do these yourself. So rather than give you the balanced equations, I will tell you that
a)S changes oxidation state from +4 on the left to +6 on the right. Mn changes its oxidation state from +7 on the elft to +2 on the right.

b)S changes oxidation state from 2 (for each S) on the left to +6 on the right.
Cl changes from zero on the left to -1 on the right.

If this doesn’t help all that much, tell me what you don’t understand about how to balance redox equation.

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posted by DrBob222
Aug 5, 2009
I’m mainly having trouble with adding H20/H+ and then balancing it!

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posted by Sunshine
Aug 5, 2009
Post your work on the first one as far as you can go and I’ll help you through it.

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posted by DrBob222
Aug 5, 2009
A)
OX
H20+SO32- –> SO42- +2H+ 2e-

RED

16H+ +2MnO4- —> 2Mn2+ 8H20

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posted by Sunshine
Aug 5, 2009

Both half reactions look balanced to me; the second one is twice what is necessary. Notice you can reduce each coefficient by 1/2 to make them 8,1,1,4.
Looks like you did a good job to me.

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posted by DrBob222
Aug 5, 2009
Here is a method I use.
SO3^- ==> SO4^-2

  1. S changes from +4 to 6. Add electrons to the appropriate side (right for this one) to balance the change in oxidation state.
  2. SO3^-2 ==> SO4^-2 + 2e
  3. Now count the charge on each side. The left is -2 and the right is -4; therefore, add
    (a) H^+ to balance the charge if it is acid solution or
    (b)OH^- to balance the charge if it is basic or neutral solution.
    This is acid so we add 2H^+ to balance the charge.
  4. SO3^-2 ==> SO4^-2 + 2e + 2H^+
  5. Now add water (usually to the opposide side) to balance the H atoms. Oxygen SHOULD balance at that point.
    H2O + SO3^-2 ==> SO4^-2 + 2e + 2H^+

For the Mn, without all the explaining, but I’ll follow the same format.

MnO4^- ==> Mn^+2
Mn goes from 7 to 2; therefore, add 5e to the left side.
MnO4^- + 5e ==> Mn^+2

Charge is -6 on the left, +2 on the right, add 8H^+ on the left to balance the charge.
8H^+ + 5e + MnO4^- ==>Mn^+2

Now add 4H2O to the right to balance the H atoms
8H^+ + 5e + MnO4^- ==> Mn^+2 + 4H2O

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posted by DrBob222
Aug 5, 2009
Thank you

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posted by Sunshine
Aug 5, 2009

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