and combine like term, and you have your answer3ClO2-(aq) + 2H2O(l) +4MnO4-(aq) +-----> 3ClO4-(aq) +4MnO2(s) +4OH Look at these series of videos, to help you get a handle on how to balance half-reactions:Chemistry Tutorial 12.1a: Determining Oxidation Numbershttp://www.youtube.com/watch?......
and you have,
3ClO2-(aq) + 6H2O(l) + 12OH -----------> 3ClO4-(aq) + 12H2O + 12e-
4MnO4-(aq) + 16H2O + 12e- ----------> 4MnO2(s) + 8H2O(l) + 16OH
Step 7: Combine the half-reactions, and combine like term, and you have your answer
3ClO2-(aq) + 2H2O(l) + 4MnO4-(aq) + -----> 3ClO4-(aq) + 4MnO2(s) + 4OH
Look at these series of videos, to help you get a handle on how to balance half-reactions:
Chemistry Tutorial 12.1a: Determining Oxidation Numbers
http://www.youtube.com/watch?v=1ojEvef9o…
Chemistry Tutorial 12.1b: Oxidized And Reduced Species
http://www.youtube.com/watch?v=bKtgtee9a…
Chemistry Tutorial 12.1c: Oxidizing And Reducing Agents
http://www.youtube.com/watch?v=l5WtsbEYl…
Chemistry Tutorial 12.1d: Half Reactions
http://www.youtube.com/watch?v=iiiq94WNl…
Chemistry Tutorial 12.1e: Balancing Ionic Reactions
http://www.youtube.com/watch?v=YaTR2a2Cp…
3. Practice solving half-reaction problems.
Hope this helps,
Bill
MnO4- ( calculate the oxidation state of Mn in this cmpd using the following formula:
x + 4(-2) = -1
x = -1 - 4(-2)
x = -1 + 8
x = 7 thus the oxidation state of Mn in this cmpd is +7
ClO2- (Calculate the oxidation state of Cl in this cmpd using the following formula:
x + 2(-2) = -1 thus x= +3
in same manner calculate the oxidation states of Mn in MnO2 ( which is +4) and Cl in ClO4 ( which is +4)
Thus Cl is losing electrons and Mn is gaining electrons
Thus u write half reactions as following:
Mn +7 + 3 Electrons = Mn +4
Cl +3 = Cl +4 + 1 electron ( multiply this equation with 3 to get ewual number of electrons on both sides to get:
3Cl +3 = 3Cl +4 + 3 electrons
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Net equation: Mn +7 + 3Cl +3 = Mn +4 + 3Cl +4
or MnO4- + 3 ClO2- = MnO2 + 3ClO4-
Now balance the oxygens on both sides by using the instructions given in the link!
