How do I balance this equation? MnO 4 - (aq) + CH 3 OH(aq) Æ Mn 2+ (aq) +HCO 2 H

Question

How do I balance this equation?

MnO4-(aq) + CH3OH(aq) Æ Mn2+(aq) +HCO2H(aq)
How do I balance this equation?

MnO4-(aq) + CH3OH(aq) Æ Mn2+(aq) +HCO2H(aq)

Explanation / Answer

first you split the equation up into two half reactions with likeparts: MnO4   ---> Mn 2+ CH3OH ---> HCO2H Then you balance each seperatly following these steps 1. balance all elements except H and O MnO4- ---> Mn 2+ Already balanced 2. Balance O by adding H2O MnO4- ---> Mn 2+ + 4H2O 3. Then balance the H's MnO4- + 8 H+ ---> Mn 2+ + 4 H2O 4. Next balance the charges by adding e- MnO4- + 8 H+ ---> Mn 2+ + 4 H2O -1   +   8 =7                 +2 therefore need 5e to the left MnO4- + 8 H+ +5e- ---> Mn 2+ + 4H2O Do this now with the other Half equation 1. balance all elementsexcept H and O CH3OH ---> HCO2H Alreadybalanced 2. Balance O by adding H2O CH3OH + H2O ---> HCO2H 3. Then balance the H's CH3OH + H2O ---> HCO2H + 4 H+ 4. Next balance the charges by adding e- CH3OH + H2O ---> HCO2H + 4 H+    +0                +4 therefore need4e to the right CH3OH + H2O ---> HCO2H + 4 H+ + 4e- 5. Balance both half reactions e's so they have the same number bymultipling CH3OH + H2O ---> HCO2H + 4 H+ + 4e-   Common multible istwenty MnO4- + 8 H+ +5e- ---> Mn 2+ + 4H2O 5CH3OH + 5H2O ---> 5HCO2H + 20H + + 20e- 4MnO4- + 32 H+ +20e- ---> 4Mn 2+ + 20H2O Then put the two together and simplify 5CH3OH + 5H2O + 4MnO4-+ 32 H+ + 20e- - - - ->5HCO2H + 20 H + + 20e- + 4Mn2+ + 20 H2O when you cross out likes on both side it all boils down to: 5CH3OH + 4MnO4- + 12H+ - - - -> 5HCO2H +4MnO4- + 11H2O hope this helps please rate me

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