? MnO4– + ?H+ + ?C2O42– --> ?Mn2+ + ?H2O + ?CO2

Question

what's the balanced equation? correct coefficients?

is it 1MnO4- + 2H+ + 1C2O42- --> 1Mn2+ + 1H20 + 2CO2 ?

c2042..? are you missing parenthesis somewhere?

oh i see, no i believe this is correct:

MnO4- + 8h+ + C2042- -> Mn2+ + 4H2O + 2CO2

o sorry

_?(MnO4)- + _? (H)+ + _?(C2O4)2- -->

_?(Mn)2+ + _?(H2O) +_? (CO2)

The two half equations are:
MnO4^- + 8H^+ 5e ==> Mn^+2 + 4H2O
C2O4^= ==> 2CO2 + 2e
multiply equation 1 by 2 and equation 2 by 5 and add. Cancel common species if any.

are you balancing in general or blancing in a basic/acidic solution? if so then i think you would add electrons like DrBob said, what year chemistry are you? cause i didn't learn the electron thing til second year.

The H^+ tell you it is balanced in acid solution. Besides, permanganate is always used to oxidize oxalate ion in an acid solution.

Here another solution if your algebra is upto speed!

? MnO4– + ?H+ + ?(C2O4)2– --> ?Mn2+ + ?H2O + ?CO2

MnO4– + xH+ + y(C2O4)2– --> Mn2+ + aH2O + bCO2

We have 4 unknowns and can then generate 4 equations.

Charge

x-2y-1=2

x-2y=3---------------------------(1

H atoms

X=2a------------------------------(2

O atoms

4+4y=a+2b---------------------(3

C atoms

2y=b-----------------------------(4

Substitute 2x(4 in (3

4+2b=a+2b

Hence a=4

From (2
x=8

from (1

8-2y=3

y=5/2

and from (4
b=5

substitute the values back into the equation.

MnO4– + 8H+ + 5/2(C2O4)2– --> Mn2+ + 4H2O + 5CO2

Or

2MnO4– + 16H+ + 5(C2O4)2– --> 2Mn2+ + 8H2O + 10CO2

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Answer 1

correct, that is the answer I got 2, 16, 5, 2, 8, 10

Answer 2

16H(+) + 2MnO4(-) + 5C2O4(-2) → 2Mn(+2) + 4H2O + 5CO2

Answer 3

c2o4 has a charge of -2

so it's only + 1e-, not 2

Answer 4

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