Permanganate ion reacts in basic solution with oxalate ion to form carbonate ion and solid manganese dioxide. Balance the skeleton ionic equation for the reaction between NaMnO4 and Na2C2O4 in basic solution: Fill in all blanks with numbers so if the term is not in the equation make it 0. Mno4^- (aq)+ C204^2- (aq)+
H^+(aq) + OH^-(aq)
H2O(l) MnO2(s)+
CO3^2 (aq)+ H^+(aq)+
OH^- (aq) + H2O(l)

2MnO4^- (aq) + 3C2O4^2- (aq) + 2H2O (l) --> 2MnO2(s) +6CO3^2 -(aq) + 4H^+ (aq)

Explanation:

First, write the half equations for the reduction of MnO4^- and the oxidation of C2O4^2- respectively. Balance it.

Reduction requires H+ ions and e- and gives out water, vice versa for oxidation.

Reduction:

MnO4^- (aq) + 4H^+ (aq) + 3e- ---> MnO2(s) + 2H2O (l)

Oxidation:

C2O4^2- (aq) + 2H2O (l) ---> 2CO3^2 -(aq) + 4H^+ (aq) + 2e-

Balance the no. of electrons on both equations so that electrons can be eliminated. we can do so by multiplying the reduction eq by 2, and oxidation eq by 3.

2MnO4^- (aq) + 8H^+ (aq) + 6e- ---> 2MnO2(s) + 4H2O (l)

3C2O4^2- (aq) + 6H2O (l) ---> 6CO3^2 -(aq) + 12H^+ (aq) + 6e-

Now combine both equations and eliminate repeating H+ and H2O.

2MnO4^- (aq) + 8H^+ (aq) + 3C2O4^2- (aq) + 6H2O (l) --> 2MnO2(s) + 4H2O (l) +6CO3^2 -(aq) + 12H^+ (aq)

turns into:

2MnO4^- (aq) + 3C2O4^2- (aq) + 2H2O (l) --> 2MnO2(s) +6CO3^2 -(aq) + 4H^+ (aq)

its going to be mutualism cuz they both benefit from the relationship.

~batmans wife dun dun