Permanganate, Mno,, can act as an oxidizing agent in both acidic and basic aqueous solutions. Balance each of the following reactions by the half-reaction method, showing all steps in the balancing process. Mno, + Al -> Mno, + Al(OH), in aqueous base i. ii. MnO, + CH,COOH -> Mn2 + CO, in aqueous acid

i) The reaction given is,

=> MnO₄^- + Al --------> MnO₂ + Al(OH)₄^-                                     In aqueous base i.e basic medium.

In the above reaction, the oxidation state of Mn was +7 initially in MnO₄^- and its +4 finally in MnO₂ (because each O is in -2 oxidation state).

And the oxidation state of Al was 0 and finally its +3 in Al(OH)₄^-.

Hence the oxidation of Al and reduction of Mn is taking place in the reaction.

Hence the half reactions taking place can be written as,

Oxidation : Al -------> Al(OH)₄^- + e^- 

Balancing : Since we have 4 OH in RHS. Hence balancing it in LHS using hydroxide ion as the medium is basic.

=> Al + 4 OH^- -------> Al(OH)₄^- + e^- 

Now we have total 4- charge in LHS. Hence making it 4- in RHS also.

=> Al + 4 OH^- -------> Al(OH)₄^- + 3 e^- 

Since both charge and number of elements are equal in both side of the reaction now. Hence the reaction is now balanced.

Reduction : MnO₄^- + e^- -------> MnO₂ 

Balancing : Since we have 2 O extra in LHS. Hence balancing it in RHS using water.

=> MnO₄^- + e^- -------> MnO₂ + 2 H₂O

Now we have 4 H extra in RHS. Hence balancing it in LHS by adding 4 water in LHS and 4 hydroxide in RHS.

=> MnO₄^- + 2 H₂O + e^- -------> MnO₂ + 4 OH^- 

Now we have total charge in RHS as 4-. Hence making it 4- in LHS also.

=> MnO₄^- + 2 H₂O + 3 e^- -------> MnO₂ + 4 OH^- 

Since both charge and number of elements are equal in both side of the reaction now. Hence the reaction is now balanced.

Adding the balanced reduction and oxidation half reactions to get overall balanced reaction by cancelling 3 electrons as,

=> MnO₄^- + 2 H₂O + Al ------> Al(OH)₄^- + MnO₂