Some Help PLease

[busupshot83]

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I was working the following problem, and I found myself stuck at the end. Here's the problem and the steps I have completed thus far:

Write a balanced net ionic equation for the following reaction in basic solution:

Step 1: Write equation

MnO4 (aq) + IO3 (aq) ---> MnO2 (s) + IO4 (aq)

Step 2: Balance unbalanced equation

Already balanced.

Steps 3 & 4: Find oxidation numbers and determine the change in oxidation numbers.

Mn starts with +7 and ends with +4; this is reduction.
I starts with +5 and ends with +7; this is oxidation.

Step 5: Find the net changes of the oxidation numbers, than multiple the oxidized and reduced atoms with least common multipliers.

Mn has a net change of 3, so multiply the I atoms by 3.
I has a net change of 2, so multiply the Mn atoms by 2.

2[MnO4 (aq)] + 3[IO3 (aq)] ---> 2[MnO2 (s)] + 3[IO4 (aq)]

Step 6: Balance for O by adding H2O to the side with the least amount of O, and then balance for H by adding H+ to the side with the least amount of H.

2H + 2[MnO4 (aq)] + 3[IO3 (aq)] ---> 2[MnO2 (s)] + 3[IO4 (aq)] + H2O

Ok, at this point, the equation is balanced in acidic solutions... but the problem asks for basic solutions. So how do I know how much OH - to add to each side?

 

[busupshot83]

Oh yeah, I'm using the "Oxidation-Number Method," not the "Half-Reaction Method."

 

[The Musketeer]

If my memory serves me correctly....

1) to make it basic, just add the same number of OH on each side so that H+ can be cancelled out with OH-
2OH + 2H + 2[MnO4 (aq)] + 3[IO3 (aq)] ---> 2[MnO2 (s)] + 3[IO4 (aq)] + H2O + 2OH

2) combine OH- and H+ where possible to form H2O
2H20 + 2[MnO4 (aq)] + 3[IO3 (aq)] ---> 2[MnO2 (s)] + 3[IO4 (aq)] + H2O + 2OH

3) simplify H20 term in the equation
H2O + 2[MnO4 (aq)] + 3[IO3 (aq)] ---> 2[MnO2 (s)] + 3[IO4 (aq)] + 2OH

4) so now that you have OH- (and no H+) in your equation, you have a basic reaction

 

[busupshot83]

If my memory serves me correctly....

1) to make it basic, just add the same number of OH on each side so that H+ can be cancelled out with OH-
2OH + 2H + 2[MnO4 (aq)] + 3[IO3 (aq)] ---> 2[MnO2 (s)] + 3[IO4 (aq)] + H2O + 2OH

2) combine OH- and H+ where possible to form H2O
2H20 + 2[MnO4 (aq)] + 3[IO3 (aq)] ---> 2[MnO2 (s)] + 3[IO4 (aq)] + H2O + 2OH

3) simplify H20 term in the equation
H2O + 2[MnO4 (aq)] + 3[IO3 (aq)] ---> 2[MnO2 (s)] + 3[IO4 (aq)] + 2OH

4) so now that you have OH- (and no H+) in your equation, you have a basic reaction

"H2O + 2[MnO4 (aq)] + 3[IO3 (aq)] ---> 2[MnO2 (s)] + 3[IO4 (aq)] + 2OH"

That is the answer according to my solution's manuel, thanks a lot Musketeer 😀! I just wasn't turning up with that answer.