How Do Manganese Redox Reactions Work with Nitric Acid and Potassium Iodate?

[Mattara]

Hello,

I've been working on some redox reactions for the oxidation of manganese.

Question:

Write the complete formulae for the redox reactions below:

i.) Manganese reacts with nitric acid (V) to form Mn²⁺ ions.
ii.) The Mn²⁺ ions reacts with potassium iodate, KIO₄ to form MnO₄^- (VII)

My Solution:

i.)

Oxidation: 3Mn (s) --> 3Mn²⁺ (aq) + 6e^- (aq)
Reduction: 6HN0₃ (aq) + 6e^- (aq) --> 6N0₃^- (aq) + 6H⁺ (aq)
Overall reaction: 3Mn + 6HN0₃ -> 6NO₃^- + 3Mn²⁺ + 6H⁺

Is this correct? If so, does the single protons regenerate the acid and/or become attached to substances like water?

ii.)

Oxidation: Mn²⁺ + X -> MnO₄^- + ?H⁺
Reduction: IO₄^- + ?H⁺ -> I? + 4H₂O
Overall reaction: Mn²⁺ + IO₄^- -> MnO₄^- + I₂^-

I'm having trouble with this part of the question. I does not have -II as one of its common oxidation number.

Any help or hints would be greatly appreciated

Best Regards,
Mattara

 

[Gokul43201]

Hello,

I've been working on some redox reactions for the oxidation of manganese.

Question:

Write the complete formulae for the redox reactions below:

i.) Manganese reacts with nitric acid (V) to form Mn²⁺ ions.
ii.) The Mn²⁺ ions reacts with potassium iodate, KIO₄ to form MnO₄^- (VII)

My Solution:

i.)

Oxidation: 3Mn (s) --> 3Mn²⁺ (aq) + 6e^- (aq)
Reduction: 6HN0₃ (aq) + 6e^- (aq) --> 6N0₃^- (aq) + 6H⁺ (aq)
Overall reaction: 3Mn + 6HN0₃ -> 6NO₃^- + 3Mn²⁺ + 6H⁺

Is this correct? If so, does the single protons regenerate the acid and/or become attached to substances like water?

Take a step back and ask yourself,

metal + acid --> salt + (?)

You do not get protons. Look at the reduction reaction you've written. The charge is not balanced. Fix that one mistake using above hint and you're good.

ii.)

Oxidation: Mn²⁺ + X -> MnO₄^- + ?H⁺

On the RHS, you have Mn, H and O. On the left, you have Mn. Ergo, X is simple H2O.

Reduction: IO₄^- + ?H⁺ -> I? + 4H₂O
Overall reaction: Mn²⁺ + IO₄^- -> MnO₄^- + I₂^-

I'm having trouble with this part of the question. I does not have -II as one of its common oxidation number.

This is a little tricky. The reaction actually depends on the conditions. Mostly, you just have to know what happens to IO4- under different conditions. Typically, it will first get reduced to IO3-. Unless someone else suggests otherwise, you could go with that.