Manganese redox reactions

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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 :smile:

Best Regards,
Mattara
 

Gokul43201

Staff Emeritus
Science Advisor
Gold Member
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Mattara said:
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.
 
Last edited:

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