Redox and balancing

I don't understand how to balance this equation:

H+ + MnO4- + I- ---> I2 + Mn2+

Why is there an H+ there? Is it just to show that the MnO4- is acidifed?

Same with this one:

Cu + H+ + NO3- ---> Cu2+ + NO2

Thanks for any help!

Reply 1

Ydina

7

Original post by jelly1081

I don't understand how to balance this equation:

H+ + MnO4- + I- ---> I2 + Mn2+

Why is there an H+ there? Is it just to show that the MnO4- is acidifed?

Same with this one:

Cu + H+ + NO3- ---> Cu2+ + NO2

Thanks for any help!

I suppose H2O water is not mentioned in the right part of equation
H+ + MnO4- + I- ->I2 + Mn2+
https://chemequations.com/en/?s=I%3A-+%2B+MnO4%3A-+%2B+H%3A%2B+%3D+I2+%2B+Mn%3A2%2B+%2B+H2O

Original post by jelly1081

I don't understand how to balance this equation:

H+ + MnO4- + I- ---> I2 + Mn2+

Why is there an H+ there? Is it just to show that the MnO4- is acidifed?

Same with this one:

Cu + H+ + NO3- ---> Cu2+ + NO2

Thanks for any help!

Yes, t first sight redox equations can be terrifying, but do not fear there is a methodology that will allow you to balance them.

There are two fundamental conditions, acidic or basic and redox pairs often behave differently in these. So yes, you do need to know the conditions.

You also need to know what is produced from a specific reagent.

H⁺ + MnO₄^- + I^- ---> I₂ + Mn²⁺

This tells you that the manganate(VII) ion produces manganese(II) ions under acidic conditions. And that iodide ions produce iodine.

So, to balance it you need to generate two half-equations, one for the manganese and one for the iodine.

Lets start with the manganese:
*********************************
H⁺ + MnO₄^- ==> Mn²⁺

To balance this in terms of atoms you are only allowed to use either hydrogen ions and water molecules (acidic conditions). GENERAL RULE

8H⁺ + MnO₄^- ==> Mn²⁺ + 4H₂O

However, it also needs to be balanced for electrical charge by adding (negative) electrons to either side.

8H⁺ + MnO₄^-+ 5e^- ==> Mn²⁺ + 4H₂O

Now the iodine (same process)
***********************************
I^- ==> I₂

2I^- ==> I₂ + 2e^-

Combining the half-equations
********************************

Before the half-equations can be added together the electrons need to be the same. This can be achieved by multiplying the fist half-equation by 2 and the second half equation by 5.

16H⁺ + 2MnO₄^- + 10e^- ==> 2Mn²⁺ + 8H₂O
10I^- ==> 5I₂+ 10e^-
------------------------------------- add together
16H⁺ + 2MnO₄^- + 10I^- ==> 2Mn²⁺ + 8H₂O + 5I₂

It is now balanced in terms of atoms and charge.

Now see if you can do the same for the copper oxidation by nitric acid.

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Reply 3

jelly1081

OP

1

Original post by charco

Yes, t first sight redox equations can be terrifying, but do not fear there is a methodology that will allow you to balance them.

There are two fundamental conditions, acidic or basic and redox pairs often behave differently in these. So yes, you do need to know the conditions.

You also need to know what is produced from a specific reagent.

H⁺ + MnO₄^- + I^- ---> I₂ + Mn²⁺

This tells you that the manganate(VII) ion produces manganese(II) ions under acidic conditions. And that iodide ions produce iodine.

So, to balance it you need to generate two half-equations, one for the manganese and one for the iodine.

Lets start with the manganese:
*********************************
H⁺ + MnO₄^- ==> Mn²⁺

To balance this in terms of atoms you are only allowed to use either hydrogen ions and water molecules (acidic conditions). GENERAL RULE

8H⁺ + MnO₄^- ==> Mn²⁺ + 4H₂O

However, it also needs to be balanced for electrical charge by adding (negative) electrons to either side.

8H⁺ + MnO₄^-+ 5e^- ==> Mn²⁺ + 4H₂O

Now the iodine (same process)
***********************************
I^- ==> I₂

2I^- ==> I₂ + 2e^-

Combining the half-equations
********************************

Before the half-equations can be added together the electrons need to be the same. This can be achieved by multiplying the fist half-equation by 2 and the second half equation by 5.

16H⁺ + 2MnO₄^- + 10e^- ==> 2Mn²⁺ + 8H₂O
10I^- ==> 5I₂+ 10e^-
------------------------------------- add together
16H⁺ + 2MnO₄^- + 10I^- ==> 2Mn²⁺ + 8H₂O + 5I₂

It is now balanced in terms of atoms and charge.

Now see if you can do the same for the copper oxidation by nitric acid.

-----------------------------------------------------------------------------------------------
Signature
Colourful Solutions IB Chemistry Testing and Notes

Thanks for the help! got everything solved and checked with my teacher. Hope you have a goodnight.