Question

How do you represent the oxidation of `Cu^+` by `MnO_4^(-)`?

Answer 1

`Cu^(+) rarr Cu^(2+) + e^-` OXIDATION
`MnO_4^(-) + 8H^(+) + 5e^(-)rarr Mn^(2+) + 4H_2O` REDUCTION

Explanation:

We can write the overall redox equation by adding `5` `xx` the oxidation equation to `1` of the reduction reaction to give:

`5Cu^+ + MnO_4^(-) + 8H^(+) rarr Mn^(2+) + 5Cu^(2+) + 4H_2O(l)`

This reaction is stoichiometrically balanced with respect to mass and charge, and it is self indicating. Why? Because permanganate ion, `MnO_4^-`, has a strong purple colour, whereas `Mn^(2+)` is almost colourless. The colour of the permanganate ion will persist at the endpoint.

Concentration of cuprous ion: `(n_(Cu^+))/(2.49xx10^-2*L)`

Moles of permanganate ion `=` `0.128*mol*cancel(L)^(-1)xx3.19xx10^(-2)cancel(L)` `=` `4.08xx10^(-3) mol`. Given the stoichiometry there were `5xx4.08xx10^(-3)mol` `=` `2.042xx10^(-2)` `mol` of cuprous ion.

SO the concentration of the original solution was `(2.042xx10^(-2)*mol)/(2.49xx10^-2*L)` `=` `0.820` `mol*L^(-1)` in `Cu^+` ion.