Question

How to do 16th question?

Answer 1

You'll have to write out the half-reactions to figure this out... except we'll write the proper phases...

`2stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) + 3stackrel(color(blue)(+1))(""^(1)"H"_2)""^(18)stackrel(color(blue)(-1))("O"_2)(aq) -> 2stackrel(color(blue)(+4))("Mn")stackrel(color(blue)(-2))("O"_2)(s) + 2stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O")(l) + 3stackrel(color(blue)(0))(""^(18)"O"_2)(g) + 2stackrel(color(blue)(-2))("O")stackrel(color(blue)(+1))("H"^(-))(aq)`

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REDUCTION HALF-REACTION

In either basic or acidic media, permanganate gets reduced to manganese (IV) oxide, and then `"Mn"^(2+)` if it goes all the way:

`stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) -> stackrel(color(blue)(+4))("Mn")stackrel(color(blue)(-2))("O"_2)(s)`

This is of course containing NORMAL `""^(16) "O"`... Here we choose basic media... balance the oxygen.

`stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) -> stackrel(color(blue)(+4))("Mn")stackrel(color(blue)(-2))("O"_2)(s) + 2stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O")(l)`

Balance hydrogens...

`4stackrel(color(blue)(+1))("H"^(+))(aq) + stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) -> stackrel(color(blue)(+4))("Mn")stackrel(color(blue)(-2))("O"_2)(s) + 2stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O")(l)`

Balance charge...

`3e^(-) + 4stackrel(color(blue)(+1))("H"^(+))(aq) + stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) -> stackrel(color(blue)(+4))("Mn")stackrel(color(blue)(-2))("O"_2)(s) + 2stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O")(l)`

That's the reduction half-reaction, which must be added to an oxidation half-reaction.

OXIDATION HALF-REACTION

Hence, we proceed to the oxidation of `""^(1)"H"_2""^(18)"O"_2` to see where those oxygen atoms go...

`stackrel(color(blue)(+1))(""^(1)"H"_2)""^(18)stackrel(color(blue)(-1))("O"_2)(aq) -> stackrel(color(blue)(0))(""^(18)"O"_2)(g)`

Here we KNOW it must become `""^(18)"O"_2` and NOT `""^(18)"O"""^(1)"H"^(-)`, as we KNOW that oxidation occurs here.

Balance the hydrogens...

`stackrel(color(blue)(+1))(""^(1)"H"_2)""^(18)stackrel(color(blue)(-1))("O"_2)(aq) -> stackrel(color(blue)(0))(""^(18)"O"_2)(g) + 2stackrel(color(blue)(+1))(""^(1)"H"^(+))(aq)`

Balance the charge...

`stackrel(color(blue)(+1))(""^(1)"H"_2)""^(18)stackrel(color(blue)(-1))("O"_2)(aq) -> stackrel(color(blue)(0))(""^(18)"O"_2)(g) + 2stackrel(color(blue)(+1))(""^(1)"H"^(+))(aq) + 2e^(-)`

OVERALL REACTION

And now we add them.

`2(cancel(e^(-)) + 4stackrel(color(blue)(+1))("H"^(+))(aq) + stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) -> stackrel(color(blue)(+4))("Mn")stackrel(color(blue)(-2))("O"_2)(s) + 2stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O")(l))`
`ul(3(stackrel(color(blue)(+1))(""^(1)"H"_2)""^(18)stackrel(color(blue)(-1))("O"_2)(aq) -> stackrel(color(blue)(0))(""^(18)"O"_2)(g) + 2stackrel(color(blue)(+1))(""^(1)"H"^(+))(aq) + cancel(2e^(-))))`
`2stackrel(color(blue)(+1))("H"^(+))(aq) + 2stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) + 3stackrel(color(blue)(+1))(""^(1)"H"_2)""^(18)stackrel(color(blue)(-1))("O"_2)(aq) -> 2stackrel(color(blue)(+4))("Mn")stackrel(color(blue)(-2))("O"_2)(s) + 4stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O")(l) + 3stackrel(color(blue)(0))(""^(18)"O"_2)(g)`

And of course, we're in basic media, so we add the usual `"OH"^(-)` to both sides...

`cancel(2stackrel(color(blue)(+1))("H"^(+))(aq) + 2stackrel(color(blue)(-2))("O")stackrel(color(blue)(+1))("H"^(-))(aq))^(2"H"_2"O"(l)) + 2stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) + 3stackrel(color(blue)(+1))(""^(1)"H"_2)""^(18)stackrel(color(blue)(-1))("O"_2)(aq) -> 2stackrel(color(blue)(+4))("Mn")stackrel(color(blue)(-2))("O"_2)(s) + cancel(4)^(2)stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O")(l) + 3stackrel(color(blue)(0))(""^(18)"O"_2)(g) + 2stackrel(color(blue)(-2))("O")stackrel(color(blue)(+1))("H"^(-))(aq)`

And our final result is then...

`bb(2stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) + 3stackrel(color(blue)(+1))(""^(1)"H"_2)""^(18)stackrel(color(blue)(-1))("O"_2)(aq) -> 2stackrel(color(blue)(+4))("Mn")stackrel(color(blue)(-2))("O"_2)(s) + 2stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O")(l) + 3stackrel(color(blue)(0))(""^(18)"O"_2)(g) + 2stackrel(color(blue)(-2))("O")stackrel(color(blue)(+1))("H"^(-))(aq))`