Question

Question #1a680

Answer 1

`n` `"factor" = 2`

Explanation:

You know that an unknown reducing agent will reduce sodium nitrate to ammonia, so start by looking at the reduction half-reaction.

Keep in mind that you don't need to balance the half-reaction because you're only interested in how the oxidation state of nitrogen changes!

`stackrel(color(blue)(+5))("N") "O"_ 3""^(-) + 8"e"^(-) -> stackrel(color(blue)(-3))("N")"H"_ 3`

Now, notice that the oxidation state of nitrogen goes from `color(blue)(+5)` on the reactants' side to `color(blue)(-3)` on the products' side, which basically means that every atom of nitrogen gains `8` electrons.

This means that the `n` factor of the sodium nitrate, which acts as an oxidizing agent in this reaction, will be equal to `8`.

Now, in every redox reaction, the number of electrons lost in the oxidation half reaction must be equal to the number of electrons gained in the reduction half-reaction.

In your case, you know that `7.65 * 10^(-3)` moles of sodium nitrate will gain a total of

`7.65 * 10^(-3) color(red)(cancel(color(black)("moles NaNO"_3))) * "8 e"^(-)/(1color(red)(cancel(color(black)("mole NaNO"_3)))) = 6.12 * 10^(-2)` `"moles e"^(-)`

Since you know that the reaction consumed `3.06 * 10^(-2)` moles of this unknown reducing agent, you can say that you will have

`1 color(red)(cancel(color(black)("mole reducing agent"))) * (6.12 * 10^(-2)color(white)(.)"moles e"^(-))/(3.06 color(red)(cancel(color(black)("moles reducing agent")))) = "2 moles e"^(-)`

Since every `1` mole of this reducing agent loses `2` moles of electrons, you can say that its `n` factor will be equal to

`ncolor(white)(.)"factor reducing agent" = 2`

SIDE NOTE I would try to balance the reduction half-reaction like this--the reaction takes place in basic medium!

`9"H"_ 2"O" + stackrel(color(blue)(+5))("N") "O"_ 3""^(-) + 8"e"^(-) -> stackrel(color(blue)(-3))("N")"H"_ 3 + 3"H"_2"O" + 9"OH"^(-)`