Question

What is the reduction half-reaction for the unbalanced redox equation `Cr_2O_7^(2-) + NH_4^+ -> Cr_2O_3 + N_2`?

Answer 1

If you already found the oxidation half-reaction, then you should have already found the reduction half-reaction. That's why they're called half-reactions. Once you have one of them, you know the other is the other kind.

So, I assume you haven't found the oxidation half-reaction either.

Reduction

`"Cr"_2"O"_7^(2-)(aq) + 8"H"^(+)(aq) + 6e^(-) -> "Cr"_2"O"_3(s) + 4"H"_2"O"(l)`

Oxidation

`2"NH"_4^(+)(aq) -> "N"_2(g) + 8"H"^(+)(aq) + 6e^(-)`

Ultimately I got the full reaction as:

`"Cr"_2"O"_7^(2-)(aq) + 2"NH"_4^(+)(aq) -> "Cr"_2"O"_3(s) + "N"_2(g) + 4"H"_2"O"(l)`

---

We have two half-reactions corresponding to each species (`"Cr"`, `"N"` species):

`"Cr"_2"O"_7^(2-)(aq) -> "Cr"_2"O"_3(s)`

`"NH"_4^(+)(aq) -> "N"_2(g)`

Note that in order for `"NH"_4^(+)` to exist in solution (`"pKa" ~~ 9.4`), the solution should be acidic (otherwise, it would exist as `"NH"_3`, above approximately pH `9.4`), so we are balancing in acidic conditions.

REDUCTION HALF-REACTION

The natural oxidation state of `"O"` ion not in a peroxide is `-2`, so by finding the charge balance, you can find the oxidation state of each `"Cr"` ion.

Work this out and convince yourself it works:

• `"Cr"_2"O"_7^(2-)`: `2x + 7xx(-2) = -2 => color(red)(x = +6)`.
• `"Cr"_2"O"_3`: `2x + 3xx(-2) = 0 => color(red)(x = +3)`.

`stackrel(color(red)(+6))("Cr"_2)stackrel(color(red)(-2))("O"_7^(2-))(aq) -> stackrel(color(red)(+3))("Cr"_2)stackrel(color(red)(-2))("O"_3)(s)`

Two ways you could understand that this is a reduction half-reaction:

1. When the number of oxygens increases, the pertinent species is oxidized. So, when the opposite happens, the pertinent species is reduced. Specifically, `+6 -> +3`.
2. When the oxidation state gets less positive (or more negative), the element was reduced. Specifically, `+6 -> +3`.

Now, let's balance this in acid. Here's how I would do it:

1. Add whole-number coefficients to balance main species (i.e. `"Cr"`, `"N"`, etc), if needed.
2. Add water to balance oxygens, if needed.
3. Add `\mathbf("H"^(+))` to balance hydrogens, if needed.
4. Add electrons to balance the charges, if needed.

`=> "Cr"_2"O"_7^(2-)(aq) -> "Cr"_2"O"_3(s) + 4"H"_2"O"(l)`

`=> "Cr"_2"O"_7^(2-)(aq) + 8"H"^(+)(aq) -> "Cr"_2"O"_3(s) + 4"H"_2"O"(l)`

`=> color(blue)("Cr"_2"O"_7^(2-)(aq) + 8"H"^(+)(aq) + 6e^(-) -> "Cr"_2"O"_3(s) + 4"H"_2"O"(l))`

Charge Balance check:

`color(green)((-2) + 8xx(+1) + (-6) = 0)`

OXIDATION HALF-REACTION

Let's identify the oxidation states. Here's this one, unbalanced.

`stackrel(color(red)(-3))("N")stackrel(color(red)(+1))("H"_4^(+)(aq)) -> stackrel(color(red)(0))("N"_2(g))`

• A compound in its elemental state, like `"N"_2(g)` or `"O"_2(g)`, has an oxidation state of `color(red)(0)`.
• The more electronegative element in `"NH"_4^(+)` has the negative oxidation state.

So, `"H"` doesn't have an oxidation state of `-1`, but `color(red)(+1)`. Therefore, `+1xx3 - 3 = 0`, and `"N"` has an oxidation state of `color(red)(-3)`.

Since we had `stackrel(color(red)(-3))("N") -> stackrel(color(red)(0))("N")`, nitrogen was oxidized, i.e. its oxidation state became more positive (or less negative).

Next, let's balance this in acid. Same strategy as before, but the nitrogens are unbalanced and we don't need to add water since there is no oxygen on either side.

`=> 2"NH"_4^(+)(aq) -> "N"_2(g)`

`=> 2"NH"_4^(+)(aq) -> "N"_2(g) + 8"H"^(+)(aq)`

`=> color(blue)(2"NH"_4^(+)(aq) -> "N"_2(g) + 8"H"^(+)(aq) + 6e^(-))`

Charge Balance check:

`color(green)(2xx(+1) = 8xx(+1) + (-6))`

OVERALL BALANCED REACTION

Thus, the overall reaction is...

`"Cr"_2"O"_7^(2-)(aq) + cancel(8"H"^(+)(aq)) + cancel(6e^(-)) -> "Cr"_2"O"_3(s) + 4"H"_2"O"(l)`
`2"NH"_4^(+)(aq) -> "N"_2(g) + cancel(8"H"^(+)(aq)) + cancel(6e^(-))`
`"------------------------------------------------------------------"`
`\mathbf(color(blue)("Cr"_2"O"_7^(2-)(aq) + 2"NH"_4^(+)(aq) -> "Cr"_2"O"_3(s) + "N"_2(g) + 4"H"_2"O"(l)))`