Question

Question #de191

Answer 1

!! LONG ANSWER !!

Start by assigning oxidation numbers to all the atoms that take part in the reaction

`stackrel(color(blue)(+1))(Cu_2) stackrel(color(blue)(-2))(S) + stackrel(color(blue)(+1))(H) stackrel(color(blue)(+5))(N) stackrel(color(blue)(-2))(O_3) -> stackrel(color(blue)(+2))(Cu) overbrace((NO_3)_2)^(color(blue)(-2)) + stackrel(color(blue)(+4))(S) stackrel(color(blue)(-2))(O_2) + stackrel(color(blue)(+4))(N) stackrel(color(blue)(-2))(O_2) + stackrel(color(blue)(+1))(H_2) stackrel(color(blue)(-2))(O)`

Notice that copper's oxidation state changes from `color(blue)(+1)` on the reactants' side, to `color(blue)(+2)` on the products' side, which means that copper is getting oxidized.

Moreover, sulfur is being oxidized as well, since its oxidation state goes from `color(blue)(-2)` to `color(blue)(+4)`.

On the other hand, nitrogen is going from `color(blue)(+5)` on the reactants' side, to `color(blue)(+4)` on the products' side, which means that nitrogen is getting reduced.

Write the oxidation and reduction half-reactions and balance using `H^(+)` and `H_2O` when needed.

• Oxidation half-reaction

`stackrel(color(blue)(+1))(Cu_2) stackrel(color(blue)(-2))(S) -> 2stackrel(color(blue)(+2))(Cu_2) + stackrel(color(blue)(+4))(S)O_2 + 8e^(-)`

You get `1e^(-)` lost from each of the two copper atoms, and `6e^(-)` lost from the sulfur atom `->` `8e^(-)` are lost altogether.

Balance the oxygen by adding two water molecules on the reactants' side, and the resulting hydrogen by adding four `H^(+)` on the products' side

`2H_2O + Cu_2S -> 2Cu_2^(2+) + SO_2 + 8e^(-) + 4H^(+)`

• Reduction half-reaction

`stackrel(color(blue)(+5))(N)O_3^(-) + 1e^(-) -> stackrel(color(blue)(+4))(N)O_2`

Once again, balance the oxygen by adding one water molecule on the products' side, and two `H^(+)` on the reactants' side

`2H^(+) + NO_3^(-) + 1e^(-) -> NO_2 + H_2O`

Now it's time to balance the electrons lost during oxidation half-reaction and gained during the reduction half-reaction. Notice that you have `8e^(-)` in the oxidation half-reaction, but only `1e^(-)` in the reduction half-reaction.

Multiply the reduction half-reaction by 8 to get

`{ (2H_2O + Cu_2S -> 2Cu_2^(2+) + SO_2 + 8e^(-) + 4H^(+)), (16H^(+) + 8NO_3^(-) + 8e^(-) -> 8NO_2 + 8H_2O) :}`

Add these two equations to get

`2H_2O + Cu_2S + 16H^(+) + 8NO_3^(-) + cancel(8e^(-)) -> 2Cu_2^(2+) + SO_2 + 8NO_2 + cancel(8e^(-)) + 4H^(+) + 8H_2O`

Eliminate compounds that are on both sides of the equation to get

`Cu_2S + 12H^(+) + 8NO_3^(-) -> 2Cu_2^(2+) + SO_2 + 8NO_2 + 6H_2O`

The complete and balanced equation will be

`Cu_2S + 12HNO_3 -> 2Cu_2(NO_3)_2 + SO_2 + 8NO_2 + 6H_2O`

Answer 2

Warning! This is a long answer. The balanced equation is

`"Cu"_2"S" + "12HNO"_3 → "2Cu"("NO"_3)_2 + "SO"_2 + "8NO"_2 + "6H"_2"O"`

We start with the unbalanced equation:

`"Cu"_2"S" + "HNO"_3 → "Cu"("NO"_3)_2 + "SO"_2 + "NO"_2 + "H"_2"O"`

Step 1. Identify the atoms that change oxidation number

`stackrel(color(blue)(+1))("Cu")_2 stackrel(color(blue)(-2))("S") + stackrel(color(blue)(+1))("H") stackrel(color(blue)(+5))("N") stackrel(color(blue)(-2))("O")_3 → stackrel(color(blue)(+2))("Cu")( stackrel(color(blue)(+5))("N") stackrel(color(blue)(-2))("O")_3)_2 + stackrel(color(blue)(+4))("S") stackrel(color(blue)(-2))("O")_2 + stackrel(color(blue)(+4))("N") stackrel(color(blue)(-2))("O")_2 + stackrel(color(blue)(+1))("H")_2 stackrel(color(blue)(-2))("O")`

Left hand side: `"Cu" = +1`; `"S" = -2`; `"H" = +1`; `"N" = +5`; `"O" = -2`
Right hand side: `"Cu" = +2`; `"N in NO"_3 = +5`; `"S" = +4`; `"O" = -2`; `"N in NO"_2 = +4`; `"H" = +1`

The changes in oxidation number are:

`"Cu"`: +1 → +2; Change =+1
`"S"`: -2 → +4; Change = +6
`"N"`: +5 → +4; Change = -1

But `"Cu"_2"S"` behaves as a unit.

The total change for `"Cu"_2"S"` is [2×(+1) + 6] = +8.

Step 2. Equalize the changes in oxidation number

You need 8 atoms of `"N"` that change oxidation number for every 1 unit of `"Cu"_2"S"`. This gives us total changes of -8 and +8.

Step 3. Insert coefficients to get these numbers

`color(red)(1)"Cu"_2"S" + "HNO"_3 → color(red)(2)"Cu"("NO"_3)_2 + color(red)(1)"SO"_2 + color(red)(8)"NO"_2 + "H"_2"O"`

Note: We can't fix a number in front of the `"HNO"_3` because some of the `"N"` atoms go to `"NO"_2` and some remain unchanged as `"NO"_3` ions.

Step 4. Balance `"N"` by adding `"HNO"_3"` molecules to the appropriate side

`color(red)(2)"Cu"_2"S" + color(blue)(12)"HNO"_3 → color(red)(2)"Cu"("NO"_3)_2 + color(red)(1)"SO"_2 + color(red)(8)"NO"_2 + "H"_2"O"`

Step 4. Balance `"O"` by adding `"H"_2"O"` molecules to the appropriate side

`color(red)(2)"Cu"_2"S" + color(blue)(12)"HNO"_3 → color(red)(2)"Cu"("NO"_3)_2 + color(red)(1)"SO"_2 + color(red)(8)"NO"_2 + color(orange)(6)"H"_2"O"`

The balanced equation is

`color(red)("Cu"_2"S" + "12HNO"_3 → "2Cu"("NO"_3)_2 + "SO"_2 + "8NO"_2 + "6H"_2"O")`