Question

Could elemental copper react with sulfuric acid?

Answer 1

Let's change the acid to hydrochloric....

Explanation:

If conc. hydrochloric were used the element reduced would be hydrogen...

`H^+Cl^(-) +e^(-)rarr 1/2H_2(g) + Cl^(-)`

And so here the element reduced would by hydrogen. Were copper to react with sulfuric acid, to all intents and purposes a non-oxidizing acid...sulfur might be reduced to `SO_2`, but you would need pretty fierce conditions....

`H_2SO_4(aq)+2H^+ +2e^(-)rarr SO_2(g)+2H_2O(l)`

Answer 2

Sulfur.

Explanation:

Joe L. is right—copper normally doesn't displace hydrogen, because copper is higher up on an activity series than hydrogen.

However, after some digging, I've found that copper, when heated, can react with a concentrated solution of sulphuric acid.

This would be the skeleton equation:

`Cu + H_2SO_4 -> CuSO_4 + SO_2 + H_2O`
(It's not the standard single-displacement reaction.)

To find out which element is reduced, we need to understand what reduction means: to gain electrons.
For example, in the reaction `O -> O^(-2)`, oxygen would be reduced.

To begin, let's balance our skeleton equation.

`Cu + 2H_2SO_4 -> CuSO_4 + SO_2 + 2H_2O`

From this equation, we know that:

• Copper, from its oxidation state of `0` as a free element, becomes `"+2"`. This is because the oxidation state of the polyatomic ion, `SO_4`, is `"-2"` and the compound's overall charge is `0`.

• Hydrogen, from its oxidation state of `"+1"` in `H_2SO_4` (because there are `2` hydrogens and `1` sulphuric acid), doesn't change in `H_2O`. It's still `"+1"`.
  One reason is because the oxidation state of oxygen is the same as sulphuric acid; another reason is because hydrogen is always either `"+1"` or `"-1"`.

• The oxidation state of oxygen, like hydrogen, remains the same throughout.

• Sulfur, from its oxidation state of `"+6"` in `SO_4^(2-)` (because there are `4` oxygens which add up to a charge of `"-8"`, and the overall charge of the ion is `"-2"`), either stays the same in `CuSO_4` (it's still in sulphate) or becomes `"+4"` in `SO_2`.
  It's `"+4"` in `SO_2` because there are `2` oxygens, adding up to a charge of `"-4"`, and the overall charge of the compound is `0`, so sulfur must balance it out and be `"+4"`.

Because the oxidation state of one of the two sulphur atoms goes from `"+6"` to `"+4"`, that sulphur atom is the one that's being reduced

To go to a lower oxidation state, it must've had to gain `2` electrons. :)