Calculate: Electrolysis of copper(II) tetraoxosulfate(VII) using a platinum electrode.

During the electrolysis of copper(II) tetraoxosulfate(VII) using a platinum electrode, the following reactions occur at the anode and cathode:

At the anode (positive electrode):
2SO4^2- → 2SO4 + 4e-
This reaction results in the oxidation of sulfate ions to sulfur dioxide gas and electrons are released.

At the cathode (negative electrode):
Cu^2+ + 2e- → Cu
This reaction results in the reduction of copper(II) ions to copper metal, which deposits on the platinum electrode.

Overall, the electrolysis of copper(II) tetraoxosulfate(VII) results in the deposition of copper metal on the platinum electrode and the release of sulfur dioxide gas. This process can be used to extract copper from its ore or to purify copper metal.

Electrolysis of Copper(II) Tetraoxosulfate(VI)

Cathode (reduction):

```
CuSO₄(aq) + 2H₂O(l) + 2e⁻ → Cu(s) + H₂SO₄(aq) + 2OH⁻(aq)
```

Anode (oxidation):

```
2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻
```

Overall reaction:

```
CuSO₄(aq) + H₂O(l) → Cu(s) + O₂(g) + H₂SO₄(aq)
```

Observations:

- Bubbles of oxygen gas are released at the anode.
- A reddish-brown deposit of copper metal forms on the cathode.
- The solution turns a lighter blue as the copper ions are removed.

Notes:

- Platinum electrodes are used because they are inert and do not react with the reactants or products.
- The electrolyte solution is copper(II) tetraoxosulfate(VI) (CuSO₄) dissolved in water.
- The applied voltage must be high enough to overcome the electromotive force (EMF) of the cell.
- Electrolysis can be used to extract metals from their ores.