Chem5 Flashcards
Why is 99% impure copper unsuitable for electrical wires?
Impurities (Ag, Au, Zn, Fe) reduce conductivity; electrolysis increases purity to 99.95%.
What is the anode made of in copper purification, and what happens there?
Impure copper (99%); oxidation dissolves Cu:
Cu⁰(s) → Cu²⁺(aq) + 2e⁻.
What is the cathode made of, and what forms there?
Pure copper wire; reduction deposits Cu:
Cu²⁺(aq) + 2e⁻ → Cu⁰(s).
What happens to less active impurities (Ag, Au) during electrolysis?
They fall as solid ‘anode mud’ since they resist oxidation.
Why do Zn and Fe remain in solution instead of depositing at the cathode?
Their ions (Zn²⁺, Fe²⁺) are harder to reduce than Cu²⁺ due to lower reduction potentials.
What happens to the mass of the anode and cathode during purification?
Anode mass decreases (dissolves); cathode mass increases (Cu deposits).
How can this process extract silver and gold from crude copper?
Ag/Au collect as anode mud, allowing their recovery.
Write the oxidation reactions for Zn and Fe at the anode.
Zn(s) → Zn²⁺(aq) + 2e⁻;
Fe(s) → Fe²⁺(aq) + 2e⁻.
Why doesn’t the Cu²⁺ concentration change in the solution?
Cu²⁺ produced at anode = Cu²⁺ reduced at cathode.
What is the role of CuSO₄ in the solution?
It provides Cu²⁺ ions for the electrolyte:
CuSO₄(aq) → Cu²⁺(aq) + SO₄²⁻(aq).
Why are Ag/Au not oxidized at the anode?
They are less reactive than Cu (higher reduction potentials).
What property determines if an impurity stays in solution or forms anode mud?
Relative reactivity: More active (Zn, Fe) oxidize; less active (Ag, Au) do not.
