Some1 please answer these questions in stuck

the reactions at the cathode:

1. the reduced active metal is water, other than that the metal will be reduced

2. H⁺ of the acid will be reduced

the reactions at the anode:

1. if the electrodes are not inert then the metal is oxidized

2. If inert then:

a. OH⁻ from the base will be oxidized

b. The halogen metal will oxidize

1. HCl

H⁺ of the acid will be reduced

Cathode: 2H⁺(aq) + 2e⁻ ⇒ H₂(g)

The halogen metal will oxidize

Anode: 2Cl⁻(aq) ⇒ Cl₂(g) + 2e⁻

2. NaNO₃

Na : active metal, water reduced

Cathode: 2H₂O + 2e ⇒ 2OH- + H₂

NO₃: oxyacid, water will be oxidized

Anode: 2H₂O ⇒ 4 H + + O₂ + 4e

3. CuCl

Cu : not active metal

Cathode: Cu²⁺ (aq) + 2e ⇒ Cu (s)

Cl : The halogen metal will oxidize

Anode: 2Cl⁻ (aq) ⇒ Cl₂ (g) + 2e