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

c. because step up transformer steps up the input voltage i.e increases the input voltage  

the number of turns are greater on the secondary side of step up transformer  

transformer does not handle dc current