Some1 please answer these questions in stuck
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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