Question

Buat hasil elektrolisis larutan dengan elektroda C
a.NaCl
b.Na2SO4
C.CuSO4
d.HCl
e.KOH

Answer 1

A. NaCl
Katode = 2H₂O + 2e⁻ --> H₂ (g) + 2OH⁻
Anode = 2Cl⁻ --> Cl₂ (g) + 2e⁻
------------------------------------------------
Reaksi= 2H₂O+2Cl⁻ --> H₂ (g) + 2OH⁻ + Cl₂ (g)

b. Na₂SO₄
Katode = 4H₂O + 4e⁻ --> 2H₂ (g) + 4OH⁻
Anode = 2H₂O --> O₂ (g) + 4H⁺ + 4e⁻
------------------------------------------------
Reaksi = 6H₂O (l) --> 2H₂ (g) + O₂ (g)+ 4H₂O (l)
            = 2H₂O (l) --> 2H₂ (g)+ O₂ (g)

c. CuSO₄
Katode = 2Cu²⁺ + 4e⁻ --> 2Cu (s)
Anode = 2H₂O --> O₂ + 4H⁺ + 4e⁻
------------------------------------------------
Reaksi = 2Cu²⁺ + 2H₂O --> 2Cu + O₂ + 4H⁺

d. HCl
Katode = 2H⁺ + 2e⁻ --> H₂ 
Anode = 2Cl⁻ --> Cl₂ + 2e⁻
-------------------------------------
Reaksi = 2H⁺ + 2Cl⁻ --> H₂ + Cl₂ 

e. KOH
Katode = 4H₂O + 4e⁻ --> 2H₂ + 4OH⁻
Anode = 4OH⁻ --> 2H₂O + O₂ + 4e⁻
--------------------------------------------
Reaksi = 2H₂O --> 2H₂ + O₂

correct me if i wrong :)