Some popular cases of electrolysis include:

1. Using platinum or carbon electrodes

When an aqueous solution of copper(II) sulfate, CuSO₄ is electrolysed using inert electrodes such as platinum or carbon.

Ionization:

             Cation     Anion
     CuSO₄     Cu²⁺      SO₄²⁻
     H₂O       H⁺        OH⁻
  

Observations: The Cu²⁺ are preferentially discharged over H⁺ at the cathode and reduced to copper metal. At the inert anode the OH⁻ is preferentially discharged leaving behind SO₄²⁻ in the solution. Hence, Copper metal deposits on the cathode; bubbles of oxygen appear at the anode;

Cathode (reduction):   2Cu²⁺(aq) + 4e⁻ → 2Cu(s)

Anode (oxidation):     4OH⁻(aq) → 2H₂O(l) + O₂(g) + 4e⁻       


Overall (net) reaction: 2Cu²⁺(aq) + 4OH⁻ → 2Cu(s) + 2H₂O(l) + O₂(g) 

Notes: Because the anode is inert (does not dissolve), copper is not supplied from the electrode. Hence, the blue colour of CuSO₄ begins to fade. The solution becomes more acidic at the anode because H⁺ and SO₄²⁻ are dissolved in the solution. Copper deposits are relatively pure.

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2. Electrolysis of Copper(II) Sulfate Using Copper Electrodes (Electrolytic Refining / Purification of Copper)

If both electrodes are copper( impure copper as the anode and a pure copper starter plate as the cathode) and the electrolyte is copper(II) sulfate, the electrolysis is used to refine copper. The anode dissolves as Cu → Cu²⁺ + 2e⁻ and these Cu²⁺ ions migrate to the cathode where they are reduced and plated as pure copper. Insoluble impurities fall off as anode sludge.

Ionization:

             Cation     Anion
     CuSO₄     Cu²⁺      SO₄²⁻
     H₂O       H⁺        OH⁻
  

Observations: At the anode, neither SO₄²⁻ nor OH⁻ is preferentially discharged. Instead, the copper anode gradually loses mass releasing Cu²⁺ which moves to the cathode. The cathode gains shiny pure copper, an anode mud collects below the anode containing impurities (Ag, Au, Pt, etc.).

Anode (oxidation):  Cu(s) → Cu²⁺(aq) + 2e⁻

Cathode (reduction):  Cu²⁺(aq) + 2e⁻ → Cu(s)

Overall (net) :       Cu → Cu²⁺ → Cu 

Notes: This is the standard industrial method for electrolytic refining of copper. The electrolyte stays rich in Cu²⁺; impurities that are less reactive than copper (e.g., Ag, Au) fall down as anode slime; more reactive impurities remain in solution. The blue colour of copper sulphate remains at ghe end of the reaction.

Electrolysing concentrated aqueous sodium chloride (brine) with carbon as the anode and platinum as the cathode yields chlorine gas at the anode and hydrogen gas at the cathode. The solution left behind becomes alkaline (NaOH formation) because OH⁻ accumulates. Platinum is not used as the anode because chlorine attacks platinum.

Ionization:

             Cation    Anion
     NaCl     Na⁺       Cl⁻
     H₂O       H⁺       OH⁻
  

Observations: At the anode, Cl⁻ is preferentially discharged over OH⁻ due to the high concentration of chloride ions. Chlorine gas (greenish-yellow, pungent) is liberated at anode. At the cathode H⁺ is preferentially discharged over Na⁺ even though Na⁺ is more concentrated due to its position in the activity series and because it is very far from Na⁺ (Here, concentration holds no effect). Hydrogen gas is liberated at cathode, and the remaining solution becomes basic (contains NaOH).

At cathode (reduction):   2H⁺(aq) + 2e⁻ → H₂(g)

At anode (oxidation):     2Cl⁻(aq) → Cl₂(g) + 2e⁻

Overall (net):            2H⁺(aq) + 2Cl⁻(aq) → H₂(g) + Cl₂(g) 

Notes: This is the basis of the chlor-alkali industry (commercial production of Cl₂, H₂ and NaOH). The relative ease of Cl⁻ discharge is influenced by concentration.

With dilute NaCl solution, the concentration of Cl⁻ is low. Thus oxygen is formed at the anode rather than chlorine; at the cathode, hydrogen is still produced .

Ionization:

             Cation    Anion
     NaCl     Na⁺       Cl⁻
     H₂O       H⁺       OH⁻
  

Observations: At the cathode, H⁺ is preferentially discharged over Na⁺ due to position in the activity series. Hence, Bubbles of hydrogen are liberated at the cathode. At the anode, OH⁻ is preferentially discharged over Cl⁻ due to position in the activity series. Hence, bubbles of oxygen are liberated at the anode; little or no chlorine odor.

At cathode (reduction): 2H⁺ (aq) + 2e⁻ → H₂(g)

At anode (oxidation):   4OH⁻(aq) → 2H₂O(l) + O₂(g) + 4e⁻       


Overall (net):          4H⁺(aq) + 4OH⁻(aq) →   2H₂(g) + 2H₂O(l) + O₂(g) 

Notes: Whether Cl₂ or O₂ evolves depends on chloride concentration, electrode material and current density (overpotential).

Electrolysis of acidified water (an aqueous solution containing excess H⁺, e.g., dilute H₂SO₄) produces hydrogen gas at the cathode and oxygen gas at the anode in the ratio 2:1 . The electrolytic cell used in this electrolysis is Hofmen's voltameter.

Ionization:

                Cation      Anion
     H₂SO₄        2H⁺         SO₄²⁻
     H₂O          H⁺          OH⁻
  

Observations: H⁺ migrate to the cathode and attracts elections to form hydrogen gas. OH⁻ is preferentially discharged at the anode over SO₄²⁻ due to its position in the activity series.The OH⁻ gives up its electrons to form water and oxygen gas. Hence, bubbles of hydrogen are liberated at cathode and oxygen at the anode

At cathode (reduction): 2H⁺(aq) + 2e⁻ → H₂(g)

At anode (oxidation):   4OH⁻(aq) → 2H₂O(l) + O₂(g) + 4e⁻       

Overall (net):          4H⁺ + OH⁻ → 2H₂(g) + 2H₂O(l) + O₂(g) 

Notes: During the electrolysis, more water is removed to establish equilibrium. Hence, the concentration of the acid increases.