Q.Mention any three factors which decides the products of electrolysis.

The cathode is the site of reduction; Na+ + e- -> Na, so sodium metal is liberated there. Answer: (b) Na.

In molten NaCl only Na+ and Cl- ions are present (no water).

• At the cathode (reduction): Na+ + e- -> Na (sodium metal).
• At the anode (oxidation): 2Cl- -> Cl2 + 2e- (chlorine gas). …

The assertion is true: electrolysis of aqueous NaCl gives H₂ at the cathode and Cl₂ at the anode. The reason given is false — chlorine does not have a higher oxidation potential than water; in fact, water is more easily oxidised. So the correct choice is (C).

Oxidation occurs at the anode and reduction at the cathode, in both electrolytic and galvanic cells.

Definition of electrodes is by the process occurring, not by sign:

• Anode: the electrode where oxidation (loss of electrons) takes place.
• Cathode: the electrode where reduction (gain of electrons) takes place. …

Electrolysis of MOLTEN NaCl (no water present) gives sodium metal at the cathode and chlorine gas at the anode - this is the Down's process used industrially to manufacture sodium metal.

In molten NaCl, the only ions present are Na+ and Cl- (there is no water to supply H+/OH-, unlike electrolysis of aqueous NaCl brine).

At the cathode (reduction):

Na+ + e- -> Na(s)

At the anode (oxidation):

2Cl- -> Cl2(g) + 2e-

Overall: 2NaCl(l) --electrolysis--> 2Na(s) + Cl2(g)

…

Electrolysis of aqueous NaCl (brine) gives H2 at the cathode, Cl2 at the anode, and leaves NaOH behind in solution — the industrial chlor-alkali process.

At the cathode, water is preferentially reduced over Na+ (Na+ would need a far more negative potential): 2H2O + 2e- → H2 + 2OH-.

…

If the opposing external voltage exceeds the cell's own emf, the reaction is driven backward — this is electrolysis, i.e. electrolytic-cell behaviour.

In a galvanic (electrochemical) cell, the spontaneous redox reaction generates a potential $E_{cell}$. If an external opposing potential $E_{ext}$ is applied:

• When $E_{ext} < E_{cell}$: the cell drives current in its normal direction (galvanic behaviour).
• When $E_{ext} = E_{cell}$: no current flows (equilibrium, used to measure emf via a potentiometer). …

Water (H+) is reduced more easily than Na+, so hydrogen gas is evolved at the cathode.

In aqueous NaCl the possible cathode reductions are Na+ + e- → Na (E° = -2.71 V) and 2H2O + 2e- → H2 + 2OH- (E° ≈ -0.83 V). The species with the higher (less negative) reduction potential is discharged, so water is reduced in preference to Na+:

2H2O + 2e- → H2(g) + 2OH-

…

During electrolysis, reduction always occurs at the cathode and oxidation always occurs at the anode.

In an electrolytic cell, the cathode is connected to the negative terminal of the external battery, so it is rich in electrons. Positively charged ions (cations) in the electrolyte migrate towards the cathode and gain electrons there, i.e. they are reduced (e.g. Cu2+ + 2e- -> Cu). Meanwhile at the anode, negatively charged ions (anions) lose electrons, i.e. oxidation occurs.

…

Water, not Mg²⁺, gets reduced at the cathode during electrolysis of aqueous MgSO₄, so the product is H₂ gas.

During electrolysis of an aqueous solution, the species that is reduced at the cathode is the one that is easiest to reduce, i.e. has the higher (less negative) standard reduction potential, among all species present (here Mg²⁺ and H₂O).

$Mg^{2+} + 2e^- \rightarrow Mg(s), \quad E° = -2.37\ V$

$2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq), \quad E° = -0.83\ V$

…

Anode → O₂, cathode → H₂ (water is discharged in preference to F⁻ and Na⁺).

Electrolysis of dilute aqueous NaF (Na⁺, F⁻, H₂O):

• Anode (oxidation): F⁻ → ½F₂ needs a huge potential (F₂/F⁻ = +2.87 V), whereas oxidation of water (2H₂O → O₂ + 4H⁺ + 4e⁻, ≈ +1.23 V) is far easier, so O₂ is released. …