Given below are two statements :
1 M aqueous solutions of each of Cu(NO₃)₂, AgNO₃, Hg₂(NO₃)₂, Mg(NO₃)₂ are electrolysed using inert electrodes. Given: E ⁰ Ag⁺/Ag = 0.80 V, E ⁰ Hg₂²⁺/Hg = 0.79 V, E ⁰ Cu²⁺/Cu = 0.24 V and E ⁰ Mg²⁺/Mg = -2.37 V.
Statement (I) : With increasing voltage, the sequence of deposition of metals on the cathode will be Ag, Hg and Cu.
Statement (II) : Magnesium will not be deposited at the cathode instead oxygen gas will be evolved at the cathode.
In the light of the above statements, choose the most appropriate answer from the options given below :

Solution

Chemistry ma’am · 4 steps

The ion with the highest reduction potential is reduced first, and water always beats Mg²⁺ at the cathode.

1. Order at the cathode. The cathode is where reduction happens. As the applied voltage rises, the ion with the highest (most positive) E° is reduced first, because it accepts electrons most readily.

   Species	E° (V)	At the cathode
   Ag⁺	+0.80	deposits 1st
   Hg₂²⁺	+0.79	deposits 2nd
   Cu²⁺	+0.24	deposits 3rd
   H₂O (to H₂)	-0.83	H₂ gas evolves
   Mg²⁺	-2.37	never deposits

2. Statement I. The order Ag, then Hg, then Cu follows the falling E° values exactly. Statement I is correct.
3. Statement II. Water is far easier to reduce than Mg²⁺, so magnesium is not deposited. But reduction of water gives 2H₂O + 2e⁻ → H₂ + 2OH⁻, so the cathode gas is hydrogen. Oxygen comes from oxidation of water, which happens only at the anode. Statement II is incorrect.
4. So. Statement I is correct and Statement II is incorrect, which is option (c).

The idea to keep. Cathode means reduction, so the most positive E° goes first. For very active metals like Mg, Na or Al, water is reduced to H₂ instead. O₂ only ever forms at the anode.

Answer from the NTA final answer key. Solution by PrepWiser.