Given below are two statements : 1 M aqueous solutions of each of Cu(NO_3)_2 , AgNO_3 , Hg_2(NO_3)_2 , Mg(NO_3)_2 are electrolysed using inert electrodes. Given: E0 Ag^+ /Ag = 0.80 V, E0 Hg_2^2+ /Hg = 0.79 V, E0 Cu^2+ /Cu = 0.24 V and E0 Mg^2+ /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 :

Statement I is correct but Statement II is incorrect

Both Statement I and Statement II are incorrect

Statement I is incorrect but Statement II is correct

Both Statement I and Statement II are correct

Electrochemistry PYQ — JEE Main 2025 · 7 Apr · Shift 2 · Q17 with Solution

View written solutionFree

Correct answer: C

Rule for discharge at cathode during electrolysis

At the cathode, the species with higher reduction potential gets reduced more easily.

Given standard reduction potentials: $E^\circ(Ag^+/Ag)=0.80\text{ V}$ $E^\circ(Hg_2^{2+}/Hg)=0.79\text{ V}$ $E^\circ(Cu^{2+}/Cu)=0.24\text{ V}$ $E^\circ(Mg^{2+}/Mg)=-2.37\text{ V}$

Therefore, ease of reduction at cathode is: $Ag^+ > Hg_2^{2+} > Cu^{2+} > Mg^{2+}$
Checking Statement (I)

Statement (I): With increasing voltage, the sequence of deposition of metals on the cathode will be Ag, Hg and Cu.

Since higher reduction potential means earlier discharge, the order of deposition is indeed: $Ag \rightarrow Hg \rightarrow Cu$

So, Statement (I) is correct.
Checking Statement (II)

Statement (II): Magnesium will not be deposited at the cathode instead oxygen gas will be evolved at the cathode.

This statement is incorrect for two reasons:
- In aqueous solution, $Mg^{2+}$ is not deposited because water gets reduced more easily than $Mg^{2+}$ .
- But at the cathode, water gives hydrogen gas, not oxygen gas.
Cathode reaction in aqueous solution: $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$

Oxygen is evolved at the anode, not at the cathode.

Hence, Statement (II) is incorrect.
Final conclusion
- Statement (I): Correct
- Statement (II): Incorrect
Therefore, the correct option is: $\boxed{\text{C}}$

View written solutionFree

Correct answer: C

Rule for discharge at cathode during electrolysis

At the cathode, the species with higher reduction potential gets reduced more easily.

Given standard reduction potentials: $E^\circ(Ag^+/Ag)=0.80\text{ V}$ $E^\circ(Hg_2^{2+}/Hg)=0.79\text{ V}$ $E^\circ(Cu^{2+}/Cu)=0.24\text{ V}$ $E^\circ(Mg^{2+}/Mg)=-2.37\text{ V}$

Therefore, ease of reduction at cathode is: $Ag^+ > Hg_2^{2+} > Cu^{2+} > Mg^{2+}$
Checking Statement (I)

Statement (I): With increasing voltage, the sequence of deposition of metals on the cathode will be Ag, Hg and Cu.

Since higher reduction potential means earlier discharge, the order of deposition is indeed: $Ag \rightarrow Hg \rightarrow Cu$

So, Statement (I) is correct.
Checking Statement (II)

Statement (II): Magnesium will not be deposited at the cathode instead oxygen gas will be evolved at the cathode.

This statement is incorrect for two reasons:
- In aqueous solution, $Mg^{2+}$ is not deposited because water gets reduced more easily than $Mg^{2+}$ .
- But at the cathode, water gives hydrogen gas, not oxygen gas.
Cathode reaction in aqueous solution: $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$

Oxygen is evolved at the anode, not at the cathode.

Hence, Statement (II) is incorrect.
Final conclusion
- Statement (I): Correct
- Statement (II): Incorrect
Therefore, the correct option is: $\boxed{\text{C}}$