Question

Given below are two statements :
1 M aqueous solution of each of $ Cu(NO_3)_2 $, $ AgNO_3 $, $ Hg_2(NO_3)_2 $; $ Mg(NO_3)_2 $ are electrolysed using inert electrodes, Given : $ E^0_{Ag^+/Ag} = 0.80V $, $ E^0_{Hg_2^{2+}/Hg} = 0.79V $, $ E^0_{Cu^{2+}/Cu} = 0.34V $ and $ E^0_{Mg^{2+}/Mg} = -2.37V $
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 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 :

• A. Both statement I and statement II are incorrect
• B. Statement I is correct but statement II is incorrect
• C. Both statement I and statement II are correct
• D. Statement I is incorrect but statement II is correct

Hint:

In electrolysis of aqueous solutions, the species with the higher reduction potential (for reduction at the cathode) or lower oxidation potential (for oxidation at the anode) will be preferentially discharged. Remember the reduction potentials of water to predict the products when metal ions with very negative reduction potentials are present.

Answer 1

The Correct Option is B

Step 1: Analyze Statement (I).
The order of deposition at the cathode is determined by the standard reduction potentials. Higher reduction potential means easier reduction and deposition. The order of reduction potentials is \( Ag^+ (0.80V)>Hg_2^{2+} (0.79V)>Cu^{2+} (0.34V)>Mg^{2+} (-2.37V) \). 
Thus, the sequence of deposition with increasing voltage will be Ag, then Hg, then Cu. Statement I is correct. 
Step 2: Analyze Statement (II).
For Magnesium: \( Mg^{2+} + 2e^- \rightarrow Mg \quad E^0 = -2.37V \) 
For water reduction at the cathode (neutral pH): \( 2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq) \quad E^0 = -0.83V \) Since the reduction potential of water is significantly higher than that of \( Mg^{2+} \), water will be reduced at the cathode, producing hydrogen gas, and Magnesium will not be deposited. 
The first part of Statement II is correct. For water oxidation at the anode: \( 2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^- \quad E^0 = +1.23V \)
Oxygen gas is evolved at the anode, not the cathode. 
The second part of Statement II is incorrect. 
Step 3: Determine the correctness of both statements.
Statement I is correct, and Statement II is incorrect. 
Step 4: Choose the appropriate option.
The option that states Statement I is correct but Statement II is incorrect is (2).