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

To solve this question, we need to analyze the electrolysis of the given solutions and the respective standard electrode potentials (\( E^0 \)) of the ions involved.

1. Explanation of Statement I:
   The sequence of deposition of metals during electrolysis is determined by their standard reduction potentials. Metals with higher (more positive) standard reduction potentials will be deposited first.
   • From the data given:
     • \( E^0_{Ag^+/Ag} = 0.80 \, \text{V} \)
     • \( E^0_{Hg_2^{2+}/Hg} = 0.79 \, \text{V} \)
     • \( E^0_{Cu^{2+}/Cu} = 0.34 \, \text{V} \)
     • \( E^0_{Mg^{2+}/Mg} = -2.37 \, \text{V} \)
   With increasing voltage, the metals will be deposited in the order of their \( E^0 \) values:
   • First, Ag will be deposited since its \( E^0 \) value is the highest at 0.80 V.
   • Next, Hg will be deposited with an \( E^0 \) of 0.79 V.
   • Finally, Cu will be deposited, as its \( E^0 \) is 0.34 V.
   • Mg will not be deposited as its \( E^0 \) is much lower at -2.37 V.
   Therefore, Statement I is correct.
2. Explanation of Statement II:
   According to electrochemical series, magnesium has a very low standard reduction potential (\( E^0 = -2.37 \, \text{V} \)); hence, it is less likely to be reduced compared to water. Instead of magnesium being deposited at the cathode, hydrogen gas will usually be evolved due to water reduction:
   • \( 2H_2O + 2e^- \rightarrow H_2 + 2OH^- \quad (E^0 = -0.83 \, \text{V}) \)
   Therefore, based on standard potentials and typical electrolysis behavior, magnesium will remain in the solution, and hydrogen gas (not oxygen gas) will be produced at the cathode. Consequently, Statement II is incorrect.

The correct choice is: Statement I is correct but statement II is incorrect.

Answer 2

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).