Dow's process of extraction of Mg involves extraction of Mg from sea water. Sea water is concentrated in sun-light and is then treated with slaked lime. Magnesium hydroxide is heated in a stream of HCl to give `MgCl_(2)` which is electrolysed to discharge Mg. The mixture is in the ratio 35% `MgCl_(2)+50% NaCl+15% CaCl_(2)`. NaCl and `CaCl_(2)` are added to lower the fusion temperature and to increases the conductance.
`Mg^(2+) +Ca(OH)_(2) to Mg(OH)_(2)+Ca^(2+)`
`Mg(OH)_(2)+2HCl to MgCl_(2)+2H_(2)O(l)`
Electrolysis of fused `MgCl_(2) hArr Mg^(2+)+2Cl`
`Mg^(2+)+2e^(-) to Mg`( At Cathode)
`2Cl^(-) to Cl_(2)+2e^(-)` (At Anode)
Mg electrolysed is protected from atmospheric oxidation by a blanket of inert gases.
Molten mixture contains `Mg^(2+), Na^(+) and Ca^(2+)` but at cathode only `Mg^(2+)` is discharged because :

Options

A. Standar reduction potential of `Mg^(2+)` is least among the three
B. Standard oxidation potential of Mg is the least among the three
C. Discharge potential of `Mg^(2+)` is highest
D. None of these

Answer

To solve the question regarding why only \( \text{Mg}^{2+} \) is discharged at the cathode during the electrolysis of the molten mixture containing \( \text{MgCl}_2 \), \( \text{NaCl} \), and \( \text{CaCl}_2 \), we can follow these steps:

### Step 1: Understand the Components of the Mixture
The molten mixture consists of:
- \( \text{Mg}^{2+} \) ions from \( \text{MgCl}_2 \)
- \( \text{Na}^+ \) ions from \( \text{NaCl} \)
- \( \text{Ca}^{2+} \) ions from \( \text{CaCl}_2 \)

### Step 2: Identify the Electrolysis Process
During electrolysis, cations migrate towards the cathode (negative electrode) where reduction occurs. The possible cations in the mixture are \( \text{Mg}^{2+} \), \( \text{Na}^+ \), and \( \text{Ca}^{2+} \).

### Step 3: Compare Reduction Potentials
The discharge of ions at the cathode depends on their reduction potentials. The reduction potential is a measure of the tendency of a species to gain electrons and be reduced. The higher the reduction potential, the more likely the ion will be reduced.

- **Standard Reduction Potentials** (approximate values):
- \( \text{Mg}^{2+} + 2e^- \rightarrow \text{Mg} \) has a reduction potential of about -2.37 V.
- \( \text{Na}^+ + e^- \rightarrow \text{Na} \) has a reduction potential of about -2.71 V.
- \( \text{Ca}^{2+} + 2e^- \rightarrow \text{Ca} \) has a reduction potential of about -2.87 V.

### Step 4: Determine Which Ion is Discharged
Since \( \text{Mg}^{2+} \) has the highest (least negative) reduction potential among the three ions, it is the most favorable for reduction. Therefore, during electrolysis, \( \text{Mg}^{2+} \) will gain electrons and be discharged as magnesium metal at the cathode.

### Step 5: Conclusion
Only \( \text{Mg}^{2+} \) is discharged at the cathode because it has the highest reduction potential compared to \( \text{Na}^+ \) and \( \text{Ca}^{2+} \). This means it is more likely to gain electrons and be reduced to magnesium metal.

### Summary of the Solution
1. The molten mixture contains \( \text{Mg}^{2+} \), \( \text{Na}^+ \), and \( \text{Ca}^{2+} \).
2. During electrolysis, cations migrate to the cathode for reduction.
3. The reduction potentials indicate that \( \text{Mg}^{2+} \) is more favorable for reduction than \( \text{Na}^+ \) and \( \text{Ca}^{2+} \).
4. Thus, only \( \text{Mg}^{2+} \) is discharged at the cathode.