One litre 1 M `CuSO_(4)` solution is electrolysed. After passing 2 F of electricity, the molarity of solution will be

Options

A. `M//3`
B. `M//2`
C. `M//4`
D. 0

Explanation

To solve the problem of determining the molarity of a 1 M CuSO₄ solution after passing 2 Faraday of electricity, we can follow these steps:

### Step 1: Understand the Electrolysis Process
When electrolysis occurs in a CuSO₄ solution, copper ions (Cu²⁺) are reduced at the cathode, gaining electrons to form solid copper (Cu). The reaction can be represented as:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu (s)} \]

### Step 2: Calculate the Amount of Copper Deposited
According to Faraday's laws of electrolysis, 1 Faraday (F) of charge is equivalent to the charge required to deposit 1 mole of monovalent ions or 0.5 moles of divalent ions. Since Cu²⁺ is a divalent ion, passing 2 Faraday of electricity will deposit:
\[ \text{Moles of Cu deposited} = \frac{2 \, \text{F}}{2 \, \text{F/mol}} = 1 \, \text{mol} \]

### Step 3: Determine Initial Amount of Cu²⁺
In a 1 L solution of 1 M CuSO₄, the initial amount of Cu²⁺ is:
\[ \text{Initial moles of Cu}^{2+} = 1 \, \text{mol} \]

### Step 4: Calculate Remaining Cu²⁺ After Electrolysis
Since 1 mole of Cu²⁺ is reduced to solid copper, the amount of Cu²⁺ left in the solution after electrolysis is:
\[ \text{Remaining moles of Cu}^{2+} = 1 \, \text{mol} - 1 \, \text{mol} = 0 \, \text{mol} \]

### Step 5: Calculate the Molarity of the Solution
Molarity (M) is defined as the number of moles of solute per liter of solution. Since there are 0 moles of Cu²⁺ left in the solution and the volume is still 1 liter, the molarity is:
\[ \text{Molarity} = \frac{\text{Moles of Cu}^{2+}}{\text{Volume of solution in L}} = \frac{0 \, \text{mol}}{1 \, \text{L}} = 0 \, \text{M} \]

### Final Answer
The molarity of the CuSO₄ solution after passing 2 Faraday of electricity is **0 M**.

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