A current of 3.7 A is passed for 6hrs. Between Ni electrodes in 0.5 L of 2 M solution of `Ni(NO_(3))_(2)`. What will be the molarity of solution at the end of electrolysis?

Explanation

To solve the problem, we need to determine the molarity of the nickel nitrate solution after passing a current through it for a specified time. Here’s a step-by-step solution:

### Step 1: Calculate the total charge passed during electrolysis
The total charge (Q) can be calculated using the formula:
\[ Q = I \times t \]
where:
- \( I \) is the current in amperes (A)
- \( t \) is the time in seconds (s)

Given:
- \( I = 3.7 \, \text{A} \)
- \( t = 6 \, \text{hours} = 6 \times 3600 \, \text{s} = 21600 \, \text{s} \)

Calculating \( Q \):
\[ Q = 3.7 \, \text{A} \times 21600 \, \text{s} = 79920 \, \text{C} \]

### Step 2: Determine the number of moles of Nickel deposited
The electrolysis of nickel nitrate involves the reduction of nickel ions (Ni²⁺) to nickel metal (Ni). The half-reaction at the cathode is:
\[ \text{Ni}^{2+} + 2e^- \rightarrow \text{Ni} \]

From this reaction, we see that 2 moles of electrons are required to deposit 1 mole of nickel.

Using Faraday's law of electrolysis:
\[ n = \frac{Q}{F \cdot z} \]
where:
- \( n \) is the number of moles of substance deposited
- \( F \) is Faraday's constant (approximately \( 96500 \, \text{C/mol} \))
- \( z \) is the number of electrons transferred per mole of substance (for Ni, \( z = 2 \))

Calculating the number of moles of nickel deposited:
\[ n = \frac{79920 \, \text{C}}{96500 \, \text{C/mol} \cdot 2} \]
\[ n = \frac{79920}{193000} \approx 0.414 \, \text{mol} \]

### Step 3: Calculate the initial moles of Ni²⁺ in the solution
The initial molarity of the nickel nitrate solution is given as 2 M, and the volume of the solution is 0.5 L.

Calculating the initial moles of Ni²⁺:
\[ \text{Initial moles of Ni}^{2+} = \text{Molarity} \times \text{Volume} = 2 \, \text{mol/L} \times 0.5 \, \text{L} = 1 \, \text{mol} \]

### Step 4: Determine the change in moles of Ni²⁺
During electrolysis, 0.414 moles of Ni²⁺ are reduced to form nickel metal. Therefore, the moles of Ni²⁺ remaining in the solution will be:
\[ \text{Remaining moles of Ni}^{2+} = \text{Initial moles} - \text{Moles deposited} \]
\[ \text{Remaining moles of Ni}^{2+} = 1 \, \text{mol} - 0.414 \, \text{mol} = 0.586 \, \text{mol} \]

### Step 5: Calculate the final molarity of the solution
The final molarity can be calculated using the remaining moles of Ni²⁺ and the volume of the solution:
\[ \text{Final Molarity} = \frac{\text{Remaining moles of Ni}^{2+}}{\text{Volume}} \]
\[ \text{Final Molarity} = \frac{0.586 \, \text{mol}}{0.5 \, \text{L}} = 1.172 \, \text{M} \]

### Conclusion
The molarity of the nickel nitrate solution at the end of electrolysis is approximately **1.172 M**.