A current of `3.6A` is a passed for 6 hrs between Pt electrodes in `0.5L` of `2M` solution of `Ni(NO_(3))_(2)`. What will be the molarity of solution at the end of electrolysis?

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the initial moles of Ni(NO₃)₂
Given:
- Molarity (C) = 2 M
- Volume (V) = 0.5 L

Using the formula:
\[ \text{Moles} = \text{Molarity} \times \text{Volume} \]
\[ \text{Moles of Ni(NO}_3\text{)}_2 = 2 \, \text{mol/L} \times 0.5 \, \text{L} = 1 \, \text{mol} \]

### Step 2: Calculate the total charge passed during electrolysis
Given:
- Current (I) = 3.6 A
- Time (t) = 6 hours = 6 × 60 × 60 seconds = 21600 seconds

Using the formula:
\[ Q = I \times t \]
\[ Q = 3.6 \, \text{A} \times 21600 \, \text{s} = 77760 \, \text{C} \]

### Step 3: Determine the moles of nickel deposited
The reaction for nickel ions is:
\[ \text{Ni}^{2+} + 2e^- \rightarrow \text{Ni} \]
This means that 2 moles of electrons are needed to deposit 1 mole of nickel.

Using Faraday's constant (F = 96500 C/mol), we can calculate the moles of nickel deposited (n):
\[ n = \frac{Q}{nF} \]
Where n = 2 (number of electrons transferred per nickel ion):
\[ n = \frac{77760 \, \text{C}}{2 \times 96500 \, \text{C/mol}} = \frac{77760}{193000} \approx 0.403 \, \text{mol} \]

### Step 4: Calculate the remaining moles of Ni(NO₃)₂
Initial moles of Ni(NO₃)₂ = 1 mol
Moles of Ni deposited = 0.403 mol

Remaining moles of Ni(NO₃)₂:
\[ \text{Remaining moles} = 1 \, \text{mol} - 0.403 \, \text{mol} = 0.597 \, \text{mol} \]

### Step 5: Calculate the final molarity of the solution
Using the formula for molarity:
\[ \text{Molarity} = \frac{\text{moles}}{\text{volume in L}} \]
\[ \text{Final Molarity} = \frac{0.597 \, \text{mol}}{0.5 \, \text{L}} = 1.194 \, \text{M} \]

### Final Answer:
The molarity of the solution at the end of electrolysis is approximately **1.194 M**.

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