A current of `965A` is passed for `1 s` through `1L` solution of `0.02N NiSO_(4)` using `Ni` electrodes. What is the new concentration of `NiSO_(4)` ?

Options

A. `0.01N`
B. `0.01M`
C. `0.003M`
D. `0.02M`

Explanation

To solve the problem, we need to determine the new concentration of NiSO₄ after passing a current through the solution. Let's break down the steps:

### Step 1: Calculate the total charge passed through the solution
The total charge (Q) passed through the solution can be calculated using the formula:
\[ Q = I \times t \]
where:
- \( I = 965 \, \text{A} \) (current)
- \( t = 1 \, \text{s} \) (time)

Substituting the values:
\[ Q = 965 \, \text{A} \times 1 \, \text{s} = 965 \, \text{C} \]

### Step 2: Calculate the number of equivalents of nickel deposited
The number of equivalents (n) can be calculated using Faraday's law:
\[ n = \frac{Q}{F} \]
where \( F = 96500 \, \text{C/mol} \) (Faraday's constant).

Substituting the values:
\[ n = \frac{965}{96500} = 0.01 \, \text{equivalents} \]

### Step 3: Calculate the number of moles of nickel deposited
The number of moles of nickel deposited can be calculated using the equivalence factor. Nickel (Ni) has a valency of 2, so:
\[ \text{Moles of Ni} = \frac{n}{\text{valency}} = \frac{0.01}{2} = 0.005 \, \text{moles} \]

### Step 4: Determine the initial concentration of NiSO₄
The initial concentration of NiSO₄ is given as 0.02 N. Since the normality (N) for NiSO₄ is equal to its molarity (M) multiplied by the valency (2):
\[ \text{Initial Molarity} = \frac{0.02 \, \text{N}}{2} = 0.01 \, \text{M} \]

### Step 5: Calculate the final concentration of NiSO₄
The volume of the solution is 1 L. The initial moles of NiSO₄ in the solution can be calculated as:
\[ \text{Initial moles of NiSO₄} = \text{Initial Molarity} \times \text{Volume} = 0.01 \, \text{M} \times 1 \, \text{L} = 0.01 \, \text{moles} \]

After the electrolysis, 0.005 moles of Ni²⁺ are deposited, which means that 0.005 moles of NiSO₄ are consumed. Therefore, the remaining moles of NiSO₄ in the solution will be:
\[ \text{Remaining moles of NiSO₄} = 0.01 - 0.005 = 0.005 \, \text{moles} \]

### Step 6: Calculate the new concentration of NiSO₄
The new concentration (C) of NiSO₄ can be calculated as:
\[ C = \frac{\text{Remaining moles of NiSO₄}}{\text{Volume}} = \frac{0.005 \, \text{moles}}{1 \, \text{L}} = 0.005 \, \text{M} \]

### Conclusion
The new concentration of NiSO₄ is:
\[ \text{Final Concentration} = 0.005 \, \text{M} \]

### Final Answer
The new concentration of NiSO₄ is **0.01 M**.