Passage of one ampere current through 0.1M `NI(NO_3)_2` solution using Ni electrodes brings in the concentration of solution to ....... in 60 seconds.

Options

A. 0.1M
B. 0.05M
C. 0.2M
D. 0.025M

Answer

To solve the problem of determining the concentration of the `Ni(NO_3)_2` solution after passing a 1 ampere current for 60 seconds, we can follow these steps:

### Step 1: Identify the Electrochemical Reaction
The electrochemical reaction occurring at the nickel electrodes is:
\[ \text{Ni}^{2+} + 2e^- \rightarrow \text{Ni} \]
This indicates that 2 moles of electrons are required to reduce 1 mole of nickel ions to solid nickel.

### Step 2: Calculate the Total Charge Passed
Using the formula:
\[ Q = I \times t \]
where:
- \( I \) is the current in amperes (1 A),
- \( t \) is the time in seconds (60 s).

Calculating the charge:
\[ Q = 1 \, \text{A} \times 60 \, \text{s} = 60 \, \text{C} \]

### Step 3: Calculate the Number of Moles of Electrons
Using Faraday's constant \( F \) (approximately 96500 C/mol), we can find the number of moles of electrons transferred:
\[ n_e = \frac{Q}{F} = \frac{60 \, \text{C}}{96500 \, \text{C/mol}} \approx 0.000621 \, \text{mol} \]

### Step 4: Determine the Moles of Nickel Deposited
Since 2 moles of electrons are required to deposit 1 mole of nickel, the moles of nickel deposited can be calculated as:
\[ n_{\text{Ni}} = \frac{n_e}{2} = \frac{0.000621 \, \text{mol}}{2} \approx 0.0003105 \, \text{mol} \]

### Step 5: Calculate the Initial Moles of Ni²⁺
The initial concentration of `Ni(NO_3)_2` is 0.1 M in a 1 L solution, so the initial moles of `Ni²⁺` are:
\[ n_{\text{initial}} = 0.1 \, \text{mol/L} \times 1 \, \text{L} = 0.1 \, \text{mol} \]

### Step 6: Calculate the Remaining Moles of Ni²⁺
After the deposition of nickel, the remaining moles of `Ni²⁺` in the solution will be:
\[ n_{\text{remaining}} = n_{\text{initial}} - n_{\text{Ni}} = 0.1 \, \text{mol} - 0.0003105 \, \text{mol} \approx 0.0996895 \, \text{mol} \]

### Step 7: Calculate the Final Concentration of Ni²⁺
The final concentration of `Ni²⁺` in the solution is:
\[ C_{\text{final}} = \frac{n_{\text{remaining}}}{\text{Volume}} = \frac{0.0996895 \, \text{mol}}{1 \, \text{L}} \approx 0.0997 \, \text{M} \]

### Final Answer
The concentration of the `Ni(NO_3)_2` solution after passing the current for 60 seconds is approximately **0.0997 M**.

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