To solve the problem of determining the concentration of the `Ni(NO_3)_2` solution after passing a 1 ampere current through it for 60 seconds, we can follow these steps:
### Step 1: Write the Cell Reaction
The cell reaction for the deposition of nickel from the solution can be written as:
\[ \text{Ni}^{2+} + 2 \text{e}^- \rightarrow \text{Ni} \]
This indicates that 1 mole of nickel ions requires 2 moles of electrons for deposition.
### Step 2: Calculate the Total Charge Passed
Using the formula for charge:
\[ Q = I \times t \]
Where:
- \( I = 1 \, \text{A} \) (current)
- \( t = 60 \, \text{s} \)
Calculating the charge:
\[ Q = 1 \, \text{A} \times 60 \, \text{s} = 60 \, \text{C} \]
### Step 3: Calculate the Amount of Nickel Deposited
The amount of nickel deposited can be calculated using the formula:
\[ W = \frac{E \times Q}{F} \]
Where:
- \( E \) is the equivalent mass of nickel.
- \( F \) is Faraday's constant (approximately \( 96500 \, \text{C/mol} \)).
The equivalent mass \( E \) of nickel can be calculated as:
\[ E = \frac{\text{Molar mass of Ni}}{n} = \frac{58.69 \, \text{g/mol}}{2} = 29.345 \, \text{g/mol} \]
Now substituting the values:
\[ W = \frac{29.345 \, \text{g/mol} \times 60 \, \text{C}}{96500 \, \text{C/mol}} \]
Calculating \( W \):
\[ W = \frac{1760.7 \, \text{g}}{96500} \approx 0.0182 \, \text{g} \]
### Step 4: Calculate the Moles of Nickel Deposited
Now, we can calculate the number of moles of nickel deposited:
\[ \text{Moles of Ni} = \frac{W}{\text{Molar mass of Ni}} = \frac{0.0182 \, \text{g}}{58.69 \, \text{g/mol}} \approx 3.1 \times 10^{-4} \, \text{mol} \]
### Step 5: Calculate the Remaining Moles of Nickel in Solution
Initially, we had a concentration of \( 0.1 \, \text{M} \) in 1 liter, which means:
\[ \text{Initial moles of Ni}^{2+} = 0.1 \, \text{mol} \]
After deposition, the remaining moles of nickel in solution:
\[ \text{Remaining moles} = 0.1 \, \text{mol} - 3.1 \times 10^{-4} \, \text{mol} \approx 0.09969 \, \text{mol} \]
### Step 6: Calculate the Final Concentration
The final concentration of nickel ions in the solution after 60 seconds is:
\[ \text{Final concentration} = \frac{\text{Remaining moles}}{\text{Volume}} = \frac{0.09969 \, \text{mol}}{1 \, \text{L}} \approx 0.09969 \, \text{M} \]
Since this is very close to the initial concentration, we can conclude that the final concentration remains approximately \( 0.1 \, \text{M} \).
### Final Answer
The concentration of the `Ni(NO_3)_2` solution after passing a 1 ampere current for 60 seconds is approximately **0.1 M**.
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