To solve the problem, we need to find the molarity of \( Zn^{2+} \) ions after some zinc has been deposited from the \( ZnSO_4 \) solution. We will follow these steps:
### Step 1: Calculate the total charge passed through the solution
We can use the formula:
\[
Q = I \times t
\]
where:
- \( I = 0.965 \, \text{A} \) (current)
- \( t = 10 \, \text{minutes} = 10 \times 60 \, \text{seconds} = 600 \, \text{s} \)
Substituting the values:
\[
Q = 0.965 \, \text{A} \times 600 \, \text{s} = 579 \, \text{C}
\]
### Step 2: Calculate the number of moles of zinc deposited
Using Faraday's law, the number of moles of zinc deposited can be calculated using:
\[
n = \frac{Q}{F \times n}
\]
where:
- \( F = 96500 \, \text{C/mol} \) (Faraday's constant)
- \( n = 2 \) (number of electrons transferred per zinc ion)
Substituting the values:
\[
n = \frac{579 \, \text{C}}{96500 \, \text{C/mol} \times 2} = \frac{579}{193000} \approx 0.003 \, \text{moles}
\]
### Step 3: Calculate the initial number of moles of \( Zn^{2+} \)
The initial concentration of \( ZnSO_4 \) is \( 0.2 \, \text{M} \) and the volume is \( 500 \, \text{ml} = 0.5 \, \text{L} \). The initial number of moles is given by:
\[
\text{Initial moles} = \text{Molarity} \times \text{Volume} = 0.2 \, \text{mol/L} \times 0.5 \, \text{L} = 0.1 \, \text{moles}
\]
### Step 4: Calculate the remaining moles of \( Zn^{2+} \)
After the deposition of zinc, the remaining moles of \( Zn^{2+} \) will be:
\[
\text{Remaining moles} = \text{Initial moles} - \text{Deposited moles} = 0.1 \, \text{moles} - 0.003 \, \text{moles} = 0.097 \, \text{moles}
\]
### Step 5: Calculate the new molarity of \( Zn^{2+} \)
The new molarity can be calculated using the formula:
\[
\text{Molarity} = \frac{\text{Remaining moles}}{\text{Volume}} = \frac{0.097 \, \text{moles}}{0.5 \, \text{L}} = 0.194 \, \text{M}
\]
### Final Answer
The molarity of \( Zn^{2+} \) after the deposition of zinc is \( 0.194 \, \text{M} \).
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