To solve the problem step by step, we will follow these steps:
### Step 1: Calculate the total charge (Q) passed through the solution
We know the formula for charge is given by:
\[ Q = I \times t \]
Where:
- \( I \) = current in amperes (A)
- \( t \) = time in seconds (s)
Given:
- \( I = 0.965 \, \text{A} \)
- \( t = 10 \, \text{minutes} = 10 \times 60 = 600 \, \text{s} \)
Calculating the charge:
\[ Q = 0.965 \, \text{A} \times 600 \, \text{s} = 579 \, \text{C} \]
### Step 2: Determine the amount of zinc deposited
The reaction for the deposition of zinc can be represented as:
\[ \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn} \]
From this reaction, we see that 2 moles of electrons (2 Faradays) are required to deposit 1 mole of zinc.
The total charge required to deposit 1 mole of zinc can be calculated using Faraday's constant:
\[ 1 \, \text{mol of e}^- = 96500 \, \text{C} \]
Thus, for 1 mole of zinc:
\[ Q = 2 \times 96500 \, \text{C} = 193000 \, \text{C} \]
Now, we can find the number of moles of zinc deposited:
\[ \text{Moles of Zn deposited} = \frac{Q \text{ (passed)}}{Q \text{ (required for 1 mole of Zn)}} = \frac{579 \, \text{C}}{193000 \, \text{C/mol}} \approx 0.003 \, \text{mol} \]
### Step 3: Calculate the initial moles of Zn²⁺ in the solution
The initial concentration of \( \text{ZnSO}_4 \) is given as 0.2 M in a volume of 500 mL.
Calculating the initial moles of \( \text{Zn}^{2+} \):
\[ \text{Moles of Zn}^{2+} = \text{Molarity} \times \text{Volume (L)} = 0.2 \, \text{mol/L} \times 0.5 \, \text{L} = 0.1 \, \text{mol} \]
### Step 4: Calculate the remaining moles of Zn²⁺ after deposition
The moles of \( \text{Zn}^{2+} \) remaining after deposition:
\[ \text{Remaining moles of Zn}^{2+} = \text{Initial moles} - \text{Moles deposited} = 0.1 \, \text{mol} - 0.003 \, \text{mol} = 0.097 \, \text{mol} \]
### Step 5: Calculate the new molarity of Zn²⁺ in the solution
The total volume of the solution remains 500 mL (or 0.5 L). Therefore, the new molarity of \( \text{Zn}^{2+} \) is:
\[ \text{Molarity of Zn}^{2+} = \frac{\text{Remaining moles}}{\text{Volume (L)}} = \frac{0.097 \, \text{mol}}{0.5 \, \text{L}} = 0.194 \, \text{M} \]
### Final Answer
The molarity of \( \text{Zn}^{2+} \) after deposition of zinc is approximately **0.194 M**.
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