A disposable galvanic cell `Zn|Zn^(2+)||Sn^(2+)|SN` is produced using `1.0 mL` of `0.5 M Zn(NO_(3))_(2)` and `1.0 mL` of `0.50 M Sn(NO_(3))_(2)`. It is needed to power a pace maker that draws a constant current of `10^(-6)` Amp to run it and requires atleast `0.50 V` to function. Calculate the value of `|Zn^(+2)|` when cell reaches `0.5V` at `298K`.
(Given: `E^(@) (Zn^(2+)//Zn) =- 0.76V: E^(@) (SN^(2+)//Sn) =- 0.14V)`.

Explanation

To solve the problem, we need to calculate the concentration of \( \text{Zn}^{2+} \) when the cell reaches a voltage of \( 0.5 \, \text{V} \) at \( 298 \, \text{K} \). We will use the Nernst equation and the given standard reduction potentials.

### Step 1: Identify the half-reactions and standard potentials
The cell consists of two half-reactions:
- Anode (oxidation): \( \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \) with \( E^\circ = -0.76 \, \text{V} \)
- Cathode (reduction): \( \text{Sn}^{2+} + 2e^- \rightarrow \text{Sn} \) with \( E^\circ = -0.14 \, \text{V} \)

### Step 2: Calculate the standard cell potential \( E^\circ_{\text{cell}} \)
Using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Substituting the values:
\[ E^\circ_{\text{cell}} = (-0.14) - (-0.76) = 0.62 \, \text{V} \]

### Step 3: Write the Nernst equation
The Nernst equation is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \left( \frac{[\text{Zn}^{2+}]}{[\text{Sn}^{2+}]} \right) \]
Where \( n \) is the number of moles of electrons transferred (which is 2 in this case).

### Step 4: Set up the equation for the cell at \( 0.5 \, \text{V} \)
We know:
- \( E_{\text{cell}} = 0.5 \, \text{V} \)
- \( E^\circ_{\text{cell}} = 0.62 \, \text{V} \)
- The initial concentrations are \( [\text{Zn}^{2+}] = 0.5 \, \text{M} \) and \( [\text{Sn}^{2+}] = 0.5 \, \text{M} \).

Substituting into the Nernst equation:
\[ 0.5 = 0.62 - \frac{0.0591}{2} \log \left( \frac{[\text{Zn}^{2+}]}{0.5} \right) \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ 0.5 - 0.62 = -0.02955 \log \left( \frac{[\text{Zn}^{2+}]}{0.5} \right) \]
\[ -0.12 = -0.02955 \log \left( \frac{[\text{Zn}^{2+}]}{0.5} \right) \]

### Step 6: Solve for the logarithm
Dividing both sides by \(-0.02955\):
\[ \log \left( \frac{[\text{Zn}^{2+}]}{0.5} \right) = \frac{-0.12}{-0.02955} \approx 4.06 \]

### Step 7: Exponentiate to find \( [\text{Zn}^{2+}] \)
Taking the antilogarithm:
\[ \frac{[\text{Zn}^{2+}]}{0.5} = 10^{4.06} \]
\[ [\text{Zn}^{2+}] = 0.5 \times 10^{4.06} \approx 0.5 \times 11513.56 \approx 5756.78 \, \text{M} \]

### Conclusion
The concentration of \( \text{Zn}^{2+} \) when the cell reaches \( 0.5 \, \text{V} \) is approximately \( 5756.78 \, \text{M} \).

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