Consider the reduction of Sn^{4+}  to Sn^{1+} at a constant rate of 5.846 mmol /h.  How much current passes through the solution?
Now imagine a circuit in which a battery generates a potential difference of 2.80 V across a battery that has 200  Ω of resistance. How much power is delivered by the battery? Assume no other resistance across the circuit

To calculate the current for the reduction of Sn\(^{4+}\) to Sn\(^{1+}\) and the power delivered by the battery, we can use the following formulas.

### 1. Current from the reduction reaction

The reduction of Sn\(^{4+}\) to Sn\(^{1+}\) involves a transfer of 3 electrons:

\[
\text{Sn}^{4+} + 3e^- \rightarrow \text{Sn}^{1+}
\]

To find the current, we can use Faraday's law of electrolysis:

\[
I = \frac{nF}{t}
\]

Where:
- \(I\) = current (A)
- \(n\) = number of moles of electrons transferred
- \(F\) = Faraday's constant (\(96485 \, \text{C/mol}\))
- \(t\) = time (in seconds)

### Given:
- Reduction rate = 5.846 mmol/h = \(5.846 \times 10^{-3} \, \text{mol/h}\)
- Convert to moles of electrons:
  \[
  n = 5.846 \times 10^{-3} \, \text{mol/h} \times 3 = 1.754 \times 10^{-2} \, \text{mol/h}
  \]
- Convert hours to seconds:
  \[
  t = 1 \, \text{h} = 3600 \, \text{s}
  \]
  
Substituting values into the equation:

\[
I = \frac{1.754 \times 10^{-2} \, \text{mol}}{3600 \, \text{s}} \times 96485 \, \text{C/mol}
\]

Calculating this:

\[
I \approx \frac{1.754 \times 10^{-2} \times 96485}{3600} \approx 0.468 \, \text{A}
\]

### 2. Power delivered by the battery

The power \(P\) delivered by the battery can be calculated using the formula:

\[
P = \frac{V^2}{R}
\]

Where:
- \(V\) = voltage (V)
- \(R\) = resistance (Ω)

### Given:
- \(V = 2.80 \, \text{V}\)
- \(R = 200 \, \Omega\)

Substituting the values:

\[
P = \frac{(2.80)^2}{200}
\]

Calculating this:

\[
P = \frac{7.84}{200} = 0.0392 \, \text{W} \, \text{or} \, 39.2 \, \text{mW}
\]

### Summary of Results
- The current passing through the solution is approximately **0.468 A**.
- The power delivered by the battery is approximately **39.2 mW**.

Would you like more details on these calculations or have any specific questions? Here are some related questions you might consider:

1. What other reactions could occur during the reduction of Sn\(^{4+}\)?
2. How does temperature affect the electrochemical processes?
3. What is the significance of Faraday's law in electrochemistry?
4. How do changes in resistance affect the current in a circuit?
5. What are some practical applications of Sn\(^{4+}\) reduction in industry?

**Tip:** Understanding the relationship between voltage, current, and resistance (Ohm's Law) can help you analyze various electrical circuits effectively.

Consider the reduction of Sn^{4+} to Sn^{1+} at a constant rate of 5.846 mmol/h. How much current passes through the solution? Now imagine a circuit in which a battery generates a potential difference of 2.80 V across a battery that has 200 Ω of resistance. How much power is delivered by the battery? Assume no other resistance across the circuit.

Explore how to calculate the current from the reduction of Sn^{4+} to Sn^{1+} at a rate of 5.846 mmol/h and the power delivered by a battery with a potential difference of 2.80 V and resistance of 200 Ω. Suitable for advanced high school students.

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