How many hours are required to produce 795 grams of lead from molten lead(II) chloride (PbCl2, 278.1 g/mol) if the current used is 116 amps? Faraday's constant is 96,485 C/mol e-. Write your answer with 3 significant digits.

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### Electrolysis Time Calculation

**Problem Statement:**

How many hours are required to produce 795 grams of lead from molten lead(II) chloride (PbCl₂, 278.1 g/mol) if the current used is 116 amps? Faraday's constant is 96,485 C/mole e⁻.

Write your answer with 3 significant digits.

**Explanation:**

This problem is related to the electrochemical production of lead via electrolysis. The following steps outline how to approach this calculation:

1. **Determine moles of lead (Pb) to be produced:**

   Given mass of Pb: 795 grams  
   Molar mass of Pb: 207.2 g/mol

   Moles of Pb (n):
   \[
   n = \frac{\text{Mass of Pb}}{\text{Molar mass of Pb}} = \frac{795 \text{ g}}{207.2 \text{ g/mol}} = 3.84 \text{ mol}
   \]

2. **Calculate the amount of charge (Q) required:**

   The reduction reaction at the cathode is:
   \[
   \text{Pb}^{2+} + 2e^- \rightarrow \text{Pb}
   \]
   Thus, 2 moles of electrons are required to produce 1 mole of Pb.

   Therefore, total moles of electrons required (n_e):
   \[
   n_e = 2 \times n = 2 \times 3.84 = 7.68 \text{ mol e}^-
   \]

   Using Faraday's constant (F):
   \[
   F = 96485 \text{ C/mol e}^-
   \]

   Total charge (Q) required:
   \[
   Q = n_e \times F = 7.68 \times 96485 \text{ C} = 741,821 \text{ C}
   \]

3. **Calculate the time (t) required:**

   Using the relationship:
   \[
   Q = I \times t
   \]
   Where, I = current in amps, t = time in seconds.

   Given current (I):
   \[
   I = 116 \text{ A}
   \]

   Rearrange to solve for time (t):
   \[
   t = \frac{Q}{I} = \
Transcribed Image Text:### Electrolysis Time Calculation **Problem Statement:** How many hours are required to produce 795 grams of lead from molten lead(II) chloride (PbCl₂, 278.1 g/mol) if the current used is 116 amps? Faraday's constant is 96,485 C/mole e⁻. Write your answer with 3 significant digits. **Explanation:** This problem is related to the electrochemical production of lead via electrolysis. The following steps outline how to approach this calculation: 1. **Determine moles of lead (Pb) to be produced:** Given mass of Pb: 795 grams Molar mass of Pb: 207.2 g/mol Moles of Pb (n): \[ n = \frac{\text{Mass of Pb}}{\text{Molar mass of Pb}} = \frac{795 \text{ g}}{207.2 \text{ g/mol}} = 3.84 \text{ mol} \] 2. **Calculate the amount of charge (Q) required:** The reduction reaction at the cathode is: \[ \text{Pb}^{2+} + 2e^- \rightarrow \text{Pb} \] Thus, 2 moles of electrons are required to produce 1 mole of Pb. Therefore, total moles of electrons required (n_e): \[ n_e = 2 \times n = 2 \times 3.84 = 7.68 \text{ mol e}^- \] Using Faraday's constant (F): \[ F = 96485 \text{ C/mol e}^- \] Total charge (Q) required: \[ Q = n_e \times F = 7.68 \times 96485 \text{ C} = 741,821 \text{ C} \] 3. **Calculate the time (t) required:** Using the relationship: \[ Q = I \times t \] Where, I = current in amps, t = time in seconds. Given current (I): \[ I = 116 \text{ A} \] Rearrange to solve for time (t): \[ t = \frac{Q}{I} = \
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