23) It took 116 hours to produce 603 g of metal X by performing electrolysis on molten XCI3 with a current of 2.00 A. Calculate the molar mass of X. a) 55.8 g/mol b) 72.6 g/mol c) 27.0 g/mol d) 204 g.mol e) 209 g/mol

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# Standard Reduction Potentials

The following table lists standard reduction potentials (E°) for various redox couples in volts (V). Standard reduction potentials are used to predict the direction of electron flow in electrochemical cells.

## Reduction Potential Table

| Couple          | E° (Volts) | Couple         | E° (Volts) |
|-----------------|------------|----------------|------------|
| F₂              | HF (H⁺)    | +3.03          | SO₄²⁻      | H₂SO₃ (H⁺) | +0.20          |
| F₂              | F⁻         | +2.87          | Sn⁴⁺       | Sn²⁺        | +0.15          |
| S₂O₈²⁻          | SO₄²⁻      | +2.05          | S          | H₂S (H⁺)    | +0.141         |
| BiO₃⁻           | Bi³⁺       | +2.0           | Hg₂Br₂     | Hg (Br⁻)    | +0.140         |
| H₂O₂            | H₂O (H⁺)   | +1.78          | AgBr       | Ag (Br⁻)    | +0.0713        |
| PbO₂            | PbSO₄ (H⁺, SO₄²⁻) | +1.685   | H⁺         | H₂          | +0.0000        |
| Ce⁴⁺            | Ce³⁺       | +1.61          | Pb²⁺       | Pb          | -0.126        |
| MnO₄⁻           | Mn²⁺ (H⁺)  | +1.491         | Sn²⁺       | Sn          | -0.136        |
| ClO₃⁻           | Cl⁻        | +1.47          | AgI        | Ag (I⁻)     | -0.152        |
| PbO₂            | Pb²⁺ (H⁺)  | +1.46          | Ni²⁺       | Ni          | -0.250        |
| Au³⁺            | Au         | +1.42          | Co²
Transcribed Image Text:# Standard Reduction Potentials The following table lists standard reduction potentials (E°) for various redox couples in volts (V). Standard reduction potentials are used to predict the direction of electron flow in electrochemical cells. ## Reduction Potential Table | Couple | E° (Volts) | Couple | E° (Volts) | |-----------------|------------|----------------|------------| | F₂ | HF (H⁺) | +3.03 | SO₄²⁻ | H₂SO₃ (H⁺) | +0.20 | | F₂ | F⁻ | +2.87 | Sn⁴⁺ | Sn²⁺ | +0.15 | | S₂O₈²⁻ | SO₄²⁻ | +2.05 | S | H₂S (H⁺) | +0.141 | | BiO₃⁻ | Bi³⁺ | +2.0 | Hg₂Br₂ | Hg (Br⁻) | +0.140 | | H₂O₂ | H₂O (H⁺) | +1.78 | AgBr | Ag (Br⁻) | +0.0713 | | PbO₂ | PbSO₄ (H⁺, SO₄²⁻) | +1.685 | H⁺ | H₂ | +0.0000 | | Ce⁴⁺ | Ce³⁺ | +1.61 | Pb²⁺ | Pb | -0.126 | | MnO₄⁻ | Mn²⁺ (H⁺) | +1.491 | Sn²⁺ | Sn | -0.136 | | ClO₃⁻ | Cl⁻ | +1.47 | AgI | Ag (I⁻) | -0.152 | | PbO₂ | Pb²⁺ (H⁺) | +1.46 | Ni²⁺ | Ni | -0.250 | | Au³⁺ | Au | +1.42 | Co²
**Electrolysis Calculation Example**

**Problem Statement:** 

It took 116 hours to produce 603 g of metal X by performing electrolysis on molten XCl₃ with a current of 2.00 A. Calculate the molar mass of X.

**Multiple Choice Answers:**
a) 55.8 g/mol  
b) 72.6 g/mol  
c) 27.0 g/mol  
d) 204 g/mol  
e) 209 g/mol  

**Solution Explanation:**

To solve this problem, follow these steps:

1. **Calculate Total Charge (Q):**
   Using the formula: \( Q = I \times t \)

   Where:
   - \( I = 2.00 \) A (Current)
   - \( t = 116 \) hours \( = 116 \times 3600 \) seconds (time converted to seconds)

   \[ Q = 2.00 \, \text{A} \times 116 \times 3600 \, \text{s} \]
   \[ Q = 835200 \, \text{Coulombs (C)} \]

2. **Calculate Moles of Electrons:**
   Using Faraday's constant (\( F = 96485 \) C/mol)

   \[ \text{Moles of electrons} = \frac{Q}{F} \]
   \[ \text{Moles of electrons} = \frac{835200 \, \text{C}}{96485 \, \text{C/mol}} \]
   \[ \text{Moles of electrons} \approx 8.65 \, \text{mol} \]

3. **Relate Moles of Electrons to Moles of Metal X:**
   From the formula XCl₃, each mole of metal X requires 3 moles of electrons (since X has a +3 charge).

   \[ \text{Moles of X} = \frac{\text{Moles of electrons}}{3} \]
   \[ \text{Moles of X} = \frac{8.65 \, \text{mol}}{3} \]
   \[ \text{Moles of X} \approx 2.88 \, \text{mol} \]

4. **Calculate Molar Mass of Metal X:**

   \[ \text{
Transcribed Image Text:**Electrolysis Calculation Example** **Problem Statement:** It took 116 hours to produce 603 g of metal X by performing electrolysis on molten XCl₃ with a current of 2.00 A. Calculate the molar mass of X. **Multiple Choice Answers:** a) 55.8 g/mol b) 72.6 g/mol c) 27.0 g/mol d) 204 g/mol e) 209 g/mol **Solution Explanation:** To solve this problem, follow these steps: 1. **Calculate Total Charge (Q):** Using the formula: \( Q = I \times t \) Where: - \( I = 2.00 \) A (Current) - \( t = 116 \) hours \( = 116 \times 3600 \) seconds (time converted to seconds) \[ Q = 2.00 \, \text{A} \times 116 \times 3600 \, \text{s} \] \[ Q = 835200 \, \text{Coulombs (C)} \] 2. **Calculate Moles of Electrons:** Using Faraday's constant (\( F = 96485 \) C/mol) \[ \text{Moles of electrons} = \frac{Q}{F} \] \[ \text{Moles of electrons} = \frac{835200 \, \text{C}}{96485 \, \text{C/mol}} \] \[ \text{Moles of electrons} \approx 8.65 \, \text{mol} \] 3. **Relate Moles of Electrons to Moles of Metal X:** From the formula XCl₃, each mole of metal X requires 3 moles of electrons (since X has a +3 charge). \[ \text{Moles of X} = \frac{\text{Moles of electrons}}{3} \] \[ \text{Moles of X} = \frac{8.65 \, \text{mol}}{3} \] \[ \text{Moles of X} \approx 2.88 \, \text{mol} \] 4. **Calculate Molar Mass of Metal X:** \[ \text{
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