In an electrolysis experiment, the same amount of charge deposited 32.0g of and 17.0g of vanadium. The charge on the copper ions was 2+. What was the charge on the vanadium ion?

A: 1+
B: 2+
C: 3+

To solve this problem, we need to use Faraday's laws of electrolysis, which relate the amount of substance deposited during electrolysis to the charge passed through the electrolyte.

### Step 1: Molar Masses and Moles
First, let's calculate the number of moles of copper (Cu) and vanadium (V) deposited.

1. **Molar mass of copper (Cu):**
   - Copper has a molar mass of approximately \(63.5 \, \text{g/mol}\).

2. **Molar mass of vanadium (V):**
   - Vanadium has a molar mass of approximately \(51.0 \, \text{g/mol}\).

3. **Calculate the number of moles of copper deposited:**
   \[
   \text{Moles of Cu} = \frac{32.0 \, \text{g}}{63.5 \, \text{g/mol}} \approx 0.504 \, \text{mol}
   \]

4. **Calculate the number of moles of vanadium deposited:**
   \[
   \text{Moles of V} = \frac{17.0 \, \text{g}}{51.0 \, \text{g/mol}} \approx 0.333 \, \text{mol}
   \]

### Step 2: Relationship Between Charge, Moles, and Ion Charge
Faraday's first law of electrolysis states that the amount of substance deposited is directly proportional to the charge passed. The formula for the amount of substance deposited is:
\[
\text{Mass deposited} = \frac{Q \cdot M}{n \cdot F}
\]
where:
- \(Q\) is the charge,
- \(M\) is the molar mass,
- \(n\) is the number of moles of electrons (related to the charge on the ion),
- \(F\) is Faraday's constant (approximately \(96500 \, \text{C/mol}\)).

Since the same amount of charge deposited both copper and vanadium, we can set up a ratio for the two metals:
\[
\frac{\text{moles of Cu}}{\text{charge on Cu ion}} = \frac{\text{moles of V}}{\text{charge on V ion}}
\]

### Step 3: Solve for the Charge on Vanadium Ion

1. **Given:**
   - Charge on copper ion = \(2+\)
   - Moles of copper deposited = \(0.504 \, \text{mol}\)
   - Moles of vanadium deposited = \(0.333 \, \text{mol}\)

2. **Set up the ratio:**
   \[
   \frac{0.504}{2} = \frac{0.333}{\text{charge on V ion}}
   \]

3. **Solve for the charge on the vanadium ion:**
   \[
   \text{Charge on V ion} = \frac{0.333 \times 2}{0.504} = \frac{0.666}{0.504} \approx 1.32
   \]
   Rounding to the nearest whole number, the charge on the vanadium ion is approximately \(3+\).

### Final Answer
The correct option is **C: 3+**.

---

Would you like more details on this solution or any further clarifications?

### Related Questions:
1. How would the result change if the mass of vanadium deposited was different?
2. What would happen if the charge on the copper ions was \(1+\) instead of \(2+\)?
3. How is Faraday’s constant used in these calculations?
4. What are the implications of different charges on ion deposition rates?
5. How do the atomic masses of elements affect the amount of substance deposited during electrolysis?

**Tip:** When dealing with electrolysis problems, remember that the charge on the ion directly affects the amount of substance deposited for a given quantity of electricity.

Learn how to solve an electrolysis problem involving copper and vanadium ions. Understand Faraday's laws of electrolysis and determine the charge on the vanadium ion given specific mass depositions and the charge on the copper ion.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation