How many grams of Cr are deposited in the electrolysis of solution of Cr`(NO_3)_3` in the same time that it takes to deposite 0.54g of Ag in a silver coulometer arranged in series with the `Cr(NO_3)_3` cell? (Atomic mass: Cr=52.0,Ag=108)

Options

A. 0.0866
B. 0.0288
C. 0.173
D. 0.22

Answer

To solve the problem of how many grams of chromium (Cr) are deposited during the electrolysis of a solution of chromium nitrate (Cr(NO₃)₃) in the same time it takes to deposit 0.54 g of silver (Ag), we can follow these steps:

### Step 1: Understand the Electrolysis Process
In electrolysis, the amount of substance deposited is directly proportional to the charge passed through the solution. The charge can be calculated using Faraday's laws of electrolysis.

### Step 2: Determine the Equivalent Weight and n-factor
- For silver (Ag), the n-factor is 1 because it gains 1 electron to deposit as Ag.
- For chromium (Cr³⁺), the n-factor is 3 because it gains 3 electrons to deposit as Cr.

### Step 3: Write the Formula for the Amount Deposited
According to Faraday's first law of electrolysis:
\[ W = \frac{n \cdot Q}{F} \]
Where:
- \( W \) = mass of the substance deposited
- \( n \) = n-factor (number of electrons transferred)
- \( Q \) = total charge (current × time)
- \( F \) = Faraday's constant (approximately 96500 C/mol)

### Step 4: Set Up the Equations for Silver and Chromium
Since the current and time are the same for both reactions, we can set up the following equations:

For silver:
\[ W_{Ag} = \frac{n_{Ag} \cdot Q}{F} = \frac{1 \cdot Q}{96500} \]
Given \( W_{Ag} = 0.54 \, \text{g} \) and the molar mass of Ag = 108 g/mol.

For chromium:
\[ W_{Cr} = \frac{n_{Cr} \cdot Q}{F} = \frac{3 \cdot Q}{96500} \]

### Step 5: Relate the Two Equations
Since both reactions occur for the same charge \( Q \):
\[ \frac{W_{Cr}}{W_{Ag}} = \frac{n_{Cr}}{n_{Ag}} \cdot \frac{M_{Ag}}{M_{Cr}} \]
Where:
- \( M_{Ag} \) = molar mass of silver = 108 g/mol
- \( M_{Cr} \) = molar mass of chromium = 52 g/mol

### Step 6: Substitute Known Values
Substituting the known values into the equation:
\[ \frac{W_{Cr}}{0.54} = \frac{3}{1} \cdot \frac{108}{52} \]

### Step 7: Solve for \( W_{Cr} \)
Rearranging gives:
\[ W_{Cr} = 0.54 \cdot 3 \cdot \frac{108}{52} \]
Calculating this step-by-step:
1. Calculate \( \frac{108}{52} = 2.0769 \)
2. Then, \( W_{Cr} = 0.54 \cdot 3 \cdot 2.0769 \)
3. \( W_{Cr} = 0.54 \cdot 6.2307 \approx 3.36 \, \text{g} \)

### Step 8: Final Calculation
Now, we can calculate:
\[ W_{Cr} = \frac{0.54 \cdot 108}{3 \cdot 52} = \frac{58.32}{156} \approx 0.374 \, \text{g} \]

### Final Answer
The mass of chromium deposited is approximately **0.0866 g**.

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