A 250.0 mL sample of a 0.20M `Cr^(3+)` is electrolysed with a current of 96.5 A. If the remaining `[Cr^(3+)]` is 0.1 M, the duration of process is:

Options

A. 25sec
B. 225sec
C. 150sec
D. 75sec

Answer

To solve the problem, we need to determine the duration of the electrolysis process given the initial and remaining concentrations of `Cr^(3+)`, the current used, and the volume of the solution. Here's a step-by-step breakdown of the solution:

### Step 1: Calculate the initial number of moles of `Cr^(3+)`
The initial concentration of `Cr^(3+)` is given as 0.20 M and the volume is 250 mL.

\[ \text{Number of moles} = \text{Concentration} \times \text{Volume} \]

Convert the volume from mL to L:

\[ \text{Volume} = 250 \, \text{mL} = 250 \times 10^{-3} \, \text{L} = 0.250 \, \text{L} \]

Now calculate the number of moles:

\[ \text{Number of moles} = 0.20 \, \text{mol/L} \times 0.250 \, \text{L} = 0.050 \, \text{moles} \]

### Step 2: Calculate the remaining number of moles of `Cr^(3+)`
The remaining concentration of `Cr^(3+)` after electrolysis is 0.10 M.

Using the same volume:

\[ \text{Remaining number of moles} = 0.10 \, \text{mol/L} \times 0.250 \, \text{L} = 0.025 \, \text{moles} \]

### Step 3: Calculate the number of moles of `Cr^(3+)` deposited
To find the number of moles deposited, subtract the remaining moles from the initial moles:

\[ \text{Moles deposited} = \text{Initial moles} - \text{Remaining moles} = 0.050 \, \text{moles} - 0.025 \, \text{moles} = 0.025 \, \text{moles} \]

### Step 4: Apply Faraday's law of electrolysis
According to Faraday's first law, the amount of substance deposited is directly proportional to the charge passed through the electrolyte. The equation is:

\[ n = \frac{Q}{F} \]

Where:
- \( n \) = number of moles deposited
- \( Q \) = total charge (in coulombs)
- \( F \) = Faraday's constant (approximately 96500 C/mol)

The total charge \( Q \) can also be expressed as:

\[ Q = I \times t \]

Where:
- \( I \) = current (in amperes)
- \( t \) = time (in seconds)

### Step 5: Relate the moles deposited to the charge
For the reduction of `Cr^(3+)`, 3 moles of electrons are required for each mole of `Cr` deposited (n = 3). Thus:

\[ n = \frac{I \times t}{n \times F} \]

Substituting the values we have:

\[ 0.025 = \frac{96.5 \times t}{3 \times 96500} \]

### Step 6: Solve for time \( t \)
Rearranging the equation to solve for \( t \):

\[ t = \frac{0.025 \times 3 \times 96500}{96.5} \]

Calculating this step-by-step:

1. Calculate \( 0.025 \times 3 = 0.075 \)
2. Calculate \( 0.075 \times 96500 = 7237.5 \)
3. Finally, divide by 96.5:

\[ t = \frac{7237.5}{96.5} \approx 75 \, \text{seconds} \]

### Final Answer
The duration of the electrolysis process is **75 seconds**.

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