A current of 5.0 A flows for 4.0 h through an electrolytic cell containing a molten slat of metal M. This result in deposition of 0.25 mol of the metal M at the cathode. The oxidation state of M in the molten salt is `+x`. The value of 'x' is (1 Faraday = 96000 C `mol^(-1)`) 3

Answer

To solve the problem step by step, we will follow the outlined process to determine the oxidation state of the metal M in the molten salt.

### Step 1: Calculate the total charge (Q) passed through the electrolytic cell.
The formula to calculate charge is:
\[ Q = I \times T \]
Where:
- \( I \) = current in amperes (A)
- \( T \) = time in seconds (s)

Given:
- \( I = 5.0 \, A \)
- \( T = 4.0 \, h \)

First, we need to convert time from hours to seconds:
\[ T = 4.0 \, h \times 60 \, \text{min/h} \times 60 \, \text{s/min} = 14400 \, s \]

Now, substituting the values into the charge formula:
\[ Q = 5.0 \, A \times 14400 \, s = 72000 \, C \]

### Step 2: Relate the charge to the number of moles of metal deposited.
The relationship between charge (Q), number of moles (n), and Faraday's constant (F) is given by:
\[ Q = n \times F \times x \]
Where:
- \( n \) = number of moles of metal deposited
- \( F \) = Faraday's constant (approximately \( 96500 \, C/mol \))
- \( x \) = number of electrons transferred per ion of metal M

Given:
- \( n = 0.25 \, mol \)
- \( F = 96500 \, C/mol \)

### Step 3: Rearranging the equation to find x.
From the equation:
\[ Q = n \times F \times x \]
We can rearrange it to solve for \( x \):
\[ x = \frac{Q}{n \times F} \]

Substituting the known values:
\[ x = \frac{72000 \, C}{0.25 \, mol \times 96500 \, C/mol} \]

### Step 4: Calculate the value of x.
Calculating the denominator:
\[ 0.25 \, mol \times 96500 \, C/mol = 24125 \, C \]

Now substituting this back into the equation for x:
\[ x = \frac{72000 \, C}{24125 \, C} \approx 2.99 \]

Rounding this value gives:
\[ x \approx 3 \]

### Conclusion:
The oxidation state of metal M in the molten salt is \( +3 \).

### Final Answer:
The value of \( x \) is 3.

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