The electrolytic cell, one containing acidified ferrous chlo- ride and another with acidified ferric chloride are connected in series. The ratio of iron deposited at cathodes in the two cells when electricity is passed through the cells will be:

Options

A. `3 :1`
B. `2:1`
C. `1:1`
D. `3:2`

Answer

To solve the problem of determining the ratio of iron deposited at the cathodes in two electrolytic cells containing acidified ferrous chloride (FeCl2) and acidified ferric chloride (FeCl3), we will follow these steps:

### Step 1: Identify the oxidation states and reactions
- In the first cell (FeCl2), the ferrous ion (Fe²⁺) is reduced at the cathode:
\[ \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \quad \text{(Equation 1)} \]
- In the second cell (FeCl3), the ferric ion (Fe³⁺) is reduced at the cathode:
\[ \text{Fe}^{3+} + 3e^- \rightarrow \text{Fe} \quad \text{(Equation 2)} \]

### Step 2: Calculate the equivalent weight of iron in both cases
- The equivalent weight of a substance is calculated using the formula:
\[ \text{Equivalent weight} = \frac{\text{Atomic weight}}{n} \]
where \( n \) is the number of electrons transferred.

- For the ferrous ion (Equation 1):
- Atomic weight of iron (Fe) = 56 g/mol
- \( n = 2 \) (for Fe²⁺)
\[ \text{Equivalent weight of Fe in FeCl2} = \frac{56}{2} = 28 \, \text{g/equiv} \]

- For the ferric ion (Equation 2):
- \( n = 3 \) (for Fe³⁺)
\[ \text{Equivalent weight of Fe in FeCl3} = \frac{56}{3} \approx 18.67 \, \text{g/equiv} \]

### Step 3: Establish the ratio of iron deposited
- The amount of iron deposited at the cathode is directly proportional to the equivalent weight of the ions involved.
- Therefore, the ratio of iron deposited from the two cells can be expressed as:
\[ \text{Ratio} = \frac{\text{Equivalent weight of Fe in FeCl2}}{\text{Equivalent weight of Fe in FeCl3}} = \frac{28}{18.67} \]

### Step 4: Simplify the ratio
- To simplify the ratio:
\[ \text{Ratio} = \frac{28}{18.67} \approx 1.5 \]
To express this as a ratio of whole numbers, we can multiply both sides by 3 to avoid decimals:
\[ \text{Ratio} = \frac{28 \times 3}{18.67 \times 3} \approx \frac{84}{56} = \frac{3}{2} \]

### Final Answer
The ratio of iron deposited at the cathodes in the two cells is:
\[ \text{Ratio} = 3:2 \]