Potassium dichromate is used as'lab reagent and dichromate ion can be used as a primary standard in volumetric analysis.
`(1)/(2)Cr_(2)O_(7)^(2-)+7H^(+)+3e^(-)iffCr^(3+)+3(1)/(2)H_(2)O,E^(@)=1.33V`
When`H_(2)O_(2)` is added to an acidified' solution of dichromate, a complicated reaction occurs. The formation of product depends on the pH value and concentration of dichromate ions. A deep-blue violet peroxo compound X is formed which decomposes rapidly in aqueous solution. This compound can be extracted in ether, where it reacts with pyridine forming the adduct `PyCro(O_(2))_(2)` In alkaline solution, `K_(2)Cr_(2)O_(7)` with 30% `H_(2)O_(2)` a red-brown compound `Y (K_(3) CrO_(8)` is formed. In ammonia solution the dark red-brown compound. `Z [(NH_(3)),CrO_(4)]` is formed. In alkaline solution, dichromate ion changes to chromate ion which is also used as reagent in quantitative analysis of cations
QThe number of moles of `Cr_(2)O_(7)^(2-)`required in acidic medium to oxidize 1 mol of ferric oxalate is

Options

A. `10 mol`
B. `0.5 mol`
C. `1.0 mol`
D. `0.1 mol`

Answer

To determine the number of moles of \( \text{Cr}_2\text{O}_7^{2-} \) required to oxidize 1 mole of ferric oxalate in acidic medium, we will follow these steps:

### Step 1: Identify the oxidation states
Ferric oxalate is represented as \( \text{Fe}_2(\text{C}_2\text{O}_4)_3 \). In this compound, iron (Fe) is in the +3 oxidation state, and oxalate (\( \text{C}_2\text{O}_4^{2-} \)) is in the -2 oxidation state.

### Step 2: Write the oxidation half-reaction
The oxalate ion can be oxidized to carbon dioxide. The balanced oxidation half-reaction for the oxidation of oxalate to carbon dioxide is:
\[ \text{C}_2\text{O}_4^{2-} \rightarrow 2 \text{CO}_2 + 2 \text{e}^- \]

### Step 3: Determine the number of moles of oxalate ions
Since 1 mole of ferric oxalate contains 3 moles of oxalate ions, the total number of moles of oxalate ions from 1 mole of ferric oxalate is:
\[ 3 \text{ moles of } \text{C}_2\text{O}_4^{2-} \]

### Step 4: Write the reduction half-reaction
The reduction half-reaction for dichromate in acidic medium is given as:
\[ \frac{1}{2} \text{Cr}_2\text{O}_7^{2-} + 7 \text{H}^+ + 3 \text{e}^- \rightarrow 2 \text{Cr}^{3+} + 3 \frac{1}{2} \text{H}_2\text{O} \]
This indicates that 1 mole of dichromate ion can accept 6 electrons.

### Step 5: Balance the overall reaction
To balance the overall reaction, we need to ensure that the number of electrons lost in the oxidation matches the number of electrons gained in the reduction. From the oxidation half-reaction, 3 moles of oxalate ions will produce:
\[ 3 \times 2 = 6 \text{ electrons} \]
Thus, we need to multiply the reduction half-reaction by 1 to balance the electrons:
\[ \text{Cr}_2\text{O}_7^{2-} + 14 \text{H}^+ + 6 \text{e}^- \rightarrow 2 \text{Cr}^{3+} + 3 \text{H}_2\text{O} \]

### Step 6: Combine the half-reactions
Now we can combine the oxidation and reduction half-reactions:
\[ \text{Cr}_2\text{O}_7^{2-} + 14 \text{H}^+ + 3 \text{C}_2\text{O}_4^{2-} \rightarrow 2 \text{Cr}^{3+} + 6 \text{CO}_2 + 7 \text{H}_2\text{O} \]

### Step 7: Conclusion
From the balanced equation, we see that 1 mole of \( \text{Cr}_2\text{O}_7^{2-} \) is required to oxidize 3 moles of oxalate ions, which corresponds to 1 mole of ferric oxalate. Therefore, the number of moles of \( \text{Cr}_2\text{O}_7^{2-} \) required to oxidize 1 mole of ferric oxalate is:
\[ \boxed{1} \]