The n-factor for `K_(2)Cr_(2)O_(7)` in acidic medium is

Options

A. `+2`
B. `+4`
C. `+6`
D. `+8`

Answer

To find the n-factor for \( K_2Cr_2O_7 \) in acidic medium, we can follow these steps:

### Step 1: Identify the species and its oxidation states
- The compound \( K_2Cr_2O_7 \) contains the dichromate ion \( Cr_2O_7^{2-} \).
- In this ion, we need to determine the oxidation state of chromium (Cr).

### Step 2: Calculate the oxidation state of chromium
- The formula for the oxidation state can be set up as follows:
\[ 2x + 7(-2) = -2 \]
where \( x \) is the oxidation state of chromium.
- Solving this equation:
\[ 2x - 14 = -2 \\ 2x = 12 \\ x = +6 \]
- Thus, the oxidation state of chromium in \( Cr_2O_7^{2-} \) is +6.

### Step 3: Determine the reduction product in acidic medium
- In acidic medium, the dichromate ion \( Cr_2O_7^{2-} \) is reduced to \( Cr^{3+} \).
- The oxidation state of chromium in \( Cr^{3+} \) is +3.

### Step 4: Calculate the change in oxidation state
- The change in oxidation state for one chromium atom is:
\[ +6 \text{ (in } Cr_2O_7^{2-}) \rightarrow +3 \text{ (in } Cr^{3+}) \\ \text{Change} = 6 - 3 = 3 \]

### Step 5: Calculate the total n-factor
- Since there are 2 chromium atoms in \( K_2Cr_2O_7 \), the total change for both chromium atoms is:
\[ 2 \times 3 = 6 \]
- Therefore, the n-factor for \( K_2Cr_2O_7 \) in acidic medium is 6.

### Final Answer
The n-factor for \( K_2Cr_2O_7 \) in acidic medium is **6**.

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