In the chemical reaction,
`K_(2)Cr_(2)O_(7)+aH_(2)SO_(4)+bSO_(2) rarr K_(2)SO_(4)+Cr_(2)(SO_(4))_(3)+CH_(2)O.` a, b and c are

Options

A. 1, 3, 1
B. 4, 1, 4
C. 3, 2, 3
D. 2, 1, 2

Answer

To solve the given redox reaction and find the values of \( a \), \( b \), and \( c \) in the equation:

\[ K_2Cr_2O_7 + aH_2SO_4 + bSO_2 \rightarrow K_2SO_4 + Cr_2(SO_4)_3 + cH_2O \]

we will follow these steps:

### Step 1: Write the unbalanced equation
The unbalanced equation is:

\[ K_2Cr_2O_7 + aH_2SO_4 + bSO_2 \rightarrow K_2SO_4 + Cr_2(SO_4)_3 + cH_2O \]

### Step 2: Identify oxidation states
- In \( K_2Cr_2O_7 \), the oxidation state of Cr is +6.
- In \( SO_2 \), the oxidation state of S is +4.
- In \( SO_4^{2-} \), the oxidation state of S is +6.
- In \( Cr_2(SO_4)_3 \), the oxidation state of Cr is +3.

### Step 3: Determine the changes in oxidation states
- Chromium changes from +6 in \( K_2Cr_2O_7 \) to +3 in \( Cr_2(SO_4)_3 \).
- Sulfur changes from +4 in \( SO_2 \) to +6 in \( SO_4^{2-} \).

### Step 4: Balance the changes in oxidation states
- For 2 Cr atoms, the change is \( 2 \times (6 - 3) = 6 \).
- For S, the change is \( 3 \times (6 - 4) = 6 \).

### Step 5: Write the half-reactions
1. Reduction half-reaction for Cr:
\[ Cr_2O_7^{2-} + 6e^- \rightarrow 2Cr^{3+} \]
2. Oxidation half-reaction for S:
\[ 3SO_2 \rightarrow 3SO_4^{2-} + 6e^- \]

### Step 6: Combine the half-reactions
The balanced equation can be written as:
\[ K_2Cr_2O_7 + 3SO_2 + 4H_2SO_4 \rightarrow K_2SO_4 + Cr_2(SO_4)_3 + 3H_2O \]

### Step 7: Identify coefficients
From the balanced equation:
- \( a = 4 \) (for \( H_2SO_4 \))
- \( b = 3 \) (for \( SO_2 \))
- \( c = 3 \) (for \( H_2O \))

### Final Answer
Thus, the values of \( a \), \( b \), and \( c \) are:
- \( a = 4 \)
- \( b = 3 \)
- \( c = 3 \)