Which one is the oxidising agent in the reaction given below
`2CrO_(4)^(2-)+2H^(+)rarrCr_(2)O_(7)^(-2)+H_(2)O`

Options

A. `H^(+)`
B. `Cr_(2)O_(7)^(-2)`
C. `Cr^(++)`
D. None

Answer

To determine the oxidizing agent in the reaction:

\[ 2CrO_4^{2-} + 2H^+ \rightarrow Cr_2O_7^{2-} + H_2O \]

we will analyze the oxidation states of the elements involved, particularly chromium (Cr) and hydrogen (H).

### Step 1: Assign Oxidation States
- In \( CrO_4^{2-} \):
- Oxygen (O) has an oxidation state of -2.
- Let the oxidation state of chromium (Cr) be \( x \).
- The equation for the oxidation state is:
\[ x + 4(-2) = -2 \implies x - 8 = -2 \implies x = +6 \]
- Thus, in \( CrO_4^{2-} \), Cr is in the +6 oxidation state.

- In \( Cr_2O_7^{2-} \):
- Again, let the oxidation state of chromium be \( y \).
- The equation for the oxidation state is:
\[ 2y + 7(-2) = -2 \implies 2y - 14 = -2 \implies 2y = 12 \implies y = +6 \]
- Thus, in \( Cr_2O_7^{2-} \), Cr is also in the +6 oxidation state.

- In \( H^+ \):
- The oxidation state of hydrogen is +1.

- In \( H_2O \):
- The oxidation state of hydrogen is +1, and oxygen is -2.

### Step 2: Identify Changes in Oxidation States
- The oxidation state of chromium does not change in this reaction; it remains +6 in both \( CrO_4^{2-} \) and \( Cr_2O_7^{2-} \).
- The oxidation state of hydrogen changes from +1 in \( H^+ \) to 0 in \( H_2O \).

### Step 3: Determine the Oxidizing Agent
- An oxidizing agent is a substance that gains electrons (is reduced) in a chemical reaction.
- Here, \( H^+ \) is being reduced to \( H_2O \) (oxidation state changes from +1 to 0), indicating that \( H^+ \) is the species that is reduced.
- Since \( H^+ \) is reduced, it acts as the oxidizing agent in this reaction.

### Conclusion
The oxidizing agent in the reaction is \( H^+ \).

---