Which substance is serving as a reducing agent in the following reaction?
`14H^(+)+Cr_(2)O_(7)^(2-)+3Ni rarr 2Cr^(3+)+7H_(2)O+3Ni^(2+)`

Options

A. `Cr_(2)O_(7)^(2-)`
B. `Ni`
C. `H^(+)`
D. `H_(2)O`

Explanation

To determine which substance is serving as a reducing agent in the given reaction:

**Reaction:**
\[ 14H^+ + Cr_2O_7^{2-} + 3Ni \rightarrow 2Cr^{3+} + 7H_2O + 3Ni^{2+} \]

### Step 1: Identify the oxidation states of the elements in the reactants and products.

1. **For \( Cr_2O_7^{2-} \):**
- Let the oxidation state of \( Cr \) be \( x \).
- The total charge of the dichromate ion is -2.
- The equation can be set up as:
\[ 2x + 7(-2) = -2 \]
\[ 2x - 14 = -2 \]
\[ 2x = 12 \]
\[ x = +6 \]
- Therefore, the oxidation state of \( Cr \) in \( Cr_2O_7^{2-} \) is +6.

2. **For \( Cr^{3+} \):**
- The oxidation state of \( Cr \) is +3.

3. **For \( Ni \):**
- In elemental form, the oxidation state of \( Ni \) is 0.
- In \( Ni^{2+} \), the oxidation state is +2.

4. **For \( H^+ \):**
- The oxidation state of \( H \) is +1.

5. **For \( H_2O \):**
- The oxidation state of \( H \) is +1 and \( O \) is -2.

### Step 2: Determine the changes in oxidation states.

- **Chromium (Cr):**
- Changes from +6 in \( Cr_2O_7^{2-} \) to +3 in \( Cr^{3+} \) (Reduction).

- **Nickel (Ni):**
- Changes from 0 in elemental \( Ni \) to +2 in \( Ni^{2+} \) (Oxidation).

### Step 3: Identify the reducing agent.

- A reducing agent is a substance that gets oxidized and, in the process, reduces another substance.
- Since \( Ni \) is oxidized (from 0 to +2), it is the reducing agent in this reaction.

### Conclusion:
The substance serving as the reducing agent in the reaction is **Nickel (Ni)**.

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