Intramolecular redox
`(NH_(4))_(2)Cr_(2)O_(7) rarr N_(2) + Cr_(2)O_(3) + 4H_(2)O` Shows a gain and loss of how many electrons

Explanation

To determine the gain and loss of electrons in the reaction:

\[ (NH_4)_2Cr_2O_7 \rightarrow N_2 + Cr_2O_3 + 4H_2O \]

we will analyze the oxidation states of the elements involved in the reaction.

### Step 1: Identify the oxidation states of each element in the reactant and products.

1. **In (NH₄)₂Cr₂O₇ (Ammonium Dichromate)**:
- Nitrogen (N) in NH₄⁺: +3 (since each H is +1 and there are 4 H's)
- Chromium (Cr): +6 (as it is in the dichromate ion)
- Oxygen (O): -2 (common oxidation state)

2. **In N₂ (Nitrogen gas)**:
- Nitrogen (N): 0 (as it is in elemental form)

3. **In Cr₂O₃ (Chromium(III) oxide)**:
- Chromium (Cr): +3
- Oxygen (O): -2

### Step 2: Calculate the change in oxidation states.

- **For Nitrogen**:
- Change from +3 in (NH₄)₂Cr₂O₇ to 0 in N₂.
- Change = 0 - (+3) = -3 (reduction, gain of 3 electrons)

- **For Chromium**:
- Change from +6 in (NH₄)₂Cr₂O₇ to +3 in Cr₂O₃.
- Change = +3 - (+6) = -3 (oxidation, loss of 3 electrons)
- Since there are 2 chromium atoms, the total change for chromium = 2 * (-3) = -6 (loss of 6 electrons)

### Step 3: Summarize the total gain and loss of electrons.

- Total gain of electrons (from nitrogen): 3 electrons
- Total loss of electrons (from chromium): 6 electrons

### Conclusion:

The overall reaction shows a gain of 3 electrons (from nitrogen) and a loss of 6 electrons (from chromium).

Thus, the total number of electrons involved in the redox process is 6 electrons (3 gained by nitrogen and 6 lost by chromium).

### Final Answer:
The reaction shows a gain of 3 electrons and a loss of 6 electrons.

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