What are the numbers of protons (H$^+$) and electrons (e$^-$), respectively, required for the reduction of [Cr$_2$O$_7$]$^{2-}$ to Cr$^{3+}$ under an aqueous acidic condition?
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Dichromate reduction is a classic redox process:
Each $\text{Cr}$ in $\text{Cr}_2\text{O}_7^{2-}$ is in the $+6$ state. Two $\text{Cr}^{+6}$ ions are reduced to two $\text{Cr}^{3+}$ ions.
The change in oxidation state per chromium is $3$, so for two chromium atoms, a total of $2 \times 3 = 6$ electrons are transferred.
This immediately identifies 6 as the number of electrons!
Step 1: Understanding the Question:
This question asks for the stoichiometric coefficients of protons ($\text{H}^+$) and electrons ($e^-$) required to balance the reduction half-reaction of the dichromate ion ($[\text{Cr}_2\text{O}_7]^{2-}$) to chromium(III) ion ($\text{Cr}^{3+}$) in an acidic aqueous medium. Step 2: Key Formula or Approach:
Use the ion-electron method to balance the reduction half-reaction:
1. Write the unbalanced key species equation.
2. Balance the chromium atoms.
3. Balance the oxygen atoms by adding water ($\text{H}_2\text{O}$) molecules.
4. Balance the hydrogen atoms by adding protons ($\text{H}^+$).
5. Balance the net charge by adding electrons ($e^-$). Step 3: Detailed Explanation:
Let's balance the half-reaction step-by-step:
• Step 1: Unbalanced species:
\[ \text{Cr}_2\text{O}_7^{2-} \rightarrow \text{Cr}^{3+} \]
• Step 2: Balance Chromium atoms:
There are 2 chromium atoms on the reactant side, so we multiply $\text{Cr}^{3+}$ by 2 on the product side:
\[ \text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+} \]
• Step 3: Balance Oxygen atoms:
There are 7 oxygen atoms on the reactant side. Add 7 molecules of $\text{H}_2\text{O}$ to the product side:
\[ \text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \]
• Step 4: Balance Hydrogen atoms:
The addition of 7 water molecules introduces 14 hydrogen atoms on the product side. Add 14 protons ($\text{H}^+$) to the reactant side:
\[ \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \]
• Step 5: Balance the electric charge:
- Charge on the reactant side: $(-2) + 14(+1) = +12$
- Charge on the product side: $2(+3) + 7(0) = +6$
- To equalize the charge, add 6 electrons ($e^-$) to the reactant side (which reduces the $+12$ charge to $+6$):
\[ \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \]
Step 4: Final Answer:
The balanced half-reaction requires 14 protons ($\text{H}^+$) and 6 electrons ($e^-$). This matches option (A).
