Question

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?

• A. 14, 6
• B. 6, 14
• C. 7, 3
• D. 7, 6

Hint:

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!

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We need to balance the reduction half-reaction of the dichromate ion in acidic media.

Step 2: Detailed Explanation:
1. Balance Atoms: $[Cr_{2}O_{7}]^{2-} \rightarrow 2Cr^{3+}$
2. Balance Oxygen with Water: $[Cr_{2}O_{7}]^{2-} \rightarrow 2Cr^{3+} + 7H_{2}O$
3. Balance Hydrogen with Protons: $[Cr_{2}O_{7}]^{2-} + 14H^{+} \rightarrow 2Cr^{3+} + 7H_{2}O$
4. Balance Charge with Electrons:
LHS: $(-2) + (+14) = +12$
RHS: $2 \times (+3) = +6$
To go from $+12$ to $+6$, we must add $6e^{-}$ to the LHS.
$[Cr_{2}O_{7}]^{2-} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_{2}O$.

Step 3: Final Answer:
Protons = 14, Electrons = 6.
This matches option (A).