Question

Given the reaction: $ \text{Cr}_2\text{O}_7^{2-} + x e^- + y H^+ \to 2 \text{Cr}^{3+} + z H_2O $ Find the values of $x$, $y$, and $z$.

• A. \( x = 6, y = 14, z = 7 \)
• B. \( x = 3, y = 6, z = 3 \)
• C. \( x = 6, y = 12, z = 6 \)
• D. \( x = 4, y = 8, z = 4 \)

Hint:

When balancing redox reactions, always balance atoms first and then charge using electrons. In acidic solutions, use \( H^+ \) and \( H_2O \) to balance hydrogen and oxygen atoms.

Answer 1

The Correct Option is C

We are given the reaction: \[ \text{Cr}_2\text{O}_7^{2-} + x e^- + y H^+ \to 2 \text{Cr}^{3+} + z H_2O \] To balance this redox reaction, we need to follow the steps: 
1. Balance the Cr atoms: There are 2 chromium atoms on the left (in \( \text{Cr}_2\text{O}_7^{2-} \)), so we need 2 chromium ions on the right, which is already the case (as \( 2 \text{Cr}^{3+} \)). 
2. Balance the Oxygen atoms: There are 7 oxygen atoms on the left in \( \text{Cr}_2\text{O}_7^{2-} \), so we need 7 oxygen atoms on the right. These come from water molecules. Therefore, we need 7 water molecules, so \( z = 7 \). 
3. Balance the Hydrogen atoms: There are 14 hydrogen atoms on the right (from 7 \( H_2O \) molecules), so we need 14 hydrogen ions on the left, so \( y = 14 \). 
4. Balance the Charges: The charge on the left is \( 2- \) from \( \text{Cr}_2\text{O}_7^{2-} \), and on the right is \( 2 \times 3+ = 6+ \). 
Thus, we need 6 electrons to balance the charges. Therefore, \( x = 6 \). 
Thus, the balanced reaction is: \[ \text{Cr}_2\text{O}_7^{2-} + 6 e^- + 14 H^+ \to 2 \text{Cr}^{3+} + 7 H_2O \] 
Thus, the values of \( x \), \( y \), and \( z \) are \( 6 \), \( 14 \), and \( 7 \) respectively, so the correct answer is \( C \).