Question

In acidic medium, dichromate behaves as an oxidizing agent represented as: $Cr_2O_7^{2-} + xH^+ + ye^- \rightarrow 2Cr^{3+} + zH_2O$. The values of $x$, $y$ and $z$ are respectively:

• A. 6, 7 and 14
• B. 7, 6 and 14
• C. 14, 6 and 7
• D. 14, 7 and 6
• E. 6, 12 and 7

Hint:

Dichromate is a powerful 6-electron oxidant in acidic solutions.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem requires balancing a redox half-reaction in an acidic medium. We need to balance the atoms (Cr, O, H) and the charge on both sides of the equation to find the stoichiometric coefficients x, y, and z.
Step 2: Key Formula or Approach:
The standard procedure for balancing half-reactions in acidic solution is: 1. Balance all elements other than O and H. 2. Balance the oxygen atoms by adding H\(_2\)O molecules to the appropriate side. 3. Balance the hydrogen atoms by adding H\(^+\) ions to the appropriate side. 4. Balance the charge by adding electrons (e\(^-\)) to the more positive side. Step 3: Detailed Explanation:
The given unbalanced half-reaction is: \[ \text{Cr}_2\text{O}_7^{2-} + \text{xH}^+ + \text{ye}^- \rightarrow 2\text{Cr}^{3+} + \text{zH}_2\text{O} \] 1. Balance Cr: There are 2 Cr atoms on the left (in Cr\(_2\)O\(_7^{2-}\)) and 2 Cr atoms on the right (in 2Cr\(^{3+}\)). The chromium atoms are already balanced. \[ \text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+} \] 2. Balance O: There are 7 oxygen atoms on the left. To balance them, we need to add 7 H\(_2\)O molecules to the right side. \[ \text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \] Comparing this with the given format, we find \( z = 7 \). 3. Balance H: The right side now has \( 7 \times 2 = 14 \) hydrogen atoms. To balance them, we need to add 14 H\(^+\) ions to the left side. \[ \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \] Comparing this with the given format, we find \( x = 14 \). 4. Balance Charge: Now, calculate the total charge on both sides. - Left side charge: \( (-2) + 14(+1) = -2 + 14 = +12 \) - Right side charge: \( 2(+3) + 7(0) = +6 \) The left side is more positive. To balance the charge, we need to add electrons to the left side. The number of electrons to add is the difference in charge: \( 12 - 6 = 6 \). \[ \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \] Comparing this with the given format, we find \( y = 6 \). Step 4: Final Answer:
The values are x = 14, y = 6, and z = 7.