Question

In acidic medium, $K_2Cr_2O_7$ shows oxidizing action as represented in the half-reaction: $Cr_2O_7^{2-} + XH^+ + Ye^- \rightarrow 2A + ZH_2O$
$X$, $Y$, $Z$, and $A$ are respectively:

• A. 8, 6, 4, and $Cr_2O_3$
• B. 14, 7, 6, and $Cr^{3+}$
• C. 8, 4, 6, and $Cr_2O_3$
• D. 14, 6, 7, and $Cr^{3+}$

Answer 1

Answer: (D)

14, 6, 7, and $Cr^{3+}$

Answer 2

The balanced reaction is:

$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$

From the reaction, we have:

$X = 14, \quad Y = 6, \quad Z = 7, \quad A = \text{Cr}^{3+}$