In acidic medium, $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ shows oxidising action as represented in the half reaction:

$$ \mathrm{Cr}_2 \mathrm{O}_7{ }^{2-}+\mathrm{XH}^{+}+\mathrm{Ye}^{\ominus} \rightarrow 2 \mathrm{~A}+\mathrm{ZH}_2 \mathrm{O} $$

$\mathrm{X}, \mathrm{Y}, \mathrm{Z}$ and $\mathrm{A}$ are respectively are :

<ul> <li><strong>Balancing the Half-Reaction</strong></li> </ul> <br/><ol> <li><strong>Chromium (Cr) :</strong> Already balanced with 2 Cr on each side.</li><br> <li><strong>Oxygen (O):</strong> Add 7 H₂O to the right:</li> </ol> <p>$$\mathrm{Cr}_2 \mathrm{O}_7{ }^{2-} + \mathrm{XH}^{+} + \mathrm{Ye}^{\ominus} \rightarrow 2 \mathrm{~A} + 7\mathrm{H}_2 \mathrm{O}$$</p> <ol> <li><strong>Hydrogen (H) :</strong> Add 14 H⁺ to the left:</li> </ol> <p>$$\mathrm{Cr}_2 \mathrm{O}_7{ }^{2-} + 14\mathrm{H}^{+} + \mathrm{Ye}^{\ominus} \rightarrow 2 \mathrm{~A} + 7\mathrm{H}_2 \mathrm{O}$$</p> <ol> <li><strong>Charge :</strong> Add 6e⁻ to the left:</li> </ol> <p>$$\mathrm{Cr}_2 \mathrm{O}_7{ }^{2-} + 14\mathrm{H}^{+} + 6\mathrm{e}^{\ominus} \rightarrow 2 \mathrm{~A} + 7\mathrm{H}_2 \mathrm{O}$$</p> <ul> <li><strong>Identifying A :</strong></li> </ul> <p>The reduction of Cr₂O₇²⁻ in acidic medium forms Cr³⁺ ions.</p> <ul> <li><strong>Final Balanced Equation :</strong></li> </ul> <p>$$\mathrm{Cr}_2 \mathrm{O}_7{ }^{2-} + 14\mathrm{H}^{+} + 6\mathrm{e}^{\ominus} \rightarrow 2 \mathrm{Cr}^{3+} + 7\mathrm{H}_2 \mathrm{O}$$</p> <strong>Answer</strong> <br/><br/><ul> <li>X = 14</li><br> <li>Y = 6</li><br> <li>Z = 7</li><br> <li>A = Cr³⁺</li> </ul>