Question -An element A in a compound ABD has oxidation number A n-. It is oxidised by Cr2O−27 in acidic medium. In the experiment 1.68×10−3 x moles ofK2Cr2O7 were used for 3.26×10−3 moles of ABD. The new oxidation number of A after oxidation is:
Options:
A .  3
B .  3-n
C .  n-3
D .  +n
Answer: Option B : B First - let us write down the reaction: Cr2O2−7+An−BD⟶Cr3++A−n+x From stoichiometry, we have: Equivalents ofCr2O2−7 = equivalents of ABD 1.68×10−3=3.26×10−3×x Solving, we get x=3. Hence, the new oxidation number is: 3-n
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