An element A in a compound ABD has an oxidation number A n It is oxidized by Cr2O72 in - Chemistry - Redox reactions

Question

An element A in a compound ABD has an oxidation number
A–n It is oxidized by Cr2O7
2–in acidic medium In an experiment 1 68 ×
10–3mol of
K2Cr2O7was required for 3 26
× 10–3mol of the compound ABD Calculate new
oxidation state of A (1) 3 – n (2) n + 3 (3) –3 (4) +n

Shruti Srivastava · Accepted Answer

Dear â€‹Utkarsh
Kaushik,

Equivalent of Cr2O72- = Equivalent of ABD or A-nmole×valency factor=mole×valency factorin acidic medium Cr2O72-  changes to Cr3+ hence n-factor or valency factro is 6 (change in oxi
 no is 3 per atom)so n-factor for A can be calculated1
68×10-3×6=3
26×10-3×n-factorn-factor=3so A-n will be oxidized so oxidation no
 increase from n to 3-n
Option 1 is correct

Regards

An element A in a compound ABD has an oxidation number A–n. It is oxidized by Cr2O7 2–in acidic medium. In an experiment 1.68 × 10–3mol of K2Cr2O7was required for 3.26 × 10–3mol of the compound ABD. Calculate new oxidation state of A. (1) 3 – n (2) n + 3 (3) –3 (4) +n

An element A in a compound ABD has an oxidation number A^–n. It is oxidized by Cr₂O₇ ^2–in acidic medium. In an experiment 1.68 × 10^–3mol of K₂Cr₂O₇was required for 3.26 × 10^–3mol of the compound ABD. Calculate new oxidation state of A. (1) 3 – n (2) n + 3 (3) –3 (4) +n