An element A in a compound ABD has oxidation number A n- . It is oxidised by C r 2 O − 2 7 in acidic medium. In the experiment 1.68 × 10 − 3 x moles of K 2 C r 2 O 7 were used for 3.26 × 10 − 3 moles of ABD. The new oxidation number of A after oxidation is: Options: A . &nbsp 3 B . &nbsp 3-n C . &nbsp n-3 D . &nbsp +n

Answer: Option B : B First - let us write down the reaction: C r 2 O 2 − 7 + A n − B D ⟶ C r 3 + + A − n + x From stoichiometry, we have: Equivalents of C r 2 O 2 − 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

An element A in a compound ABD has oxidation number A n-. It is oxidised by Cr

Discussion Forum : Redox Reactions

Question -

An element A in a compound ABD has oxidation number A ^n-. It is oxidised by $C r_{2} O_{7}^{- 2}$ in acidic medium. In the experiment 1.68 $\times$ $10^{- 3}$ x moles of $K_{2} C r_{2} O_{7}$ were used for 3.26 $\times$ $10^{- 3}$ 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:
$C r_{2} O_{7}^{2 -} + A^{n -} B D ⟶ C r^{3 +} + A^{- n + x}$
From stoichiometry, we have:
Equivalents of $C r_{2} O_{7}^{2 -}$ = equivalents of ABD
$1.68 \times 10^{- 3} = 3.26 \times 10^{- 3} \times x$
Solving, we get $x = 3$ .
Hence, the new oxidation number is: 3-n

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