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Introduction to Oxidation and Reduction

In an oxidati...

Question

In an oxidation-reduction reaction, dichromate $({C r}_{2} O_{7}^{- 2})$ ion is reduced to ${C r}^{+ 3}$ ion. The equivalent weight of $K_{2} {C r}_{2} O_{7}$ in this reaction is:

\frac{M o l e c u l a r w e i g h t}{3}

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\frac{M o l e c u l a r w e i g h t}{6}

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\frac{M o l e c u l a r w e i g h t}{1}

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\frac{M o l e c u l a r w e i g h t}{2}

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Solution

The correct option is B $\frac{M o l e c u l a r w e i g h t}{6}$
In $C r_{2} O_{7}^{- 2}$ , the oxidation state of $C r$ is $+ 6$ and it is reduced to $C r^{+ 3}$ .

$C r_{2} O_{7}^{2 -} \to 2 C r^{3 +} + {6 e}^{-}$

So, the valency factor $= 2 \times 3 = 6$

So, equivalent weight $= \frac{M o l e c u l a r w e i g h t}{V a l e n c y f a c t o r}$ $= \frac{M}{6}$

Hence, option B is correct.

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Similar questions

H_{3} P O_{2} = \frac{m o l e c u l a r w e i g h t}{1}

H_{3} P O_{3} = \frac{m o l e c u l a r w e i g h t}{2}

H_{3} P O_{4} = \frac{m o l e c u l a r w e i g h t}{3}

N H_{3}

N_{2} \to N H_{3}

N_{2}

= \frac{Molecular weight}{Number of electrons lost or gained per molecule}

{3 C H}_{3} {C H}_{2} O H + 2 K_{2} {C r}_{2} O_{7} + 8 H_{2} S O_{4} ⟶ 3 {C H}_{3} C O O H + 2 {C r}_{2} {(S O_{4})}_{3}

+ {2 K}_{2} S O_{4} + 11 H_{2} O

K_{2} {C r}_{2} O_{7}

2 {N H}_{3} ⟶ N_{2} + 2 {N H}_{3}

{N H}_{3}

= \frac{M o l e c u l a r w e i g h t}{y}

y

{N a}_{2} S_{2} O_{3}

K_{2} {C r}_{2} O_{7}

K_{2} {C r}_{2} O_{7}

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