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In the reaction Cr2 O7^(2-) 14 H^(+) + ...

In the reaction ` Cr_2 O_7^(2-) 14 H^(+) + 6e^(-) rarr 2 Cr^(3+) + 7 H_2O`, the equivalent weight fo `K_2 Cr_2 O_7` will be .

A

`M//6`

B

`M//3`

C

`M//12`

D

`M//9`

Text Solution

Verified by Experts

The correct Answer is:
A
`overset(+6)(C)r_(2)O_(7)^(2-) rarr 2 overset(+3)(C)r^(3+)`
There are 2 Cr atoms in `Cr_(2)O_(7)^(2-)`. Thus, we must calculate the change in oxidation numbers for two Cr atoms on both sides. Hence, the oxidation numbers decreases from `2(+6)=+12` to `2(+3)=+6`
Equivalent weight
`=("Formula weight (M)" ) /(("Total change in oxidation number of element" ),("oxidized"//"reduced per mole of compound"))`
`=M/6`
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