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Equivalent weight of FeS(2) in the half ...

Equivalent weight of `FeS_(2)` in the half reaction
`FeS rarr Fe_(2)O_(3) + SO_(2)` is :

A

`M//10`

B

`M//11`

C

`M//8`

D

`M//7`

Text Solution

Verified by Experts

The correct Answer is:
D

`S^(2-) rarr S^(4+) + 6e`
`2Fe^(2+) rarr (Fe^(3+))_(2) + 2e`
`FeS Fe_(2)O_(3) + SO_(2)`
involves a change of `7` electrons per mole.
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P BAHADUR-ELECTROCHEMISTRY-Exercise (9) ADVANCED NUMERICAL PROBLEMS
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