InterviewSolution
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InterviewSolution
| 1. |
`12.53 mL` of `0.0509MSeO_(2)` reacted with `25.52 mL 0.1M CrSO_(4)` solution. In the raeaction `Cr^(2+)` was oxidised to `Cr^(3+)`. To what oxidation state selenium was converted in the reaction? Write the redox change for `SeO_(2)`. |
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Answer» `Se^(4+) + (a-4)e rarr Se^(a+)` `n = a-4` ("where `n` is valence factor") or `Cr^(2+) rarr Cr^(3+) + e` `"Meq. of" SeO_(2) = "Meq. of" CrSO_(4)` `0.0509 xx 12.53 = 0.1 xx 1 xx 25.52` `therefore n = 4` `a-4=4` or `a = 0` Thus redox change is : `Se^(4+) + 4e rarr Se^(0)` |
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