`MnO_(4)^(-) + 8H^(+) + 5e^(-) rightarrow Mn^(2+) + 4H-(2)O`, If `H^(+)` concentration is decreased from 1 M to `10^(-4)` M at `25^(@)C`, whereas conc

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`MnO_(4)^(-) + 8H^(+) + 5e^(-) rightarrow Mn^(2+) + 4H-(2)O`, If `H^(+)` concentration is decreased from 1 M to `10^(-4)` M at `25^(@)C`, whereas conc

`MnO_(4)^(-) + 8H^(+) + 5e^(-) rightarrow Mn^(2+) + 4H-(2)O`,
If `H^(+)` concentration is decreased from 1 M to `10^(-4)` M at `25^(@)C`, whereas concentration of `Mn^(2+)` and `MnO_(4)^(-)` remains 1M, then:
A. The potential decreases by 0.38 V with decrease in oxidising power
B. The potential increases by 0.38 V with increase in oxidising power
C. The potential decreases by 0.25 V with decrease inn oxidising power
D. the potential decreases by 0.38 V without affecting oxidising power

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - A
`MnO_(4)^(-)+8H^(+)+5e^(-)toMn^(2+)+4H_(2)O`
`E_(1)=E^(@)-(0.0591)/(5)log(([Mn^(2+)])/([MnO_(4)^(-)]xx1^(8)))`
`E_(2)=E^(@)-(0.0591)/(g)log(([Mn^(2+)])/([MnO_(4)^(-)]xx(10^(-4))^(8)))=-(0.591)/(5)xx32=-0.37824`
`E_(1)-E_(2)=0.38"volt"`

4 views Answered 2023-02-05 15:29