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In acid medium, MnO(4)^(-) is an oxidisi...

In acid medium, `MnO_(4)^(-)` is an oxidising agent ?
`MnO_(4)^(-) + 8H^(+)+ 5e^(-) rightarrow Mn^(2+) + 4H_(2)O`
If `H^(+)` ion concentration is doubled, electrode potential of the half - cell `MnO_(4)^(-), Mn^(2+)`/ pt will:

A

increase by `28.46` m V

B

decrease by `28.46` m V

C

increase by `142.3` m V

D

decrease by `142.30` m V

Text Solution

Verified by Experts

The correct Answer is:
A
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In acid medium, MnO_(4)^(c-) is an oxidizing agent. MnO_(4)^(c-)+8H^(o+)+5e^(-) rarr Mn^(2+)+4H_(2)O If H^(o+) ion concentration is doubled, electrode potential of the half cell MnO_(4)^(c-), Mn^(2+)|Pt will

In acid medium, MnO_(4)^(c-) is an oxidizing agent. MnO_(4)^(c-)+8H^(o+)+5e^(-) rarr Mn^(2+)+4H_(2)O If H^(o+) ion concentration is doubled, electrode potential of the half cell MnO_(4)^(c-), Mn^(2+)|Pt will

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:

The value of n in : MnO_(4)^(-)+8H^(+)+n erarr Mn^(2+)+4H_(2)O is

The shape of MnO_(4)^(-) ion and the hybridisation of Mn in MnO_(4)^(-) is

MnO_(4)^(-) + 8H^(+) + 5e^(-) rightarrow Mn^(2+) + 4H_(2)O , E^(@) = 1.51V MnO_(2) + 4H^(+) + 2e^(-) righarrow Mn^(2+) + 2H_(2)O E^(@) = 1.23V E_(MnO_(4)^(-)|MnO_(2)

The shapes of MnO_(4)^(-) ion and the hybridization of Mn in MnO_(4)^(-) is :

MnO_(4)^(-) is a good oxidising agent in different mediums changing to MnO_(4)^(-)toMn^(2+) toMnO_(4)^(2-) toMnO_(2) toMn_(2)O_(3) Changes in oxidation number respectively are

MnO_(4)^(-)+H^(+)+Br^(-) to Mn^(3+)(aq.)+Br_(2)uarr

MnO_(4)^(-)+H^(+)+Br^(-) to Mn^(3+)(aq.)+Br_(2)uarr

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