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Given : (i)MnO(4)^(-)+8H^++5e^(-)toMn^...

Given :
(i)`MnO_(4)^(-)+8H^++5e^(-)toMn^(2+)+4H_2O , E^0=x_1V`
(ii)`MnO_(2)+4H^++2e^(-)toMn^(2+)+2H_2O , E^0=x_2V`
Find `E^(@)` for the following reaction :
`MnO_(4)^(-)+4H^++3e^(-)toMnO^(2)+2H_2O `

A

`x_(2)-x_(1)`

B

`x_(1)-x_(2)`

C

`(5x_(1)-2x_2)/3`

D

`(2x_(1)-5x_2)/3`

Text Solution

Verified by Experts

The correct Answer is:
C
`DeltaG_1^@=-5x_1F , " " DeltaG_2^@=-2x_2F`
`DeltaG_3^@=DeltaG_1^@-DeltaG_2^@ , " " DeltaG_3^@=2x_2F-5x_1F`
`therefore x=(5x_1-2x_2)/3`
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Balance the following ionic equation. MnO_(4)^(-)+H^(+)+Br^(-)toMn^(2+)+Br_(2)+H_(2)O

Read the following passage for the evaluation of E^(@) when different number of electrons are involved. Consider addition of the following half reactions (1) Fe^(3+)(aq)+3e^(-)rarr Fe(s) E_(1)^(@)=0.45V (2) Fe(s)rarr Fe^(2+)(aq)+2e^(-) E_(2)^(@)=-0.04V (3) Fe^(3+)(aq)+e^(-)rarr Fe^(2+)(aq) E_(3)^(@)= ? Because half-reactions (1) and (2) contains a different number of electrons, the net reaction (3) is another half-reaction and E_(3)^(@) can't be obtained simply by adding E_(1)^(@) and E_(3)^(@) . The free - energy changes however, are additive because G is a state function : Delta G_(3)^(@)=Delta G_(1)^(@)+Delta G_(2)^(@) For the reactions (1) MnO_(4)^(-)+8H^(+)+5e^(-)rarr Mn^(2)+4H_(2)O E^(@)=1.51 V (2) MnO_(2)+4H^(+)+2e^(-)rarr Mn^(2+)2H_(2)O E^(@)=1.32 then for the reaction (3) MnO_(4)^(-)+4H^(+)+3e^(-)rarrMnO_(2)+2H_(2)O, E^(@) 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)

For the reactions MnO_4^_+8H^++5e^(-) rarrMn^(2+)+4H_2O,E^0=1.51V MnO_2^_+4H^++2e^(-) rarrMn^(2+)+2H_2O,E^0=1.23V then the reaction MnO_4^_+4H^++3e^(-) rarrMnO_(2)+2H_2O,E^0 is -

Complete the reactions : SO_(2) + MnO_(4)^(−) + H_(2)O →

The equation 3MnO_4^(2-)+4H^(+)to2MnO_4^(-)+MnO_2+2H_2O represents

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