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MnO(4)^(-) + 8H^(+) + 5e^(-) rightarrow ...

`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)`

A

1.70 V

B

0.91 V

C

1.37 V

D

0.548 V

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The correct Answer is:
A
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Explore conceptually related problems

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

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 -

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