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[MnO(4)^(-)+8H^(+)+5e^(-)rarr Mn^(2+)+4H...

[MnO_(4)^(-)+8H^(+)+5e^(-)rarr Mn^(2+)+4H_(2)O;E^(0)=1.51V],[MnO_(2)+4H^(+)+2e^(-)rarr Mn^(2+)+2H_(2)O;E^(0)=-1.23V],[in^(o)MnO_(4)^(-)/MnO_(2)=?]

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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)

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)

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

For the reactions {:(MnO_(4)^(-)+8H^(+)+5e^(-)rarr Mn^(2+)4H_(2)O","E^(o) = + 1.51 V),(MnO_(2)+4H^(+)+ 2e^(-) rarr Mn^(2+)+2H_(2)O"," E^(o)= + 1.23 V ):} then for the reaction : MnO_(4)^(-) + 4H^(+) + 3e^(-) rarr MnO_(2)+2H_(2)O","E^(o)

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 -

MnO_(4)^(-)+8H^(+)+n e^(-)rarrMn^(2+)+4H_(2)O The value of n is

E^(o) of some elements are given as : {:(I_(2)+2e^(-)rarr 2I^(-),,,,E^(o)=0.54V),(MnO_(4)^(-)+8H^(o+)+5e^(-)rarr Mn^(2+)+4H_(2)O,,,,E^(o)=1.52V),(Fe^(3+)+e^(-)rarr Fe^(2+),,,,E^(o)=0.77V),(Sn^(4+)+2e^(-)rarr Sn^(2+),,,,E^(o)=0.1V):} a. Select the strongest reductant and weakest oxidant among these elements. b. Select the weakest reductant and strongest oxidant among these elements.