`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)

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

Using the following E^(@) values for electrode potentials, calculate triangle G^(@) in kJ for the indicated reaction: 5Ce^(4+) (aq) + Mn^(2+) (aq) + 4H_(2)O(l) rarr 5Ce^(3+) (aq) + MnO_(4)^(-) (aq) +8H^(+) (aq) MnO_(4)^(-) (aq) +8H^(+) (aq) +5e^(-) rarr Mn^(2+) (aq) + 4H_(2) O(l), E^(@) = +1.51V Ce^(4+) (aq) + e^(-)rarr Ce^(3+)(aq), E^(@) = +1.61V

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 -