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Given the listed standard electrode pote...

Given the listed standard electrode potentials, what is `E^(@)` for the cell:
`4BiO^+(aq)+3N_2H_5^+(aq)to4Bi(s)3N_2(g)+4H_2O)l)+7H^+(aq)`
`N_2(g)+5H^+(aq)+4e^(-)toN_2H_5^+(aq),E^(@)=-0.23V`
`BiO^+(aq)+2H^+(aq)+3e^(-)toBi(s)+H_2O(l),E^(@)=+0.32V`

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Use the following standard electrode potentials, calculate DeltaG^(@) in kJ // mol for the indicated reaction : 5Ce^(4+)(aq)+Mn^(2+)(aq)+4H_(2)O(l)to5Ce^(3+)(aq)+MnO_(4)^(-)(aq)+8H^(+)(aq) MnO_(4)^(-)(aq)+8H^(+)(aq)+5e^(-)toMn^(2+)(aq)+4H_(2)O(l), E^(@)=+1.51 V Ce^(4+)(aq)+e^(=)toCe^(3+)(aq)" "E^(@)=+1.61 V

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Use the following standard electrode potentials, calculate DeltaG^(@) in kJ // mol for the indicated reaction : 5Ce^(4+)(aq)+Mn^(2+)(aq)+4H_(2)O(l)to5Ce^(3+)(aq)+ MnO_(4)^(-)(aq)+8H^(+)(aq) MnO_(4)^(-)(aq)+8H^(+)(aq)+5e^(-)toMn^(2+)(aq)+4H_(2)O(l), E^(@)=+1.51 V Ce^(4+)(aq)+e^(=)toCe^(3+)(aq)" "E^(@)=+1.61 V

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

The standard reduction potentials at 298K for the following half cells are given : ZN^(2+)(aq)+ 2e^(-)Zn(s) , E^(@) = - 0.762V Cr^(3+)(aq)+ 3e^(-) Cr(s) , E^(@) = - 0.740V 2H^(+) (aq)+2e^(-) H_(2)(g) , E^(@)= 0.000V Fe^(3+)(aq)+ e^(-) Fe^(2+)(aq) , E^(@) = 0.770V which is the strongest reducing agent ?