Question #471de

By definition, the Standard Potential is calculated using the expression

=> #E^o = E_(ca)^o - E_(an)^o#,

where both #E^@# half-reaction potentials are tabulated reductions.

From the Table of Reduction Potentials, one should visualize the reaction with the more positive reduction potential as the reduction half-reaction and the reaction with the more negative reduction potential as the oxidation half-reaction.

Substituting the values given into the above equation give the Standard Potential for the combination of half-reactions selected.

For the half-reactions in this post...

#F_2^o + 2e^(-) => 2F^(-) ;color(white)(l) E_(ca)^o = "+2.87 V"#
#Co ^(+2) + 2e^(-) => Co^o; E_(an)^o = "-0.28 V"#

Expressing the more-positive value, #E_(ca)^o#, for a reduction half-reaction, gives

# F_2^o + 2e^"-" => 2F^-# (Reduction)

and expressing the more-negative value, #E_(an)^o#, for an oxidation half-reaction, gives

#Co^o => Co ^(+2) + 2e^-# (Oxidation)

Adding the two reactions such that charge transfer is equal between the two reaction gives the net oxidation-reduction reaction ...

#" "" "F_2^o + cancel(2e^-) => 2F^-; color(white)(mmml)E_(ca)^@ = "+2.87 V"#
#-" " ul(Co^o => Co ^(+2) + cancel(2e^-); color(white)(mmll)E_(an)^@ = "-0.28 V")#

#Co^o + F_2^o => Co^(+2) + 2F^(-) ; E^o = "+3.15 V"#