Question

Question #471de

Answer 1

`E^o = E_(ca)^o - E_(an)^o = "+2.87 V - (-0.28 V) = 3.15 V"`

Explanation:

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"`