# Question 471de

Jul 16, 2017

${E}^{o} = {E}_{c a}^{o} - {E}_{a n}^{o} = \text{+2.87 V - (-0.28 V) = 3.15 V}$

#### Explanation:

By definition, the Standard Potential is calculated using the expression

=> ${E}^{o} = {E}_{c a}^{o} - {E}_{a n}^{o}$,

where both ${E}^{\circ}$ 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, ${E}_{c a}^{o}$, for a reduction half-reaction, gives

${F}_{2}^{o} + 2 {e}^{\text{-}} \implies 2 {F}^{-}$ (Reduction)

and expressing the more-negative value, ${E}_{a n}^{o}$, for an oxidation half-reaction, gives

$C {o}^{o} \implies C {o}^{+ 2} + 2 {e}^{-}$ (Oxidation)

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

$\text{ "" "F_2^o + cancel(2e^-) => 2F^-; color(white)(mmml)E_(ca)^@ = "+2.87 V}$
-" " 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"#