From the following information, calculate the solubility product of silver bromide ?
`{:(AgBr(s)+e^(-)hArr Ag(s)+Br^(-)(aq.)E^(@)=0.07),(Ag^(+)(aq.)+e^(-)hArr Ag(s)E^(@)=0.80 V):}`

Options

A. `4 xx 10^(-7)`
B. `4. xx 10^(-17)`
C. `4 xx 10^(-13)`
D. `4 xx 10^(-10)`

Answer

C

Explanation

We can even use reducation potentials to calculate the equilibrium constant of reactions that are not oxidation-reductions if they are the combination of two half-reactions listed in the `emf` series. Some examples of such reactions are the dissociation of complex ions intop the metal ion and its ligands, and the dissolutionof a sparingly soluble solid in water.
The reaction of interest that defines the `K_(eq.)` is
`AgBr(s)hArr Ag^(+)(aq.)+Br^(-)(aq.)`
One of the half-reactions in questions include solild silver bromide:
`AgBr(s)+e^(-)hArr Ag(s)+Br^(-)(aq.)E^(@)=0.07 V`
this half-reaction is the reduction of silver form the `+1` to the `0` oxidation state. to obtain an overall reaction, we must combine. It with another half-reaction that includes the same change in oxidation state. We find in question the half-reaction.
`Ag^(+)(aq.)+e^(-)hArr Ag(s)E^(@)=0.80V`
The overall reaction that we are tryping to form must have `Ag^(+)` on the right. This half reaction has `Ag^(+)` on the left, so it is the one that we must reaverse. Therefore, we change the sign of the `E^(@)` of this half-reaction. In this tabular form:
`{:(AgBr(s)+e^(-)hArrAg(s)+Br^(-)(aq.)" "E^(@)=0.07 V),(Ag(s)hArr Ag^(+)(aq.)+e^(-)" "E^(@)=-0.80 V),(bar(AgBr(s) hArr Ag^(+)(aq.).Br^(-)(aq.)E^(@)=0.73 V)):}`
The number of moles of electrons transferred, `n`, is `1`, the number of electrons that cancel when the half-reaction are combind. We calulate `K_(sp)` by using Equation 3.
`E_("cell")^(@) = (0.0592 V)/(n) log K`
`0.73 V = (0.592 V)/(1) log K_(sp)`
Solving for `log K_(sp)` we find
`log K_(sp) = - 12.4`
Now take the antilog of both sides
`K_(sp) = "antilog" (-12.4)`
`= 4 xx 10^(-13)`
The small equilibrium constant is consistant with the negative `E^(@)`.