Question

From the following information, calculate the solubility product of \[AgBr\] .
\[AgB{r_{(s)}} + {e^ - } \to A{g_{(s)}} + B{r^ - }_{(aq)}:{E^o} = 0.07V\]
 \[A{g^ + }_{(aq)} + {e^ - } \to A{g_{(s)}}:{E^o} = 0.80V\]
A. $4 \times {10^{ - 13}}$
B. $4 \times {10^{ - 10}}$
C. $4 \times {10^{ - 17}}$
D. $4 \times {10^{ - 7}}$

Answer 1

Hint: We need to remember that the solubility product constant will explain the saturated solution of ionic compounds in a solution or it can be the equilibrium constant for the dissolution of a solid substance into aqueous solution. It is denoted as ${K_{sp}}$ . Now, let us know about electrode potential. It can also be termed as oxidation electrode potential or reduction electrode potential depending upon the reaction undergone by the electrode if loss of electron (oxidation) takes place at the electrode termed as oxidation potential. If the electrons have the tendency to gain electrons then it is called as reduction electrode potential. It is measured in volts (V) or millivolts (mV).

Complete step by step answer:
We can write the dissolution of \[AgBr\] in cell will be,
In anode \[A{g_{(s)}} + B{r^ - }_{(aq)} \to AgB{r_{(s)}}\]
In cathode \[A{g^ + }_{(aq)} + {e^ - } \to A{g_{(s)}}\]
Overall reaction $A{g^ + }_{(aq)} + B{r^ - }_{(aq)} \to AgB{r_{(s)}}$
In \[AgBr\] , cell potential of a cell $\left( {{E_{cell}}} \right) = 0$
Given the standard potential $\left( {{E^0}} \right)$ of cathode is \[0.07V\] and the standard potential ($E^0$) of anode is \[0.08V\] .Then the total cell electrode potential $E_{cell}^0$ can be calculated by
$E_{cell}^0 = E_{cathode}^0 - E_{anode}^0$
Now apply the E0 values in the above equation,
\[{E^0}_{cell} = 0.80V - 0.07V\]
The total cell electrode potential,\[{E^0}_{cell} = 0.73V\]
 From electrode potential we can calculate \[{K_{SP}}\] by the formula
${E_{cell}} = {E^0} - \dfrac{{0.059}}{n}\log {K_{SP}}$
 n= number of moles of electrons transferred
From the above reaction , n=1
Since it is equilibrium reaction ${E_{cell}} = 0$
 $0 = 0.73 - \dfrac{{0.059}}{1}\log {K_{sp}}$
Now bring the total cell electrode potential value i.e) 0.73 to the left hand side,
$0.73 = - \dfrac{{0.059}}{1}\log {K_{sp}}$
Now let us keep \[log{K_{sp}}\] on one side and values on the other side like
 \[log{K_{SP}}{\text{ }} = - \left( {0.73} \right)\dfrac{1}{{0.059}}\]
By solving the above step,
\[log{K_{SP}} = - 12.37\]
Now to find the value of\[{K_{sp}}\] , bring log to the right hand side log becomes antilog,
\[{K_{sp}} = anti\log ( - 12.37)\]
Then the value is, \[{K_{SP}} = 4 \times {10^{ - 13}}\]

So, the correct answer is Option A.

Note: We have to remember that at equilibrium condition cell potential is equal to zero. Under standard conditions, the standard electrode potential occurs in an electrochemical cell say the temperature = 298K, pressure = 1atm, concentration = 1M.solubility product can be calculated directly from the molarity or by concentration of the products i.e) it is product of concentration of the product of the reaction.