Question

The $p{K_{sp}}$of AgI is 16.07. If the ${E^0}$value for $A{g^ + }/Ag$is 0.7991 V, If the ${E^0}$for the half cell reaction $AgI + {e^ - } \to Ag + {I^ - }$, in Volts is 149 $ \times {10^{ - x}}$, then what is the value of x ?

Answer 1

Hint: For this question, we need to find the value of ${E^0}$for the half cell reaction in Volts and then by equating it to the value given; we can find out the value of x.
We have the formula for finding Standard emf of the cell is :
$E_{{I^ - }|AgI|Ag}^0 = E_{A{g^ + }|Ag}^0 + 0.0592\log {K_{sp}}$
Using this formula, we can calculate the value in volts and then equate the value to the given, the value of x will come.

Complete step by step answer :
For such numerical type questions, first let us write what is given to us and what we need to find out. We have the reaction for the cell is as -
$AgI + {e^ - } \to Ag + {I^ - }$
Given :
$p{K_{sp}}$of AgI solution = 16.07
${E^0}$( $A{g^ + }/Ag$) = 0.7991 V
${E^0}$for the half cell reaction $AgI + {e^ - } \to Ag + {I^ - }$= 149 $ \times {10^{ - x}}$V
To find : The value of x
We know that $p{K_{sp}}$= - log ${K_{sp}}$
We have the formula for finding Standard emf of the cell is :
$E_{{I^ - }|AgI|Ag}^0 = E_{A{g^ + }|Ag}^0 + 0.0592\log {K_{sp}}$
On filling the values in above formula, we can find the value -
\[E_{{I^ - }|AgI|Ag}^0 = 0.7991 - 0.0592 \times 16.07\]
\[E_{{I^ - }|AgI|Ag}^0\]= - 0.149 V
From the question, we have value 149 $ \times {10^{ - x}}$.
So, by equating we can find out the value of ‘x’ as -
149 $ \times {10^{ - x}}$ = - 0.149 V
149 $ \times {10^{ - x}}$ = 149 $ \times {10^{ - 3}}$

Thus, the value of ‘x’ comes out to be 3.

Note: It must be noted that when we remove points. This means we have multiplied by 1000 on the right-hand side and we have to equally divide by 1000 to get the same value. The value ${10^{ - 3}}$ means division by 1000. The negative sign in the power of numerator means when we will get that value to the denominator, the value will become positive.