Question

If $6.539 \times {10^{ - 2}}$g of metallic zinc is added to 100mL saturated solution of AgCl. Find the value of ${\log _{10}}\dfrac{{[Z{n^{2 + }}]}}{{{{[A{g^ + }]}^2}}}$.

Answer 1

Hint: First write the oxidation and reduction half reactions in order to find the overall reaction. Then use the Nernst equation in order to find the required value. ${E_{cell}}$ for this reaction is 1.56 Volt.

Complete step by step solution:
We need to find the value of ${\log _{10}}\dfrac{{[Z{n^{2 + }}]}}{{{{[A{g^ + }]}^2}}}$ for this reaction. We are not given the amount of silver ions present in the solution. So, we will need to find the cell reaction first and then we will use the Nernst equation to find the required log value.
- We are given that metallic zinc is added to the solution of AgCl. So, metallic zinc is in zero oxidation state and zinc will oxidize and silver will reduce. So, the half reactions can be written as:
\[\begin{gathered}
{\text{Oxidation : Z}}{{\text{n}}_{(s)}} \to Z{n_{(aq)}}^{2 + } + 2{e^ - }{\text{ }}{{\text{E}}^0} = 0.80V \\
{\text{Reduction : A}}{{\text{g}}_{(aq)}}^ + + {e^ - } \to A{g_{(s)}}{\text{ }}{{\text{E}}^0} = 0.80V \\
\end{gathered} \]
So, from these reactions, we can also write the overall reaction as
\[Z{n_{(s)}} + 2A{g_{(aq)}}^ + \to Z{n_{(aq)}}^{2 + } + 2A{g_{(s)}}\]
Now, we can write the equilibrium constant of this reaction as
\[K = \dfrac{{[{\text{Product]}}}}{{{\text{[Reactant]}}}} = \dfrac{{[Z{n^{2 + }}]{{[Ag]}^2}}}{{{{[A{g^ + }]}^2}[Zn]}} = \dfrac{{[Z{n^{2 + }}]}}{{{{[A{g^ + }]}^2}}}\]
Now, for this cell, we can write for the potential of the cell that
\[{E^0}_{cell} = {E_{cathode}} - {E_{anode}}\]
\[{E^0}_{cell} = 0.80 - 0.80 = 0V\]
Now, we can write the Nernst equation for this reaction as
\[{E_{cell}} = {E^0}_{cell} - \dfrac{{0.0591}}{n}{\log _{10}}\dfrac{{[Z{n^{2 + }}]}}{{{{[A{g^ + }]}^2}}}{\text{ }}....{\text{(1)}}\]
Here, n is the number of electrons involved in the reaction which is 2 for this reaction. ${E_{cell}}$ for this reaction is 1.56V. So, we can write the equation (1) as
\[1.56 = 0 - \dfrac{{0.0591}}{2}{\log _{10}}\dfrac{{[Z{n^{2 + }}]}}{{{{[A{g^ + }]}^2}}}\]
So, we can write that
\[{\log _{10}}\dfrac{{[Z{n^{2 + }}]}}{{{{[A{g^ + }]}^2}}} = \dfrac{{1.56 \times 2}}{{0.0591}} = 52.79\]

Thus, the answer is 52.79.

Note: In order to find the standard potential of the cell, do not get confused between the potential of cathode and potential of the anode of the cell. Here, we are using the square on the square bracket of silver ions because there are two silver ions produced in the reaction.