Question

The voltage of cell
\[Pb(s)|PbS{O_4}(s)|NaHS{O_4}(0.60M)||P{b^{2 + }}(2.50 \times {10^{ - 5}}M)|Pb(s)\]
is \[E = + 0.061V\] and \[{K_2} = [{H^ + }][SO_4^{2 - }]/[HSO_4^ - ]\]. Thus, the dissociation constant for \[HSO_4^ - \] is approximately \[{10^{ - M}}\]. Find the value of \[M\].
Given, \[Pb(s) + SO_4^{2 - }(aq) \to PbS{O_4}(s) + 2{e^ - }\]; \[(E^\circ = 0.356)\], \[E^\circ (P{b^{2 + }}/Pb) = 0.126V\]

Answer 1

Hint: First, calculate the cell potential of the chemical reaction. Now, use the Nernst equation to calculate the dissociation constant for \[HSO_4^ - \]. The dissociation constant \[{K_2}\] is a ratio of molarity of products to the molarity of reactants. You will get the value of \[\log {K_2}\]. Calculate the antilog to get the final value of \[{K_2}\]. At last, compare the value of \[{K_2}\] with \[{10^{ - M}}\] to get the value of \[M\].

Complete step by step solution:
To calculate the cell potential, we will use the following equation:
\[E{^\circ _{cell}} = E{^\circ _{cathode}} - E{^\circ _{anode}}\]
\[E{^\circ _{cell}} = - 0.126 + 0.356\]
\[E{^\circ _{cell}} = 0.230V\]
Since, \[{E_{cell}} = + 0.061V\], we will use the Nernst equation to calculate the value of \[{K_2}\]. The Nernst equation in given by,
\[{E_{cell}} = E{^\circ _{cell}} - \dfrac{{0.059}}{n}\log \dfrac{{[products]}}{{[reac\tan ts]}}\] …\[(i)\]
Where, \[{E_{cell}} \to \] max potential which can be generated when no current is flowing.
\[E{^\circ _{cell}} \to \] cell potential
\[n\] is the number of electrons gained or lost during any reaction. Here, two electrons are gained by \[P{b^{2 + }}\] to convert into \[Pb\]. Hence, \[n = 2\].
Equation \[(i)\] can be further written as,
\[0.061 = 0.23 - \dfrac{{0.059}}{2}\log \dfrac{{[HSO_4^ - ]}}{{[{H^ + }][SO_4^{2 - }][P{b^{2 + }}]}}\]
In the equation above, we wrote \[{K_2}\] because the ration of \[[HSO_4^ - ]\] and \[[{H^ + }][SO_4^{2 - }]\] is same and also it is given that \[{K_2} = [{H^ + }][SO_4^{2 - }]/[HSO_4^ - ]\].
\[0.061 = 0.23 - \dfrac{{0.059}}{2}\log \dfrac{{{K_2}}}{{2.50 \times {{10}^{ - 5}}}}\]
on further solving,
\[\dfrac{{0.059}}{2}\log \dfrac{{{K_2}}}{{2.50 \times {{10}^{ - 5}}}} = 0.23 - 0.061\]
\[\dfrac{{0.059}}{2}\log \dfrac{{{K_2}}}{{2.50 \times {{10}^{ - 5}}}} = 0.169\]
On simplifying,
\[\log \dfrac{{{K_2}}}{{2.50 \times {{10}^{ - 5}}}} = \dfrac{{0.169 \times 2}}{{0.059}}\]
\[\log \dfrac{{{K_2}}}{{2.50 \times {{10}^{ - 5}}}} = 5.72\]
And,
\[\dfrac{{{K_2}}}{{2.50 \times {{10}^{ - 5}}}} = 304.90\]
\[{K_2} = 304.90 \times 2.50 \times {10^{ - 5}}\]
Which reduces to,
\[{K_2} = 762.25 \times {10^{ - 5}}\]
Since, \[0.76\] becomes \[1\] when it is rounded it to its nearest integer. Thus, we can write,
\[{K_2} \simeq {10^{ - 2}}\]
Hence, the value of \[{K_2}\] is \[{10^{ - 2}}\] and the value of \[M\] becomes \[2\] on comparing with \[{10^{ - M}}\].

Note:
Do not forget to compare the value of \[{K_2}\] with \[{10^{ - M}}\] because that is what we had to calculate. In hurry, it might slip out of your mind, so always keep it in mind to compare the values.
The value of dissociation constant \[{K_2}\] has no units; it is simply a ratio of molarity of products to their reactants. The dissociation constant \[{K_2}\] can also be written as \[{K_d}\].