Solveeit Logo
Login

Question

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)|P...

The voltage of cell
Pb(s)∣PbSO4(s)∣NaHSO4(0.60M)∣∣Pb2+(2.50×10−5M)∣Pb(s)Pb(s)|PbS{O_4}(s)|NaHS{O_4}(0.60M)||P{b^{2 + }}(2.50 \times {10^{ - 5}}M)|Pb(s)
is E=+0.061VE = + 0.061V and K2=[H+][SO42−]/[HSO4−]{K_2} = [{H^ + }][SO_4^{2 - }]/[HSO_4^ - ]. Thus, the dissociation constant for HSO4−HSO_4^ - is approximately 10−M{10^{ - M}}. Find the value of MM.
Given, Pb(s)+SO42−(aq)→PbSO4(s)+2e−Pb(s) + SO_4^{2 - }(aq) \to PbS{O_4}(s) + 2{e^ - }; (E∘=0.356)(E^\circ = 0.356), E∘(Pb2+/Pb)=0.126VE^\circ (P{b^{2 + }}/Pb) = 0.126V

Explanation

Solution

First, calculate the cell potential of the chemical reaction. Now, use the Nernst equation to calculate the dissociation constant for HSO4−HSO_4^ - . The dissociation constant K2{K_2} is a ratio of molarity of products to the molarity of reactants. You will get the value of log⁡K2\log {K_2}. Calculate the antilog to get the final value of K2{K_2}. At last, compare the value of K2{K_2} with 10−M{10^{ - M}} to get the value of MM.

Complete step by step solution:
To calculate the cell potential, we will use the following equation:
Ecell∘=Ecathode∘−Eanode∘E{^\circ _{cell}} = E{^\circ _{cathode}} - E{^\circ _{anode}}
Ecell∘=−0.126+0.356E{^\circ _{cell}} = - 0.126 + 0.356
Ecell∘=0.230VE{^\circ _{cell}} = 0.230V
Since, Ecell=+0.061V{E_{cell}} = + 0.061V, we will use the Nernst equation to calculate the value of K2{K_2}. The Nernst equation in given by,
Ecell=Ecell∘−0.059nlog⁡[products][reactan⁡ts]{E_{cell}} = E{^\circ _{cell}} - \dfrac{{0.059}}{n}\log \dfrac{{[products]}}{{[reac\tan ts]}} …(i)(i)
Where, Ecell→{E_{cell}} \to max potential which can be generated when no current is flowing.
Ecell∘→E{^\circ _{cell}} \to cell potential
nn is the number of electrons gained or lost during any reaction. Here, two electrons are gained by Pb2+P{b^{2 + }} to convert into PbPb. Hence, n=2n = 2.
Equation (i)(i) can be further written as,
0.061=0.23−0.0592log⁡[HSO4−][H+][SO42−][Pb2+]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 K2{K_2} because the ration of [HSO4−][HSO_4^ - ] and [H+][SO42−][{H^ + }][SO_4^{2 - }] is same and also it is given that K2=[H+][SO42−]/[HSO4−]{K_2} = [{H^ + }][SO_4^{2 - }]/[HSO_4^ - ].
0.061=0.23−0.0592log⁡K22.50×10−50.061 = 0.23 - \dfrac{{0.059}}{2}\log \dfrac{{{K_2}}}{{2.50 \times {{10}^{ - 5}}}}
on further solving,
0.0592log⁡K22.50×10−5=0.23−0.061\dfrac{{0.059}}{2}\log \dfrac{{{K_2}}}{{2.50 \times {{10}^{ - 5}}}} = 0.23 - 0.061
0.0592log⁡K22.50×10−5=0.169\dfrac{{0.059}}{2}\log \dfrac{{{K_2}}}{{2.50 \times {{10}^{ - 5}}}} = 0.169
On simplifying,
log⁡K22.50×10−5=0.169×20.059\log \dfrac{{{K_2}}}{{2.50 \times {{10}^{ - 5}}}} = \dfrac{{0.169 \times 2}}{{0.059}}
log⁡K22.50×10−5=5.72\log \dfrac{{{K_2}}}{{2.50 \times {{10}^{ - 5}}}} = 5.72
And,
K22.50×10−5=304.90\dfrac{{{K_2}}}{{2.50 \times {{10}^{ - 5}}}} = 304.90
K2=304.90×2.50×10−5{K_2} = 304.90 \times 2.50 \times {10^{ - 5}}
Which reduces to,
K2=762.25×10−5{K_2} = 762.25 \times {10^{ - 5}}
Since, 0.760.76 becomes 11 when it is rounded it to its nearest integer. Thus, we can write,
K2≃10−2{K_2} \simeq {10^{ - 2}}
Hence, the value of K2{K_2} is 10−2{10^{ - 2}} and the value of MM becomes 22 on comparing with 10−M{10^{ - M}}.

Note:
Do not forget to compare the value of K2{K_2} with 10−M{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 K2{K_2} has no units; it is simply a ratio of molarity of products to their reactants. The dissociation constant K2{K_2} can also be written as Kd{K_d}.