Question

The standard cell voltage for the cell $Pb\left| P{{b}^{2+}} \right.\left\| S{{n}^{2+}} \right.\left| Sn \right.$ is $-0.01V$. If the cell exhibits ${{E}_{cell}}=0$, then the value of $\log \left( \dfrac{[S{{n}^{2+}}]}{[P{{b}^{2+}}]} \right)$ should be:
A. 0.33
B. 0.5
C. 1.5
D. -0.5

Answer 1

Think about the Nernst equation that relates the concentrations of the ionic species, the n-factor, the ideal gas constant, the standard cell voltage and the cell potential of the cell. Consider the conversion of the natural logarithm to the base 10 logarithm.

Compete step by step solution:
We know that the Nernst equation for the natural logarithm is defined as follows:
${{E}_{cell}}=E_{cell}^{o}-\dfrac{RT}{nF}\ln \left( \dfrac{[product]}{[reactant]} \right)$
Here, ${{E}_{cell}}$ is the cell potential which is defined as 0 in this question, $E_{cell}^{o}$ is the standard cell potential which is defined as $-0.01V$ in the question, $R$ is the ideal gas constant, $T$ is the temperature, $F$ is Faraday’s constant, and $n$ is the n-factor of the reaction.
To find the reactants and products of the overall reaction along with the n-factor, let us formulate the oxidation and reduction halves of the reaction from the given information.
From the notation $Pb\left| P{{b}^{2+}} \right.\left\| S{{n}^{2+}} \right.\left| Sn \right.$ we can say that the lead atoms oxidized from $Pb$ to $P{{b}^{2+}}$, and tin gets reduced from $S{{n}^{2+}}$ to $Sn$. So, the oxidation and reduction half reactions respectively will be:

& Pb\to P{{b}^{2+}}+2{{e}^{-}} \\\ & S{{n}^{2+}}+2{{e}^{-}}\to Sn \\\ \end{aligned}$$ From this, we can deduce the general reaction to be: $$Pb+S{{n}^{2+}}\to P{{b}^{2+}}+Sn$$ Now, we can say that the product is $P{{b}^{2+}}$ and the reactant is $S{{n}^{2+}}$. Here, we can see that an exchange of 2 electrons is taking place, so the n-factor of this reaction will be 2. The modified Nernst equation which shows log instead of ln is as follows: $${{E}_{cell}}=E_{cell}^{o}-\dfrac{0.0591}{n}\log \left( \dfrac{[P{{b}^{2+}}]}{[S{{n}^{2+}}]} \right)$$ Now, we will put all the values in the equation above and solve for $\log \left( \dfrac{[S{{n}^{2+}}]}{[P{{b}^{2+}}]} \right)$. $$\begin{aligned} & {{E}_{cell}}=E_{cell}^{o}-\dfrac{0.0591}{n}\log \left( \dfrac{[P{{b}^{2+}}]}{[S{{n}^{2+}}]} \right) \\\ & 0=-0.01-\dfrac{0.0591}{2}\log \left( \dfrac{[P{{b}^{2+}}]}{[S{{n}^{2+}}]} \right) \\\ & 0.01=-\dfrac{0.0591}{2}\log \left( \dfrac{[P{{b}^{2+}}]}{[S{{n}^{2+}}]} \right) \\\ & 0.3384=-\log \left( \dfrac{[P{{b}^{2+}}]}{[S{{n}^{2+}}]} \right) \\\ & \log \left( \dfrac{[S{{n}^{2+}}]}{[P{{b}^{2+}}]} \right)=0.33 \\\ \end{aligned}$$ **Hence, the correct answer to this question is ‘A. 0.33’.** **Note:** In the equation the concentrations of the reactants and products are to be considered but we will not consider the concentrations of $Sn$ and $Pb$ since they are present in their solid and metallic form. The concentration of solid substances is considered to be 1.