Question

Calculate ${K_c}$ for $H{g^{2 + }} + Hg \rightleftharpoons Hg_2^{2 + }$.
Give that $E_{2Hg/H{g^{2 + }}}^\circ = - 0.788V$ and $E_{Hg/2H{g^{2 + }}}^\circ = - 0.920V$.
Write the answer as the nearest integer after dividing by 100.

Answer 1

Calculate the value of standard cell potential, $E_{cell}^\circ$. The number of electrons involved in the above reaction is $2$. Use it in calculating the value of $\log {K_c}$ using the Nernst equation. Finally, solve the logarithmic value by taking its antilog.

Complete step by step solution:
First, we have to calculate the standard cell potential. We will calculate it using the following equation:
$E_{cell}^\circ = E_{cathode}^\circ - E_{anode}^\circ$ …$(i)$
Here, at anode,
$Hg \to H{g^{2 + }} + 2{e^ - }$
Thus, $E_{anode}^\circ = 0.788V$
Similarly, at cathode,
$2H{g^{2 + }} + 2{e^ - } \to Hg_2^{2 + }$
Thus, $E_{cathode}^\circ = 0.920V$
Putting the value of $E_{anode}^\circ$ and $E_{cathode}^\circ$ in equation $(i)$, we get,
$E_{cell}^\circ = 0.920 - 0.788$
$E_{cell}^\circ = 0.132V$
Now to calculate the value of ${K_c}$, we will use the Nernst equation,
$E_{cell}^\circ = \dfrac{{0.059}}{n}\log ({K_c})$
Where $n$ is the number of electrons gained or lost during reaction and its value here is $2$. Thus, on substituting the values, we get,
$0.132 = \dfrac{{0.059}}{2}\log ({K_c})$
$\log ({K_c}) = \dfrac{{2 \times 0.132}}{{0.059}} = 4.474$
On solving for$\log ({K_c})$, we get,
${K_c} = 29512.09$
On dividing the value of ${K_c}$ by $100$, we get, ${K_c} = 295.12$. Now, rounding off to its nearest integer, the value becomes ${K_c} = 295$.
Thus, the value of ${K_c}$ is $295$.

Additional Information:
The value of ${K_c}$ can be used to calculate Gibb’s energy and Gibb’s free energy. Nernst equation is slightly modified to calculate ${K_c}$. Here, the value of max cell potential becomes $0$ and $Q$ in replaced by ${K_c}$ as ${K_c}$ can only be calculated for those reactions which are in equilibrium. For non-equilibrium reaction, we use $Q$ in place of ${K_c}$. $Q$ and ${K_c}$ both usually represent the ratio of products to their reactants.

Note:
In the question, the negative sign for $E_{anode}^\circ = 0.788V$ and $E_{cathode}^\circ = 0.920V$ just tells us that the redox couple is a stronger reducing agent that ${H^ + }/{H_2}$ couple. The antilog is calculated by taking the base $10$ instead of base $e$. Hence, before calculating the antilog, always check whether the base is $10$ or $e$.