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.

Hint: 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 \[e\].