Question

An excess of liquid mercury is added to an acidified solution of $1.0\, \times \,{10^{ - 3}}M\,F{e^{3 + }}$. It is found that 5% of $F{e^{3 + }}$remains at equilibrium at 25°C. Calculate $E_{Hg_2^{2 + }/Hg}^o$ assuming that the only reaction that occurs is
$2Hg + 2F{e^{3 + }}\, \to \,Hg_2^{2 + } + 2F{e^{2 + }}$
(Given, ${E_{F{e^{3 + }}/F{e^{2 + }}}}$= 0.77 volt).

Answer 1

Hint: Since we know that Nernst equation is: ${E_{cell}}\, = \,E_{cell}^0\, - \,\dfrac{{0.059}}{n}\log Q$ and at equilibrium, ${E_{cell}}$ =0, so the final equation becomes $E_{cell}^0\, = \,\dfrac{{0.059}}{n}\log Q$ and then with the value of $E_{cell}^0$ we can calculate $E_{Hg_2^{2 + }/Hg}^o$ by using the formula $E_{cell}^0\, = \,E_{cathode}^0 - E_{anode}^0$ .

Complete step-by-step answer:
Firstly, let us understand about the Nernst equation.
The redox reactions that are reversible in nature can be expressed in the 'Nernst equation', which helps in the determination of the concentrations of the reduced and oxidized elements at the equilibrium.
Now writing down the reaction which takes place between $F{e^{3 + }}$ions and $Hg$:
$2F{e^{3 + }} + \,2Hg\,\, \to \,\,2F{e^{2 + }} + \,Hg_2^{2 + }$
Now at time t=0,
$F{e^{3 + }}$ ions concentration originally : $1.0 \times {10^{ - 3}}\,M$
At equilibrium, at 25°C:
$F{e^{3 + }}$ ions concentration: 5% of the original concentration.
$\Rightarrow \,\dfrac{5}{{100}} \times {10^{ - 3\,}}\, = \,5 \times {10^{ - 5}}\,M$
Now let us find the concentration of $F{e^{2 + }}$ ions which are converted from $F{e^{3 + }}$ions :
$\left[ {F{e^{2 + }}} \right] = {\left[ {F{e^{3 + }}} \right]_{initial}} + {\left[ {F{e^{3 + }}} \right]_{equilibrium}}$
i.e. we can find the concentration of $F{e^{2 + }}$ions by calculating the difference between the initial and the equilibrium concentration of $F{e^{3 + }}$ions.
$\Rightarrow (1.0 \times {10^{ - 3}})\, - \,(5 \times {10^{ - 5}})\, = \,0.95 \times {10^{ - 3}}$
Similarly the concentration of $Hg_2^{2 + }$ ions at equilibrium = $\dfrac{{0.95 \times {{10}^{ - 3}}}}{2} = 0.475 \times {10^{ - 3}}M$
Now as we know, At equilibrium ${E_{cell}}$ = 0
So Nernst equation is: $\,E_{cell}^0\, = \,\dfrac{{0.059}}{n}\log \dfrac{{\left[ {Hg_2^{2 + }} \right]\,{{\left[ {F{e^{2 + }}} \right]}^{2 + }}}}{{{{\left[ {F{e^{3 + }}} \right]}^2}}}$
So putting concentration values and n=2, we get:
$\,E_{cell}^0\, = \,\dfrac{{0.059}}{2}\log \dfrac{{\left[ {0.475 \times {{10}^{ - 3}}} \right]\,{{\left[ {0.95 \times {{10}^{ - 3}}} \right]}^{2 + }}}}{{{{\left[ {5 \times {{10}^{ - 5}}} \right]}^2}}}$
On solving further, we get: $E_{cell}^0$ = -0.0276 V
Since $E_{cell}^0$is nothing but the difference of reduction potential between cathode and anode.
Mathematically: $E_{cell}^o = E_{cathode}^0\, - \,E_{anode}^0$
Or, $E_{cell}^0$= $E_{F{e^{3 + }}/F{e^{2 + }}}^0 - E_{Hg_2^{2 + }/Hg}^0$
Also it is given to us in the question that ${E_{F{e^{3 + }}/F{e^{2 + }}}}$= 0.77 volt
So we have, -0.0276=0.77-$E_{Hg_2^{2 + }/Hg}^0$
$\Rightarrow$ $E_{Hg_2^{2 + }/Hg}^0$= 0.7976 V

Note: There are many applications of the Nernst equation one of which is in calculating ion concentration. It is also used to calculate the potential of an ion across a membrane. Moreover the equation can also be used to determine the pH value.