Question

Calculate the equilibrium constant for the reaction, $2F{e^{2 + }} + 3{I^ - } \rightleftharpoons 2F{e^{3 + }} + {I_3}^ - $.The standard reduction potential in acidic conditions are $0.77\,V$ and $0.54\, V$ respectively for $F{e^{3 + }}/F{e^{2 + }}$ and ${I^{3 - }}/{I^ - }$ couples.

Answer 1

Hint:Here in this question we will use the concepts of Nernst equation which relates the reduction potential of an electrochemical reaction for half-cell or full cell reaction to the standard electrode potential, temperature and activities.

Complete answer:
First let us discuss about nernst equation and its application:
The Nernst equation can be written as:
For a general electrochemical reaction of the type:
$aA + bB \to cC + dD$
$E = {E^0} - \dfrac{{RT}}{{nF}}\ln Q$
$\Rightarrow E = {E^0} - \dfrac{{RT}}{{nF}}\ln \dfrac{{{{[C]}^c}{{[D]}^d}}}{{{{[A]}^a}{{[B]}^b}^{}}}$
\[{K_c} = 5.88 \times {10^7}\]
Now we know that at equilibrium $E = 0$ and $\dfrac{{{{[C]}^c}{{[D]}^d}}}{{{{[A]}^a}{{[B]}^b}^{}}} = {K_c}$
So we get ${E^0} = \dfrac{{2.303RT}}{{nF}}\log {K_c}$
Let’s solve the problem for given data:
$2F{e^{2 + }} + 3{I^ - } \rightleftharpoons 2F{e^{3 + }} + {I_3}^ - $,$F{e^{3 + }}/F{e^{2 + }}$=$0.77\,V$, ${I^{3 - }}/{I^ - }$=0.54V
Given reduction potential of \[F{e^{3 + }}\] to \[F{e^{2 + }}\] = $0.77\,V$, oxidation potential = -$0.77\,V$ and \[{I^ - }\] to \[{I^{3 - }}\]=$0.54\,V$ i.e oxidation potential = -$0.54\,V$.
Cell potential = \[
  ( - 0.54) - ( - 0.77) \\\]
\[ = 0.23V
 \]
Now we know that, \[{E^0} = \dfrac{{0.0591}}{2}{\log _{10}}{K_c}\]
\[\Rightarrow 0.23 \times 2 = 0.0591{\log _{10}}{K_c}\]
\[\Rightarrow7.77 = {\log _{10}}{K_c}\]
\[\Rightarrow{K_c} = anti\log (7.77)\]
\[\Rightarrow{K_c} = 5.88 \times {10^7}\]

Note:
We have to take care while calculating the electrode potential of a full cell. We have to subtract their oxidation potential according to the given reaction and we can also use the other method to calculate the potential of the cell.