Question

Calculate the equilibrium constant for the reaction, $2F{e}^{2+}+3I^{-}\rightleftharpoons 2F{e}^{3+}+{I_3}^{-}$.The standard reduction potential in acidic conditions are $0.77V$ and $0.54V$ 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:
$aA+bB\to cC+dD$
$E=E^0-{\displaystyle \frac{RT}{nF}}\ln Q$
$\Rightarrow E=E^0-{\displaystyle \frac{RT}{nF}}\ln {\displaystyle \frac{{[C]}^c{[D]}^d}{{[A]}^a{{[B]}^b}^{}}}$
$$K_c=5.88\times {10}^7$$
Now we know that at equilibrium $E=0$ and ${\displaystyle \frac{{[C]}^c{[D]}^d}{{[A]}^a{{[B]}^b}^{}}}=K_c$
So we get $E^0={\displaystyle \frac{2.303RT}{nF}}\log K_c$
Let’s solve the problem for given data:
$2F{e}^{2+}+3I^{-}\rightleftharpoons 2F{e}^{3+}+{I_3}^{-}$,$F{e}^{3+}/F{e}^{2+}$=$0.77V$, $I^{3-}/I^{-}$=0.54V
Given reduction potential of $$F{e}^{3+}$$ to $$F{e}^{2+}$$ = $0.77V$, oxidation potential = -$0.77V$ and $$I^{-}$$ to $$I^{3-}$$=$0.54V$ i.e oxidation potential = -$0.54V$.
Cell potential = $$(-0.54)-(-0.77)$$
$$=0.23V$$
Now we know that, $$E^0={\displaystyle \frac{0.0591}{2}}{\log }_{10}{K}_c$$
$$\Rightarrow 0.23\times 2=0.0591{\log }_{10}{K}_c$$
$$\Rightarrow 7.77={\log }_{10}{K}_c$$
$$\Rightarrow K_c=anti\log (7.77)$$
$$\Rightarrow K_c=5.88\times {10}^7$$

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.