Question

Construct galvanic cells from the following pairs of half cells and calculate their emf at $25^\circ C$.
$\;F{e^{3 + }}\left( {0.1M} \right),F{e^{2 + }}\left( {1M} \right)\left( {Pt} \right)\;\;\;E_{F{e^{3 + }},F{e^{2 + }}}^0 = 0.77\;volt$
And $AgCl\left( s \right),C{l^ - }\left( {0.001M} \right)|\;Ag\;\;\;\;E_{AgCl,C{l^ - }}^0 = 0.22\;volt$

Answer 1

Hint:In order to construct a galvanic cell, first of all, we should understand where oxidation happens and where does reduction. Then we can easily construct the galvanic cell.

Complete answer:
We have given two cells $F{e^{3 + }},F{e^{2 + }}$and their reduction potential is 0.77 volt. Also, we are adding $Pt$ to an inert electrode that is necessary when there is no solid metal on the sides. The other half cell $AgC{l_2}\left( s \right),C{l^ - }\left( {0.001M} \right)|\;Ag$ with an electrode potential of 0.22 volt. Since the electrode potential of $F{e^{3 + }},F{e^{2 + }}$ is high it is more likely to reduce and hence we consider this as our anode. Therefore the cell $AgC{l_2}\left( s \right),C{l^ - }\left( {0.001M} \right)|\;Ag$ becomes our anode.
Now let’s look into the half-cell reaction
At cathode $F{e^{3 + }}$ gets reduced to give $F{e^{2 + }}$
$F{e^{3 + }} + {e^ - } \to F{e^{2 + }}$
At the anode, Ag gets oxidised to $AgC{l}$
$Ag + C{l^ - } \to {\text{AgCl}} + {{\text{e}}^ - }$
Since here oxidation happens the reduction potential get reversed
Therefore, $E_{AgCl,C{l^ - }}^0 = - 0.22\;volt$
Thus the final reaction becomes
$F{e^{3 + }} + Ag + C{l^ - } \to F{e^{2 + }} + AgCl$
Thus this is our galvanic cell.
Now to find their emf,
First, we have to find their emf of standard cell
$E_{cell}^0 = E_{cathode}^0 - E_{anode}^0$
i.e.,$E_{cell}^0 = E_{F{e^{3 + }},F{e^{2 + }}}^0 - E_{AgCl,C{l^ - }}^0$
Substituting the value, we get
$E_{cell}^0 = 0.77 - \left( { - 0.22} \right) = 0.99$
Now, we can use the Nernst equation for finding the emf of the cell
Nernst equation for finding single electrode potential.
 ${E_{cell}} = E_{cell}^0 - \left[ {\dfrac{{RT}}{{nF}}} \right]ln\left( Q \right)$
Where,
${E_{cell}}$ is the cell potential of the cell
$E_{cell}^0$ is the cell potential under standard conditions
$R$ is the universal gas potential
$T$ is the temperature
$n$ is the number of electrons transferred in the reaction
$F$ is the Faraday constant
$Q$ is the reaction constant
At standard conditions, we can use the formula
${E_{cell}} = E_{cell}^0 - \dfrac{{0.059}}{n}{\log _{10}}\left( Q \right)$
From the reaction, we can get $Q$
$\Rightarrow {E_{cell}} = E_{cell}^0 - \dfrac{{0.059}}{n}{\text{lo}}{{\text{g}}_{10}}\left( {\dfrac{{\left[ {F{e^{2 + }}} \right]}}{{\left[ {F{e^{3 + }}} \right]\left[ {C{l^ - }} \right]}}} \right)$
From the final reaction, we got $n = 1$,
And given that concentration of $\left[ {F{e^{3 + }}} \right] = 0.1M,\;\left[ {F{e^{2 + }}} \right] = 1M\;and\;\left[ {C{l^ - }} \right] = 0.001M$
Substituting everything to the reaction we get
$\Rightarrow {E_{cell}} = 0.99 - \dfrac{{0.059}}{1}{\text{lo}}{{\text{g}}_{10}}\left( {\dfrac{1}{{0.1 \times 0.001}}} \right)$
$ \Rightarrow {{\text{E}}_{{\text{cell}}}} = 0.99 + 0.059{\text{lo}}{{\text{g}}_{10}}\left( {{{10}^{ - 1}} \times {{10}^{ - 3}}} \right)$ [Since${\text{lo}}{{\text{g}}_{10}}\left( {\dfrac{1}{a}} \right) = - {\text{lo}}{{\text{g}}_{10}}\left( a \right)$]
$ \Rightarrow {{\text{E}}_{{\text{cell}}}} = 0.99 + 0.059{\text{lo}}{{\text{g}}_{10}}\left( {{{10}^{ - 4}}} \right)$
Applying the properties of log
$ \Rightarrow {{\text{E}}_{{\text{cell}}}} = 0.99 - 4 \times 0.059{\text{lo}}{{\text{g}}_{10}}\left( {10} \right) = 0.99 - 0.236$
That is ${{\text{E}}_{{\text{cell}}}} = 0.75volt$

The emf of a cell constructed has an emf of 0.75volt at $25^\circ C$.

Note:
While constructing the galvanic cell, we should also use a salt bridge with an inert electrolyte mixed with substances like agar-agar or gelatine which makes it jelly-like. It mainly completes the electrical circuit and maintains the electrical neutrality of electrolytes in two half-cells by allowing the movement of ions from one solution to the other without mixing of the two solutions.