Question

At equimolar concentration of $F{e^{2 + }}$ and $F{e^{3 + }}$ what must be $\left[ {A{g^ + }} \right]$ so that the voltage of the galvanic cell made from $A{g^ + }/Ag$ and $F{e^{3 + }}/F{e^{2 + }}$ electrode becomes zero.
$F{e^{2 + }} + A{g^ + } \to F{e^{3 + }} + Ag$
${E^ \circ }_{F{e^{3 + }}/F{e^{2 + }}} = 0.77V$
${E^ \circ }_{A{g^ + }/Ag} = 0.80V$
$\dfrac{{2.303RT}}{F} = 0.06$
A.$anti\log ( - 0.5)$
B.$anti\log (1.03)$
C.$anti\log (0.1)$
D.$anti\log ( - 0.02)$

Answer 1

Whenever voltage of any galvanic cell becomes zero the concentration of the species involved in the reaction becomes equal to their equilibrium concentration and thus Nernst equation can also be used to find out the value of equilibrium constant of any reaction
Formula used:
${E_{cell}} = {E^ \circ }_{cell} - \dfrac{{2.303RT}}{{nF}}{\log _{10}}\dfrac{{\left[ {F{e^{3 + }}} \right]}}{{\left[ {F{e^{2 + }}} \right]\left[ {A{g^ + }} \right]}}$
Here
${E_{cell}}$ is electrode potential of cell
${E^ \circ }_{cell}$ is electrode potential of cell define at standard temperature of $298K$ and at standard concentration of $1mol{L^{ - 1}}$
$R$is universal gas constant
$F$ is faraday’s constant having value 96500 $Cmo{l^{ - 1}}$
$T$ is temperature in kelvin
$n$ is no of moles of electrons gained by one mole of oxidized state to get changed into the reduced state in the process of reduction occurring at the electrode

Complete step by step answer:
After seeing the chemical equation it is cleared that iron is getting oxidize and silver is getting reduce
Hence for the given cell
${E^ \circ }_{cell} = {E^ \circ }_{Ag + /Ag} - {E^ \circ }_{F{e^{3 + }}/F{e^{2 + }}}$
${E^ \circ }_{cell} = 0.80 - 0.77$
${E^ \circ }_{cell} = 0.03V$
Now as it is mention that we have equimolar concentration of$F{e^{3 + }}$ and $F{e^{2 + }}$
That means
$\left[ {F{e^{3 + }}} \right] = \left[ {F{e^{2 + }}} \right]$
Also $\dfrac{{2.303RT}}{F} = 0.06$ (given)
${E_{cell}} = 0$ (given)
And n for the reaction is one since there is net transfer of one moles of electrons

Now as we know
${E_{cell}} = {E^ \circ }_{cell} - \dfrac{{2.303RT}}{{nF}}{\log _{10}}\dfrac{{\left[ {F{e^{3 + }}} \right]}}{{\left[ {F{e^{2 + }}} \right]\left[ {A{g^ + }} \right]}}$
Putting all the values in the Nernst equation mentioned above we get
$- 0.03 = - 0.06{\log _{10}}\dfrac{1}{{\left[ {A{g^ + }} \right]}}$
Finally by taking antilog on both sides we get $\left[ {A{g^ + }} \right]$ as antilog $( - 0.5)$ making option “A” as the correct option.

Additional Information: A cell is characterized by its voltage. A particular kind of cell generates the same voltage irrespective of the size of the cell. The only thing that depends on the cell voltage is the chemical composition of the cell, given the cell is operated at ideal conditions.

Note: In the given question as we see the cell emf was zero hence the free energy change would also be zero for this reaction thus at these conditions the cell exists at an equilibrium state .