Question

In which of the following cases , increase in concentration of ion cause increase in ${E_{cell}}$ ?
A. $Pt({H_2}){H^ + }(aq)$
B. $Pt$ |Quinhydrone|${H^ + }(aq)$
C. $Ag$ |$A{g^ + }(aq)$
D. $Ag,AgCl$| $C{l^ - }(aq)$

Answer 1

Hint: ${E_{cell}}$ represents the difference between the electrode potentials of the two half cells . When we increase the concentration of ions we observe a change in the value of cell potential .

Complete step by step answer:
We will determine this with the help of Nernst equation.
Nernst equation is the relationship between electrode potential, cell potential and the concentration of reacting species.
Now, we must be fully aware of the terms electrode potential and cell potential.
So, let us understand this with the help of an example.
When we dip a zinc rod in $ZnS{O_4}$ solution some zinc metal dissolves and some $Z{n^{2 + }}$ ions get deposited as $Zn$ . So a potential difference is created between the metal and its salt solution. This is known as electrode potential. It is represented as ${E_{cell}}$ .
In the Nernst equation, we deal with the standard electrode potential, that is, the metal and the salt solution are taken at $298K$ temperature and have the concentration of unity .It is represented as ${E^ \circ }_{cell}$ .
Cell potential is the potential between both the electrodes of a galvanic cell.
The nernst equation for the galvanic cell is-
${E_{cell}} = {E^ \circ }_{cell} - \dfrac{{0.059}}{n}\log [K]$
where, n=no. of electrons used in the reaction
K = concentration of products/concentration of reactants.
For the cell represented in option D,
${E_{cell}} = {E^ \circ }_{cell} - \dfrac{{0.059}}{1}\log \dfrac{1}{{C{l^ - }}}$
that is,
${E_{cell}} = {E^ \circ }_{cell} + 0.059\log [C{l^ - }]$
In this cell, an increase in concentration of ions causes an increase in cells.

So, the correct answer is Option D .

Note:
In Nernst equation , the term involving log is the same as the expression for equilibrium constant when there is a minus sign between the two terms on the right hand side .