Question

Calculate the EMF of the cell, \[Zn - Hg({c_1}M)|Z{n^{2 + }}(aq)|Hg - Zn({c_2}M)\] at $ {25^ \circ }C $ , if the concentration of the zinc amalgam are: $ {c_1} = 10g $ per $ 100g $ of mercury and $ {c_2} = 1g $ per $ 100g $ of mercury.

Answer 1

Hint: The given cell is a concentration cell where the concentration differs which are already provided. The half cells have the same electrode which means that the standard electrode potential of the cell will be zero.

Complete step by step answer:
The above given cell is an example of a concentration cell which means that the two half cells have the same electrode but their concentration is different. In case of concentration cells the cell dilutes the more concentrated and dilutes the more concentrated cell creating a voltage as the cell reaches equilibrium. This process is achieved by the transfer of electrons from the lower concentration cell to the higher concentration cell. Now, we will first write the half-cell reaction for the given cell representation. We know that in case of electrochemical cell oxidation takes place at anode and reduction takes place at cathode, so our half-cell reactions will be:
Anode half-cell $ Zn - Hg({c_1}M) \to Z{n^{2 + }} + 2{e^ - } $ and for this standard electrode potential is $ E_{Zn/Z{n^{2 + }}}^ \circ $ .
Cathode half-cell $ Z{n^{2 + }} + 2{e^ - } \to Zn - Hg({c_2}M) $ and for this standard electrode potential is $ E_{Z{n^{2 + }}/Zn}^ \circ $ .
Now for this we can write the Nernst equation which is:
 $ {E_{cell}} = E_{cell}^ \circ + \dfrac{{0.059}}{n}\log \dfrac{{[{c_2}M]}}{{[{c_1}M]}} $ at $ {25^ \circ }C $
And the $ E_{cell}^ \circ = E_{Zn/Z{n^{2 + }}}^ \circ + E_{Z{n^{2 + }}/Zn}^ \circ = E_{Zn/Z{n^{2 + }}}^ \circ - E_{Zn/Z{n^{2 + }}}^ \circ = 0 $ because $ E_{Zn/Z{n^{2 + }}}^ \circ = - E_{Z{n^{ + 2}}/Zn}^ \circ $
So, $ E_{cell}^ \circ = 0 $ and $ n $ is the number of electrons involved in reaction, so $ n = 2 $ . Now put all these values in the Nernst equation we can get $ {E_{cell}} $ .
 $ {E_{cell}} = \dfrac{{0.059}}{2}\log \dfrac{1}{{10}} = - 1.114V $ .
So our $ {E_{cell}} $ is $ - 1.114V $ which was asked in the question.

Note:
For any given cell representation always write the half-cell reaction first and those should be completely balanced and for normal cells write the complete reactions also. While doing the calculations always remember the units in which calculations are done.