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Question: For the cell \[{\text{Mg}},M{g^{2 + }}{\text{ '' A}}{{\text{g}}^{2 + }}{\text{, Ag }}\]. Calculate t...

For the cell Mg,Mg2+ ” Ag2+, Ag {\text{Mg}},M{g^{2 + }}{\text{ '' A}}{{\text{g}}^{2 + }}{\text{, Ag }}. Calculate the equilibrium at 25OC{25^O}C and the maximum work that can be obtained during the operation of the cell
(E Mg,Mg2+=2.37 and Ag2+,Ag=0.80VE{\text{ Mg}},M{g^{2 + }} = 2.37{\text{ and A}}{{\text{g}}^{2 + }},{\text{Ag}} = 0.80V).

Explanation

Solution

The equilibrium constant of the system is defined as the value of reaction quotient of the chemical reaction at their chemical equilibrium. Most of the dynamic chemical reactions attain this equilibrium constant after some time has elapsed.

Complete answer:
Above we studied the definition of chemical equilibrium constant, now let’s learn what is KK. KKIs the equilibrium constant and it represents the relation between the reaction and product concentration over a range of temperature and in its specific units of measurement.
Now, to calculate the KK , the equilibrium constant, we have to derive the equation in terms of the emf (electromotive force) of the cell.
Thus we know the formula. Putting the values in equation,
Eocell  =  0.0592  logK{E^o}cell\; = \;\dfrac{{0.059}}{2}\;logK
  0.80 + 2.37 = 0.0295  logK\;0.80{\text{ }} + {\text{ }}2.37{\text{ }} = {\text{ }}0.0295\;logK
logK  = 107.45logK\; = {\text{ }}107.45
K  = 2.86  ×  10107K\; = {\text{ }}2.86\; \times \;{10^{107}}
So, we obtained our result i.e. our desired equilibrium constant, KK isK  = 2.86  ×  10107K\; = {\text{ }}2.86\; \times \;{10^{107}}.
Let’s move to the next part of the question, which is to find the maximum work that can be obtained during the operation of the cell.
This work done can also be calculated by the formula consisting of an emf of the cell and faraday constant that has fixed value and plus the number of electrons transferring.
Hence, putting the values,
Wmax  = nFEocell{W_{max}}\; = {\text{ }}nF{E^o}cell
Wmax= 2×96500×3.17{W_{max}} = {\text{ }}2 \times 96500 \times 3.17
Wmax= 6.118  ×105  J{W_{max}} = {\text{ }}6.118\; \times {10^5}\;J

So, Our Wmax= 6.118  ×105  J{W_{max}} = {\text{ }}6.118\; \times {10^5}\;J

Note:
EMF or electromotive force is defined as the maximum potential difference between anode and cathode of the galvanic or voltaic cell. It tells us about the ion gaining or losing capacity of an element.