Question

For the cell Mg(s)/Mg2+(aq)//Ag+(aq)/Ag(s) calculate the equilibrium constant of the cell reaction ay25*C and maximum work that can be obtained by operating the cell reaction at 25*C and maximum work that can be obtained by operating the cell E*Mg2+=-2.37V and E*Ag+/Ag=+0.80.​

Answer 1

Answer : The value of $K_c$ at 298 is, $1.14\times 10^{11}$  and the maximum work is, -631110 J/mole

Solution :

The balanced cell reaction will be,  

$Mg(s)+2Ag^{+}(aq)\rightarrow Mg^{2+}(aq)+2Ag(s)$

Here magnesium (Mg) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

$E^0_{[Mg^{2+}/Mg]}=-2.37V$

$E^0_{[Ag^{+}/Ag]}=0.80V$

$E^0=E^0_{[Ag^{+}/Ag]}-E^0_{[Mg^{2+}/Mg]}$

$E^0=0.80V-(-2.37V)=3.27V$

Now we have to calculate the Gibbs free energy.

Formula used :

$\Delta G^o=-nFE^o$

where,

$\Delta G^o$ = Gibbs free energy = maximum work = ?

n = number of electrons = 2

F = Faraday constant = 96500 C/mole

$E^o$ = standard e.m.f of cell = 3.27 V

Now put all the given values in this formula, we get the Gibbs free energy.

$\Delta G^o=-(2\times 96500\times 3.27)=-631110J/mole$

Now we have to calculate the value of equilibrium constant.

Formula used :

$\Delta G^o=-2.303\times RT\times \log K_c$

where,

R = universal gas constant = 8.314 J/K/mole

T = temperature = [tex]25^oC=273+25=298K [/tex]

$K_c$ = equilibrium constant = ?

Now put all the given values in this formula, we get the value of $K_c$

$-631110J/mole=-2.303\times (8.314 J/K/mole)\times (298K)\times \log K_c$

$K_c=1.14\times 10^{11}$

Therefore, the value of $K_c$ at 298 is, $1.14\times 10^{11}$