Question

First we have to calculate the Gibbs free energy.
Formula used :
( Delta G^{o}=-n F E^{o} )
where,
( Delta G^{o}= ) Gibbs free energy ( =? )
( mathrm{n}= ) number of electrons ( =2 )
( mathrm{F}= ) Faraday constant ( =96500 mathrm{C} / mathrm{mole} )
( E^{o}= ) standard e.m.f of cell ( =0.295 mathrm{V} )
Now put all the given values in this formula, we get the Gibbs free
energy.
( Delta G^{o}=-(2 times 96500 times 0.295)=-56935 J / ) mole
Now we have to calculate the value of equillbrium constant.
Formula used :
( Delta G^{o}=-2.303 times R T times log K_{c} )
where,
( mathrm{R}= ) universal gas constant ( =8.314 mathrm{J} / mathrm{K} / mathrm{mole} )
( mathrm{T}= ) temperature ( =298 mathrm{K} )
( K_{c}= ) equilibrium constant ( =? )
Now put all the given values in this formula, we get the value of ( K_{c} )
( -56935 J / ) mole ( =-2.303 times(8.314 . J / K / text { mole }) times(298 K) times log K_{c} )
( K_{c}=1.0 times 10^{10} )
Therefore, the value of ( K_{c} ) at 298 is, ( 1.0 times 10^{10} )

# The standard e.m.f. of a galvanic cell involving cell reaction with n = 2 is found to be 0.295 V at 25°C. The equilibrium constant of the reaction would be (Given F= 96500 C mol-1, R= 8.314 J K-1 mol-1) (1) 2.0 * 1011 (2) 4.0 × 1012 (3) 1.0 × 102 (4) 1.0 × 1010

Solution