The standard e.m.f. of a galvanic c...
Question  # 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

NEET/Medical Exams
Chemistry
Solution 385 4.0 (1 ratings)  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} ) Quick and Stepwise Solutions Just click and Send OVER 20 LAKH QUESTIONS ANSWERED Download App for Free