The standard free energy change ( G^ ) for the following reaction (in kJ ) at 25^ C is 3 Ca ( s )+2 Au ³⁺ (aq. 1 M ) 3 Ca ²⁺( aq , 1 M )+2 Au ( s ) (given : E _ Au ³⁺ / Au ^ o =+1.50 ~V , E _ Ca ²⁺ / Ca ^0=-2.87 ~V , 1 ~F =96500 C mol ⁻¹ )

Step-by-step solution

aligned & 3 Ca ( ~s )+2 Au ³⁺( aq 1 M ) 3 Ca ²⁺( aq 1M )+2 Au ( ~S ) & E ^0= E _ L + E _ R & = E _ Ca / Ca ²⁺ + E _ Au ³⁺ / Au & =2.87+1.50 & E ^ o =4.37 ~V aligned Total no. of electron envolved by balancing the electron. aligned & n =6 & G ^ o =- nFE ^ o & =-6 96500 4.37 ~mol C ~mol ⁻¹ ~V & =-2530230 CV [1 ~J =1 CV ] & =-2530.23 ~kJ [1 ~kJ =10^3 ~J ] & G ^ o =-2.53 10^3 ~kJ aligned