Cu (s)+ Sn ²⁺(0.01 M ) Cu ²⁺(0.1 M )+ Sn (s) The gibbs free energy change for above mention reaction at NTP is x 10⁻¹ ~kJ / mol . The value of x is [Nearest integer] [ . Given E_ Cu ²⁺ / Cu ^ =0.34 ~V , E_ Sn ²⁺ / / sn ^ =-0.14 ~V F=96500 C / mol ]

Step-by-step solution

Given, aligned & Cu (s)+ Sn ²⁺(0.01 M ) Cu ²⁺(0.1 M )+ Sn (s) & E_ cell ^ =E_ cathode ^ -E_ anoode ^ & =-0.14-(0.34)=-0.48 ~V & E_ cell =E_ cell ^*- 0.059 n [ Cu ²⁺ ] [ Sn ²⁺ ] [n=2] & =-0.48- 0.059 2 [0.1] 0.01 & =-0.48-0.0290 [10]=-0.5095 ~V aligned Gibb's free energy, aligned G & =-n F E_ cell & =+2(96500) 0.5095 & =98.33 10^3 ~J / mol aligned or 983 10⁻¹ ~kJ / mol Hence, the value of x is 983 .