3050/CH6/EX6.4/Ex6_4.sce


1
2
3
4
5
6
7
8
9
10
11

//calculating potential
//Example 6.4
clc
clear
Ag=1
Ksp=8.7*(10^-17)
Agplus=sqrt(Ksp)
X=Ag/Agplus
E1=0.799//E(Ag+/Ag)
E2=E1-(0.0591*log10(X))
printf('Thus E(Ag+/Ag) = %2.3f V',E2)