Question

calculate the EMF of the following cell Mg/Mg²+ (0.1m) // Ag+(1×10-⁴/ Ag given that, E°(Ag+/Ag) = +0.80 v E°(Mg²+/Mg) = -2.37 v nn. what will be the effect on EMF is connect of AG Plus I am is increased to 1×10-³​

Answer 1

Explanation:

E=E0-0.059/n* log(p/r)

E0=0.80-(-2.37)

E0 =3.17

E=3.17-0.059/2log(0.1/[1*10^-4]^2)

=3.17-0.059/2log(10^7)

=3.17-0.059*7/2

=3.17-0.2062

E=2.964

if Ag+=1*10^-3

E=3.17-0.059/2log(0.1/10^-6)

=3.17-0.059/2log(10^5)

=3.17-0.059/2*5

=3.17-0.147

E =3.023