Question

Can anyone please solve this for me?

Answer 1

Answer:

0.936V

Explanation:

Since reduction potential of cathode of Ag is greater than that of Ni;

hence Ag+/Ag is the cathode reaction whereas Ni/Ni2+ is the anode half cell reaction.

Eo= reduction potential of (cathode-anode)=0.80+0.25=1.05V

E=Eo-$\frac{0.0591}{n}$log(Qc)

E=1.05-$\frac{0.0591}{2}$log($\frac{0.08}{0.00001}$)

 =1.05-(0.029)(3.9)

 =0.936V

Hope it helps if so then mark this solution as BRAINLIEST:)