What is the overall voltage for a redox reaction with the half reactions #Ag^+ + e^(-) -> Ag(s)# and #Cu(s) -> Cu^(2+) + 2e^-#?

1 Answer
Mar 29, 2018

Answer:

#E=0.460V+ 0.0296Log(([Ag^(2+)])^2/([Cu^(2+)]))#

Explanation:

the overall voltage dipends from the metals, but also by the concentration of the ions in the solution and eventually fron the anions that can reduce the activity of the cations, making coordination bonds:
# E_(Ag)= E°_(Ag) + 0,0592 Log([Ag^(2+)])=0,800V+ 0,0592 Log([Ag^+]#

# E_(Cu)= E°_(Cu) + 0,0592 Log([Cu^+])=0,340V+ 0,0592/2 Log([Cu^(2+)]#

#Delta E= E_(Ag)-E_(Cu)=(0.800-0.340)V + 0.0296Log(([Ag^(2+)])^2/([Cu^(2+)]))#=

#0.460V+ 0.0296Log(([Ag^(2+)])^2/([Cu^(2+)]))#