Verifying Nernst Equation Calculation: Pb2+ and Ag+

[higherme]

Can anyone check if I am doing this right?

Given:
Pb2+ + 2e- ---> Pb(s) E standard = -0.13V
Ag+ + 1e- ---> Ag(s) E standard = 0.80V

[Pb2+] = 0.05M
[Ag+] = 0.5 M these are non standard concentrations

Temp = 298K

Using the Nernst equation, find E


My answer:

Pb oxidized and Ag is reduced ( is this right?... because Ag has the higher reduction potential compared to Pb)

E = E standard - (RT/nF) ln Q

Q = [products]^p/[Reactants]^r

the reaction is: 2Ag+ + Pb(s) + 2e- ---> 2Ag(s) + Pb2 + 2e- (the electrons cancels out)

therefore, Q = (0.05M) / (0.5)^2

Q= 0.20

the E standard = 0.80 - (-0.13) = 0.80 + 0.13 = 0.93V
E = E standard - (RT/nF) ln Q
E = 0.93 - (8.314*298K/2*9.649E4) ln (0.20)
E= 207.56 V

Can anyone check the calculations for me and see if I am doing this right please?

Thank you!

 

[Gokul43201]

Can anyone check if I am doing this right?

Given:
Pb2+ + 2e- ---> Pb(s) E standard = -0.13V
Ag+ + 1e- ---> Ag(s) E standard = 0.80V

[Pb2+] = 0.05M
[Ag+] = 0.5 M these are non standard concentrations

Temp = 298K

Using the Nernst equation, find E


My answer:

Pb oxidized and Ag is reduced ( is this right?... because Ag has the higher reduction potential compared to Pb)

E = E standard - (RT/nF) ln Q

Q = [products]^p/[Reactants]^r

the reaction is: 2Ag+ + Pb(s) + 2e- ---> 2Ag(s) + Pb2 + 2e- (the electrons cancels out)

therefore, Q = (0.05M) / (0.5)^2

Q= 0.20

the E standard = 0.80 - (-0.13) = 0.80 + 0.13 = 0.93V
E = E standard - (RT/nF) ln Q
E = 0.93 - (8.314*298K/2*9.649E4) ln (0.20)

Looks good up to here.

E= 207.56 V

That's way off. Chug those numbers again.