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higherme
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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!
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!