Test 2 Question 2

[Caroline C 1G]

Question 2 of Test 2 asked "Calculate K for the following redox reaction: Sn^2+(aq)+Pb(s) yields Pb^2+(aq)+Sn(s). I calculated E knot and got -.01V. Was it assumed that the reaction was in equilibrium since we were calculating K, and that therefore, E=0 when using the equation E=Eknot-(.05916/n)logQ?

[Gwen Peng 1L]

Yes it was assumed it was in equilibrium so you could've solved using the equation E standard=(0.05916/n)*log(K) or E standard=(RT/nF)*ln(K)
Hope this helps :)

[Caroline C 1G]

Yes it was assumed it was in equilibrium so you could've solved using the equation E standard=(0.05916/n)*log(K) or E standard=(RT/nF)*ln(K)
Hope this helps :)

Would the .05916/n term be negative?

[Gwen Peng 1L]

I don't believe so because E= E standard - (0.05916/n)*log(K) and if E= 0 then you would move E standard to the other side to have -E standard= -(0.05916/n)*log(K) and the negatives would cancel.