If the Fe2+(aq) + 2e---> Fe(s) was used as a reference for the electrode potential table instead of the standard hydrogen half cell what would the potential of the following half cell be?
Ni2+(aq) + 2e---> Ni(s)
A) -0.19 V
B) -0.71 V
C) +0.19 V
D) +0.71 V
My tables may not be the same as your tables. I'll use mine and you can check yours. When you post questions like this you should ALWAYS LOOK UP THE VALUES in your text/notes. The best way to do this type problem is to draw a diagram. That's difficult to do on this site but here's the best I can do. With reference to H2 the diagram shows:
2H^+ + 2e ---> H2..................0
Ni^2+ + 2e ==> Ni.................-0.23
Fe^2+ + 2e ==> Fe................-0.41
What I've done is looked up IN MY REFERENCES the reduction potential for H, Ni and Fe. I've arranged them in order. So now if Fe is the standard you can see that the Ni potential is ABOVE Fe; therefore, Ni potential will be + the difference. The difference is 0.41-0.23 = 0.18 and it is 0.18 ABOVE Fe so it will be 0.18 referenced to Fe potential.