Answer to Question #71184 in General Chemistry for Kate

Question

Calculate the reduction potential of the Cu2+/Cu electrode when [Cu2+] = 1.0 ´ 10-8 M.
a.
+0.10 V
b.
+0.37 V
c.
+0.33 V
d.
+0.34 V
e.
+0.35 V

Answer 1

It should be used the Nernst equation:
E = E0 + (0.05916V )/z log[Cu2+]
For Cu2+/Cu electrode:
E0 = +0.34 V and z = 2
Therefore: E = 0.34 + 0.03 (-8) = +0.10 V
Answer:
E = +0.10 V