the following values may be useful when solving this tutorial. Constant Value EC
ID: 996612 • Letter: T
Question
the following values may be useful when solving this tutorial. Constant Value ECu 0.337 V EZn -0.763 V R 8.314 Jmol1K1 F 96,485 C/mol T 298 K Part A In the activity, click on the Ecell and Keq quantities to observe how they are related. Use this relation to calculate Keq for the following redox reaction that occurs in an electrochemical cell having two electrodes: a cathode and an anode. The two half-reactions that occur in the cell are Cu2+(aq)+2eCu(s) and Zn(s)Zn2+(aq)+2e The net reaction is Cu2+(aq)+Zn(s)Cu(s)+Zn2+(aq) Use the given standard reduction potentials in your calculation as appropriate. Express your answer numerically to three significant figures.
Explanation / Answer
The half-reactions are:
Cu2+ + 2e- = Cu(s)
Zn(s) = Zn2+ + 2e-
The overall reaction is:
Cu2+ + Zn(s) = Zn2+ + Cu(s)
Information:
1) Determine which reaction occurs in the anode and which reaction occurs in the anode. As a rule, reduction occurs in the cathode and oxidation occurs in the anode:
Cu2+ + 2e- = Cu(s) (Reduction = Cathode)
Zn(s) = Zn2+ + 2e- (Oxidation = Anode)
2) Calculate E0cell using the formula:
E0cell = E0cathode - E0anode
Where: E0cathode = E0 (Cu2+/Cu) and E0anode = E0 (Zn2+/Zn).
E0cell = 0.337V - (-0.763V)
E0cell = 1.100V
3) To calculate Keq use the formula:
E0cell = (RT/nF)lnKeq
Where n = the number of electrons in the half reactions (n = 2).
Solve for Keq:
Keq = e(nFEcell) / (RT)
Keq = e(2)(96485C/mol)(1.100V)/(8.314 J/molK)(298K)
Keq = 1.62 x 1037 (3 significant figures)
Answer: The Keq for the reaction is 1.62 x 1037
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