ECu 0.337V EZn -0.763 V R 8.314 Jmol1K1 F 96,485 C/mol T 298 K Part A In the act

Question

ECu 0.337V

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

Cu2+ + 2 e Cu(s) +0.337

Zn2+ + 2 e Zn(s) 0.7618

Cu is reducing, Zn oxidizing

Ecell = Ered - Eox

Ecell = (0.337) - (-0.7618) = 1.0988 V

3 sig fig -- 1.10 V

For dg:

dG = -nf*ECell

dG = -2*96500*(1.1) = -212,300 J = -212.3 kJ

For K

dG = -RT*ln(K)

-212,300 = -8.314*298*ln(K)

K = exp(212300/(8.314*298)) = 1.637*10^37

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