Question

Constant	Value
Ecu	0.337
Ezn	-7.63
R	8.314
F	96,485
T	298

In the activity, click on the E∘cell 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)+2e−→Cu(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.

Answer 1

Cu⁺² + 2e- ===> Cu +0.337 V

Zn⁺² + 2e- ===> Zn -0.763 V

E⁰ CelI = E⁰ (cathode) – E⁰(anode)

            =+0.337 V – ( - 0.763V) = 1.10V

E⁰ = RT/nF*(lnQ)

R = 8.314 J⋅mol−1⋅K−1

F= 96,485 C/mol

T = 298 K

Q=K; putting values for each standard above equation can be written as

E⁰ cell = 0.0592/n log K

Where n is the number of electrons transferred, in this case n = 2

log K = n * E⁰ cell/0.0592

         = (2) (1.10)/0.0592

         = 37.126

Keq = antilog (37.162)

        = 10^(37.162)

       = 1.45 x 10^37