Standard cell potentials are determined with 1.0 M solutions of ions. In this ex

Question

Standard cell potentials are determined with 1.0 M solutions of ions. In this experiment, we will use 0.10 M solutions of Zn2+, Cu2+, Pb2+, and Ag+ to minimize the amount of hazardous waste generated. In this series of questions, you will figure out what effect this will have on the voltages you observe.

Consider a cell consisting of a Cu2+/Cu couple and a Ni2+/Ni couple.

(a) Starting with 1.0 M solutions of ions (standard conditions), evaluate Q in the Nernst Equation.
Q =  

(b) What is log Q for this cell?
log Q =  


(c) The cell potential in this case should be which of the following?

greater than E°

less than E°

   equal to E°

(d) Starting with 0.1 M solutions of ions (at room conditions), evaluate Q in the Nernst Equation.
Q =  

(e) What is log Q for this cell?
log Q =  

(f) The cell potential in this case should be which of the following?

greater than E°

less than E°  

  equal to E°

Explanation / Answer

a)

Cu2+/Cu --> 0.1

Ni2+/Ni --> 0.1

Q = (0.1)/(0.1) = 1

Q = 1

b)

Log(1) = 0

c)

equal to E°

since

E = E° -0.0592/n*logQ

And logQ = 0 so the product is 0

E = E°

d)

Same as in a

e)

LogQ = log(1) = 0

f)

should be equal to that

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