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?

a. greater than E°

b. less than E°    

c. 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?

a. greater than E°

b. less than E°    

c. equal to E°

Explanation / Answer

a)
From web:
Eo(Cu2+/Cu) = 0.34 V
Eo(Ni2+/Ni) = -0.25 V

So, Here Cu2+/Cu will acts as cathode.

Eo = Eo cathode - Eo anode
      = 0.34 - (-0.25)
      = 0.59 V

Reaction taking place will be:
Ni + Cu2+   -----> Ni2+    +   Cu
Here number of electron transfered, n=2

uSe:
Q = [Ni2+/Cu2+]
     = 1/1
      =1

b)
logQ = log (1)
           =0

c)
usE:
E= Eo - 0.059/2 log Q
since log Q=0
E = EO
Answer: Equal to Eo

d)
uSe:
Q = [Ni2+/Cu2+]
     = 0.1/0.1
      =1

e)
logQ = log (1)
           =0

f)
usE:
E= Eo - 0.059/2 log Q
since log Q=0
E = EO
Answer: Equal to Eo

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