Write a 125 word summary of what you have learned in "Introducing the Microscope". Be sure to include steps on how to focus, how
2 answers:
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Answer :
(a) The concentration of is, 0.0337 M
(b) The concentration of is,
Solution :
<u>(a) As per question, lead is oxidized and copper is reduced.</u>
The oxidation-reduction half cell reaction will be,
Oxidation half reaction:
Reduction half reaction:
The balanced cell reaction will be,
Here lead (Pb) undergoes oxidation by loss of electrons, thus act as anode. Copper (Cu) undergoes reduction by gain of electrons and thus act as cathode.
First we have to calculate the standard electrode potential of the cell.
Now we have to calculate the concentration of .
Using Nernest equation :
where,
n = number of electrons in oxidation-reduction reaction = 2
= 0.507 V
Now put all the given values in the above equation, we get:
Therefore, the concentration of is, 0.0337 M
<u>(b) As per question, lead is reduced and copper is oxidized.</u>
The oxidation-reduction half cell reaction will be,
Oxidation half reaction:
Reduction half reaction:
The balanced cell reaction will be,
Here Copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.
First we have to calculate the standard electrode potential of the cell.
Now we have to calculate the concentration of .
Using Nernest equation :
where,
n = number of electrons in oxidation-reduction reaction = 2
= 0.507 V
Now put all the given values in the above equation, we get:
Therefore, the concentration of is,