Write the equation for each metal combination indicating which metal is being ox
ID: 1010713 • Letter: W
Question
Write the equation for each metal combination indicating which metal is being oxidized and which metal is being reduced. Using the Table above calculate the E E^o_cell for Zn(s)|Zn^2+(aq)||Cu^2+(aq)|Cu(s) Show your work. Using the Nernst equation calculate the E for the cell when Zn^2+ (aq) = 0.10 M and Cu^2+ (aq) = 0.10M Show your work. How does your value in answer #3 compare to what you measured? Using the Nernst equation calculate the E for the cell when Zn^2+ (aq) = 0.10 M and Cu^2+ (aq) = 0.001 M Show your work. How does your value in answer #5 above compare to what you measured? E^o_cell = E^o_cathode - E^o_anode Nernst Equation E = E^o_cell - 0.059/n Times log QExplanation / Answer
1) we know that
oxidation takes place at anode
reduction takes place at cathode
so
a) Zn + Cu+2 --> Zn+2 + Cu
oxidized --> Zn
reduced --> Cu+2
b)
3Zn + 2Al+3 --> 3Zn+2 + 2Al
oxidized --> Zn
reduced --> Al+3
c)
Zn + Pb+2 --> Zn+2 + Pb
oxidized --> Zn
reduced --> Pb+2
d) Pb + Cu+2 --> Pb+2 + Cu
oxidized --> Pb
reduced --> Cu+2
e)
3Pb + 2Al+3 --> 3Pb+2 + 2Al
oxidized --> Pb
reduced --> Al+3
f)
3 Cu + 2 Al+3 --> 3 Cu+2 + 2Al
oxidized --> Cu
reduced --> Al+3
g)
Zn + Cu+2 --> Zn+2 + Cu
oxidized --> Zn
reduced --> Cu+2
2)
we know that
Eo cell = Eo cathode - Eo anode
Eo cell = Eo Cu+2/Cu - Eo Zn+2/Zn
Eo cell = 0.34 + 0.76
Eo cell = 1.1
so
the value of Eo cell is 1.1
3)
the reaction is
Zn (s) + Cu+2 (aq) --> Zn+2 (aq) + Cu (s)
the reaction quotient is given by
Q = [Zn+2] / [Cu+2]
Q = 0.1 / 0.1
Q = 1
no2
E = Eocell - (0.059/n) log Q
E = 1.1- (0.059/2) log 1
E = 1.1 V
so
the value of Ecell is 1.1 V
4)
the obtained value is 0.835 V
so
the obtained value is less than the theoretical value
5)
Q = [Zn+2] / [Cu+2]
Q = 0.1 / 0.001
Q = 100
now
E = 1.1 - (0.059/2) log 100
E = 1.041
so
the E cell value is 1.041 V
6)
the obtained value is 0.782 V
so
the obtained value is less than the theoretical value
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