HOMEWORK FOR SUBMISSION
REDOX AND REDUCTION POTENTIALS
Answer Key
1)
Complete and balance the following equations. Show your working. (4 marks each part.)
a)
MnO4-(aq) + H2S(aq) 6 Mn2+(aq) + SO42-(aq) in acidic solution.
b)
i)
Write half equations:
MnO4- 6 Mn2+
H2S 6 SO42-
ii)
Balance O with H2O:
MnO4- 6 Mn2+ + 4H2O
H2S + 4H2O 6 SO42-
iii)
Balance H with H+:
MnO4- + 8H+ 6 Mn2+ + 4H2O
H2S + 4H2O 6 SO42- + 10H+
iv)
Balance charge with e-:
MnO4- + 8H+ +5e- 6 Mn2+ + 4H2O ....(1)
H2S + 4H2O 6 SO42- + 10H+ + 8e- ...(2)
v)
LCM of number of e's is LCM of 5 & 8, ie. 40. Hence we multiply (1) by 8 and (2)
by 5 to make the number of electrons in each equal to 40.
8MnO4- + 64H+ +40e- 6 8Mn2+ + 32H2O ....(3)
5H2S + 20H2O 6 5SO42- + 50H+ + 40e- ....(4)
vi)
Adding (3) and (4), we get:
8MnO4- + 64H+ +40e- + 5H2S + 20H2O 6 5SO42- + 8Mn2+ + 32H2O + 5SO42- + 50H+
+ 40e-
vii)
Cancelling e's, H+'s and H2O's and putting in states, we have:
8MnO4-(aq) + 14H+(aq) + 5H2S(aq) 6 5SO42-(aq) + 8Mn2+(aq) + 12H2O(l) + 5SO42(aq)
OCl-(aq) + I-(aq) 6 IO3-(aq) + Cl-(aq) in basic solution. Steps as before except (iva).
i)
OCl- 6 ClI- 6 IO3PAGE 1 OF 4
2)
ii)
OCl- 6 Cl- + H2O
I- + 3H2O 6 IO3-
iii)
OCl- +2H+ 6 Cl- + H2O
I- + 3H2O 6 IO3- + 6H+
iv)
OCl- +2H+ + e- 6 Cl- + H2O
I- + 3H2O 6 IO3- + 6H+ + 6e-
iva)
Since the reaction is in basic solution, add OH- to both sides to convert any H+ to
H2O:
OCl- +2H2O + e- 6 Cl- + H2O + 2OH- ....(1)
I- + 3H2O + 6OH- 6 IO3- + 6H2O + 6e-...(2)
v)
Adding (1) x 6 to (2), cancelling e's, OH-'s and H2O's, and putting in states, we
have:
3OCl-(aq) + I-(aq) 6 3Cl-(aq) + IO3-(aq)
The following standard cell is constructed;
Pt(s)*Fe2+(aq),Fe3+(aq)5Cd2+(aq)*Cd(s)
a)
Write the half equation for the reaction corresponding to the anode.
With reference to the cell, oxidation formally occurs in the LH half-cell and so this is
formally the anode:
Fe2+(aq) 6 Fe3+ + e- (1 mark)
b)
Write the half equation for the reaction corresponding to the left hand half-cell.
Cd2+(aq) + 2e- 6 Cd(s) (1 mark)
c)
Write the balanced ionic equation for the overall reaction corresponding to the cell diagram.
2Fe2+(aq) + Cd2+(aq) 62 Fe3+ + Cd(s) (1 mark)
d)
Calculate the cell e.m.f.
E± cell = E± R - E± L = -0.403 - (+0.771) = -1.174 v (1 mark)
e)
What reaction would you expect to proceed spontaneously in the cell? Explain your
answer.
Since the cell e.m.f. is negative, the reaction expected would be the reverse of that in (c)
above:
2Fe3+ + Cd(s) 6 2Fe2+(aq) + Cd2+(aq) (1)
f)
What are the implications of the word "standard" with respect to this cell?
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Address the question: The concentrations of the Fe2+(aq), Fe3+(aq) and Cd2+(aq) must be
1 M, the Pt(s) and Cd(s) used for the electrodes must be pure and the temperature is 25EC.
(3 marks)
g)
3)
Draw a fully labelled diagram of the cell. Make sure you indicate the direction of flow of
electrons, the anode, the cathode as well as all other important features. (5 marks)
Using E± values from your textbook relevant to the following species, answer the questions
below:
Ag+
Cu2+
Pb2+
Ag
Cu
Pb
Fe2+
Zn2+
Mg2+
Fe
Zn
Mg
Justify each of your answers to the following questions by referring to E± values:
The relevant standard reduction potentials are:
Ag+/Ag
0.800v
Fe2+/Fe
2+
Cu /Cu
0.340v
Zn2+/Zn
Pb2+/Pb
-0.126v
Mg2+/Mg
0.409v
-0.763v
-2.375v
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a)
Which species is the strongest oxidant? (1 mark)
Ag+ is the strongest oxidant since Ag+/Ag has the highest E± value.
b)
Which is the weakest oxidant? (1 mark)
Mg2+ is the weakest oxidant since Mg2+/Mg has the lowest E± value.
c)
Which species is the strongest reductant?
Mg is the strongest reductant since Mg2+/Mg has the lowest E± and so the reverse reaction
(Mg(s) 6 Mg2+(aq) + 2e-) goes most readily.
d)
Which is the weakest reductant?
Ag is the weakest reductant since Ag+/Ag has the highest E± and so the reverse reaction
(Ag(s) 6 Ag+(aq) + e-) goes least readily.
e)
Lead rods are placed in solutions of each of AgNO3, Cu(NO3)2, Fe(NO3)2 and Mg(NO3)2.
In which solutions could you expect a coating of another metal on the lead rod? (3 marks)
For lead to become coated with silver the following half reactions must occur:
a)
Ag+(aq) + e-6 Ag(s) (Ag+ acting as an oxidant.)
b)
Pb(s) 6Pb2+(aq) + 2e- (Pb being oxidized.)
Since the E value for Ag+/Ag is larger than that for Pb2+/Pb, Ag+ is a sufficiently powerful
oxidant to oxidize Pb to Pb2+ ie. lead would become coated with silver in a solution of
silver nitrate.
The same argument can be used to show that lead would become coated with copper in a
solution of copper(II) nitrate, but no coating would form in iron(II) nitrate or magnesium
nitrate solutions, since the relevant E± would be less than that for Pb2+/Pb.
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