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Skills Worksheet
Problem Solving Electrochemistry The potential in volts has been measured for many different reduction halfreactions. The potential value is measured against the standard hydrogen electrode, which is assigned a value of zero. For consistency, these half-reactions are always written in the direction of the reduction. A half-reaction that has a positive reduction potential proceeds in the direction of the reduction when it is coupled with a hydrogen electrode. A reaction that has a negative reduction potential proceeds in the oxidation direction when it is coupled with a hydrogen electrode. Table 1 gives some common standard reduction potentials. TABLE 1
Reduction half-reaction
Standard electrode potential, E 0 (in volts)
Reduction half-reaction
Standard electrode potential, E 0 (in volts)
MnO 4 8H 5e ^ 2 Mn 4H2O
1.50
Fe3 3e ^ Fe
0.04
Au3 3e ^ Au
1.50
Pb2 2e ^ Pb
0.13
1.36
Sn
2e ^ Sn
0.14
1.23
Ni2 2e ^ Ni
0.26
MnO2 4H 2e ^ Mn2 2H2O
1.22
Cd2 2e ^ Cd
0.40
Br2 2e ^ 2Br
1.07
Fe2 2e ^ Fe
0.45
Cl2 2e ^ 2Cl
Cr2O2 7 14H 6e ^ 3 2Cr 7H2O
Hg
2
2
2e ^ Hg
Ag e ^ Ag
0.85
S 2e ^ S
0.48
0.80
Zn2 2e ^ Zn
0.76
2
2e ^ 2Hg
0.80
Al
Fe3 e ^ Fe2
0.77
Mg2 2e ^ Mg
Hg2 2
MnO 4
e ^
MnO2 4
I2 2e ^ 2I 2
Cu
2e ^ Cu
S 2H(aq) 2e ^ H2S(aq)
2H (aq) 2e ^ H2
3
3e ^ Al
1.66 2.37
0.56
Na e ^ Na
2.71
0.54
Ca2 2e ^ Ca
2.87
0.34
Ba
0.14
K e ^ K
2
0.00
2e ^ Ba
Li e ^ Li
2.91 2.93 3.04
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Problem Solving continued You can use reduction potentials to predict the direction in which any redox reaction will be spontaneous. A spontaneous reaction occurs by itself, without outside influence. The redox reaction will proceed in the direction for which the difference between the two half-reaction potentials is positive. This direction is the same as the direction of the more positive half-reaction. General Plan for Solving Electrochemical Problems
1
2a
Balanced ionic equation for the redox reaction Write the individual halfreactions for both oxidation and reduction.
Balanced equation for reduction Reduction occurs at the cathode.
3a
Balanced equation for oxidation
Oxidation occurs at the anode.
3b
Cathode Look up the reduction potential for the halfreaction in Table 1.
4a
2b
Anode
Look up the reduction potential for the halfreaction in Table 1.
4b
E0cathode
E0anode
Substitute and solve.
5
E0cell E0cathode E0anode
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Problem Solving continued
Sample Problem 1 Calculate the cell potential to determine whether the following reaction is spontaneous in the direction indicated. Cd2(aq) 2I(aq) → Cd(s) I2(s)
Solution ANALYZE What is given in the problem?
reactants and products
What are you asked to find?
whether the reaction is spontaneous
Items
Data
Reactants
Cd2(aq) 2I(aq)
Products
Cd(s) I2(s)
E 0cathode
?V
E 0anode
?V
E 0cell
?V
PLAN What steps are needed to determine whether the reaction is spontaneous? Separate into oxidation and reduction half-reactions. Find reduction potentials for each. Solve the equation for the cell potential to determine if the reaction is spontaneous.
1 Cd2(aq) 2I(aq) 3 Cd(s) I2(s) write the individual half-reactions
2a
2b
Cd2(aq) 2e 3 Cd(s)
2I(aq) 3 I2(s) 2e
reduction occurs at the cathode
oxidation occurs at the anode
3a
3b
Cathode
Anode
look up the reduction potential for the half-reaction in Table 1
look up the reduction potential for the half-reaction in Table 1
4a
4b
E0cathode
E0anode substitute and solve
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Problem Solving continued Write the given equation. Cd2(aq) 2I(aq) → Cd(s) I2(s) The oxidation number of cadmium decreases; it is reduced. Cd2(aq) 2e → Cd(s) The oxidation number of iodine increases; it is oxidized. 2I(aq) → I2(s) 2e Cadmium is the cathode, and iodine is the anode. from Table 1
E0cathode 0.40 V from Table 1
E0anode 0.54 V given above
given above
E0cell E0cathode E0anode
Determine spontaneity. If the cell potential is positive, the reaction is spontaneous as written. If the cell potential is negative, the reaction is not spontaneous as written, but the reverse reaction is spontaneous. COMPUTE E 0cell 0.40 V 0.54 V 0.94 V The reaction potential is negative. Therefore, the reaction is not spontaneous. The reverse reaction would have a positive potential and would, therefore, be spontaneous. Cd2(aq) 2I(aq) → Cd(s) I2(s) not spontaneous Cd(s) I2(s) → Cd2(aq) 2I(aq) spontaneous EVALUATE Are the units correct? Yes; cell potentials are in volts.
Is the number of significant figures correct? Yes; the number of significant figures is correct because the half-cell potentials have two significant figures.
Is the answer reasonable? Yes; the reduction potential for the half-reaction involving iodine was more positive than the potential for the reaction involving cadmium, which means that I2 has a greater attraction for electrons than Cd2. Therefore, I2 is more likely to be reduced than Cd2. The reverse reaction is favored.
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Problem Solving continued
Practice Use the reduction potentials in Table 1 to determine whether the following reactions are spontaneous as written. Report the E 0cell for the reactions.
1. Cu2 Fe → Fe2 Cu ans: 0.79 V; spontaneous
2. Pb2 Fe2 → Fe3 Pb ans: 0.90 V; nonspontaneous
3. Mn2 4H2O Sn2 → MnO 4 8H Sn ans: 1.64 V; nonspontaneous
4. MnO2 4 Cl2 → MnO4 2Cl ans: 0.80 V; spontaneous
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Problem Solving continued 2 5. Hg2 2 2MnO4 → 2Hg 2MnO4 ans: 0.24 V; spontaneous
6. 2Li Pb → 2Li Pb2 ans: 2.91 V; nonspontaneous
7. Br2 2Cl → 2Br Cl2 ans: 0.29 V; nonspontaneous
8. S 2I → S2 I2 ans: 1.02 V; nonspontaneous
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Problem Solving continued
Sample Problem 2 A cell is constructed in which the following two half-reactions can occur in either direction. Zn2 2e ^ Zn Br2 2e ^ 2Br Write the full ionic equation for the cell in the spontaneous direction, identify the reactions occurring at the anode and cathode, and determine the cell’s voltage.
Solution ANALYZE What is given in the problem?
the reversible half-reactions of the cell
What are you asked to find?
the equation in the spontaneous direction; the voltage of the cell
Items
Data
Half-reaction 1
Zn2 2e ^ Zn
Half-reaction 2
Br2 2e ^ 2Br
Reduction potential of 1
0.76 V
Reduction potential of 2
1.07 V
Full ionic reaction
?
Cell voltage
?
PLAN What steps are needed to determine the spontaneous reaction of the cell and the cell voltage? Determine which half-reaction has the more positive reduction potential. This will be the reduction half-reaction; it occurs at the cathode. Reverse the other half-reaction so that it becomes an oxidation half-reaction; it occurs at the anode. Adjust the half-reactions so that the same number of electrons are lost as are gained. Add the reactions together. Compute the cell voltage by the formula E 0cell E 0cathode E 0anode, using the reduction potentials for the reaction at each electrode.
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Problem Solving continued Br2 2e ^ 2Br
Zn2 2e ^ Zn
look up the reduction potential for the half-reaction in Table 1
look up the reduction potential for the half-reaction in Table 1
E0half-reaction
E0half-reaction
the half-cell with the larger reduction potential is the cathode
the half-cell with the smaller reduction potential is the anode
Cathode
Anode
reduction occurs at the cathode
oxidation occurs at the anode
Br2 2e 3 2Br
Zn 3 Zn2 2e combine to write the full ionic equation
Br2 Zn 3 2Br Zn2 substitute and solve
E0cell E0cathode E0anode
First, look up the reduction potentials for the two half-reactions in Table 1. from Table 1
0 1.07 V EBr 2
from Table 1
0 0.76 V EZn
Br2 has the larger reduction potential; therefore, it is the cathode. Zn has the smaller reduction potential; therefore, it is the anode. The cathode half-reaction is Br2 2e → 2Br The anode half-reaction is Zn → Zn2 2e The full-cell equation is Br2 2e → 2Br Zn → Zn2 2e Br2 Zn → 2Br Zn2 Substitute the reduction potentials for the anode and cathode into the cell potential equation, and solve the equation. E0Br2
E0Zn
E0cell E0cathode E0anode
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Problem Solving continued COMPUTE
Br2 Zn → 2Br Zn2 E 0cell 1.07 V (0.76 V) 1.83 V
EVALUATE Are the units correct? Yes; the cell potential is in volts.
Is the number of significant figures correct? Yes; the half-cell potentials were given to two decimal places.
Is the answer reasonable? Yes; you would expect the reaction to have a positive cell potential because it should be spontaneous.
Practice If a cell is constructed in which the following pairs of reactions are possible, what would be the cathode reaction, the anode reaction, and the overall cell voltage?
1. Ca2 2e ^ Ca Fe3 3e ^ Fe ans: cathode: Fe3 3e 3 Fe, anode: Ca 3 Ca2 2e, E 0cell 2.83 V
2. Ag e ^ Ag S 2H 2e ^ H2S ans: cathode: Ag e 3 Ag, anode: H2S 3 S 2H 2e, E 0cell 0.66 V
3. Fe3 e ^ Fe2 Sn2 2e ^ Sn ans: cathode: Fe3 e 3 Fe2, anode: Sn 3 Sn2 2e, E 0cell 0.91 V
4. Cu2 2e ^ Cu Au3 3e ^ Au ans: cathode: Au3 3e 3 Au, anode: Cu 3 Cu2 2e, E 0cell 1.16 V
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Problem Solving continued
Additional Problems Use reduction potentials to determine whether the reactions in the following 10 problems are spontaneous.
1. Ba Sn2 → Ba2 Sn 2. Ni Hg2 → Ni2 Hg 2 3. 2Cr3 7H2O 6Fe3 → Cr2O2 7 14H 6Fe
4. Cl2 Sn → 2Cl Sn2 5. Al 3Ag → Al3 3Ag 2 6. Hg2 → 2Hg S 2 S
7. Ba 2Ag → Ba2 2Ag 8. 2I Ca2 → I2 Ca 2 9. Zn 2MnO 2MnO2 4 → Zn 4 10. 2Cr3 3Mg2 7H2O → Cr2O2 7 14H 3Mg
In the following problems, you are given a pair of reduction half-reactions. If a cell were constructed in which the pairs of half-reactions were possible, what would be the balanced equation for the overall cell reaction that would occur? Write the half-reactions that occur at the cathode and anode, and calculate the cell voltage.
11. Cl2 2e ^ 2Cl Ni2 2e ^ Ni 12. Fe3 3e ^ Fe Hg2 2e ^ Hg 2 13. MnO 4 e ^ MnO4 Al3 3e ^ Al 2 14. MnO 4H2O 4 8H 5e ^ Mn S 2H 2e ^ H2S
15. Ca2 2e ^ Ca Li e ^ Li 16. Br2 2e ^ 2Br 2 MnO 4H2O 4 8H 5e ^ Mn 17. Sn2 2e ^ Sn Fe3 e ^ Fe2 18. Zn2 2e ^ Zn 3 7H2O Cr2O2 7 14H 6e ^ 2Cr 19. Ba2 2e ^ Ba Ca2 2e ^ Ca 20. Hg2 2 2e ^ 2Hg Cd2 2e ^ Cd
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