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Electrochemistry
Chapter 11
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Electrochemistry – ch 11
1. Consider the following reaction:
a. What substance is getting reduced?
b. How many moles of electrons are transferred?
2 VO2+ + 4 H+ + Cd → 2 VO2+ + 2 H2O + Cd2+
Electrochemistry – ch 11
2. Balance the following reaction under basic conditions:
Fe(CN)64- + Ce4+ → Ce(OH)3 + Fe(OH)3 + CO32- + NO3-
Electrochemistry – ch 11
3. Which is the strongest oxidizing agent?
a. Mn2+
b. Brc. Br2
d. Ag+
Electrochemistry – ch 11
4. Which is the strongest reducing agent?
a. Na+
b. Al
c. Zn2+
d. Fe. Mn
Electrochemistry – ch 11
5. Which of the following are true about galvanic cells (aka. voltaic
cells)?
a. Spontaneously produce a current
b. A current must be provided in order to run
c. Oxidation occurs at the cathode
d. Have possible plating out of metals at the cathode
e. The current flows from anode to cathode
f. The concentrations at the electrodes are 1M
Electrochemistry – ch 11
6. You want to design a galvanic cell with silver and gold electrodes.
Show or describe how you would set it up? Assign the electrodes,
write the overall reaction, and calculate the standard cell potential.
As the cell operates what happens to the masses of the silver and
gold electrodes?
Electrochemistry – ch 11
7. Using reduction potentials answer the following:
a. Is Cl2 able to reduce Cr3+?
b. Is Pb2+ able to oxidize Ni?
c. Will Au dissolve in an HCl solution?
d. Will Zn dissolve in an HCl solution?
e. What can oxidize Al but not Fe?
f. What can reduce Ag+ but not I2?
Electrochemistry – ch 11
8. Consider the following reaction at 75 °C.
Pb2+(aq) + 2 Cr2+(aq)  Pb(s) + 2 Cr3+(aq)
a. Calculate the standard potential
b. Calculate the standard change in free energy
c. Calculate the equilibrium constant
d. If the above cell is allowed to operate spontaneously, will the
voltage increase, decrease or remain the same?
e. If the above cell is allowed to operate spontaneously, what will
happen to the concentration and mass at the electrodes?
f. Calculate the initial cell voltage for the above reaction if the
initial concentrations are [Pb2+] = 0.25 M, [Cr2+] = 0.20M
and [Cr3+] = 0.005 M.
Electrochemistry – ch 11
9. Consider the following cell:
Cu (s) | Cu2+(aq)(0.001 M) || Br2(l) | Br-(aq) (? M), Pt(s)
a.
b.
c.
d.
e.
Assign the electrodes.
Determine the direction of electron flow.
Describe the flow of ions in the salt bridge
Calculate the standard potential.
At 25 °C the measured cell voltage is 0.975 V. Calculate the
concentration of the bromide ion.
Electrochemistry – ch 11
10. Consider the following cell:
Al(s) | Al3+ (1.0 M) | | Pb2+(1.0 M) | Pb (s)
Calculate the cell potential after the reaction has operated long
enough for the [Al3+] to have changed by 0.6 M at 25 °C.
Electrochemistry – ch 11
11. Consider two electrodes connected by a wire. One side has 0.0001
M Fe2+/Fe (s) and the other side has 10 M Fe2+/Fe(s).
a. Assign the electrodes
b. Determine the direction of electron flow
c. Calculate the standard cell voltage at 25 °C.
d. Calculate the cell voltage at 25 °C.
e. What are the concentrations at equilibrium?
Electrochemistry – ch 11
12. What mass of Co forms from a solution of Co2+ when a current of
15 amps is applied for 1.15 hours?
Electrochemistry – ch 11
13. A solution of iron chloride underwent electrolysis for 2 hrs at 10
amps yielding 20.84 g of iron. What is the oxidation state of the
iron in the iron chloride?
Electrochemistry – ch 11
14. How long will it take (in min) to plate out 10.0 g of Bi from a
solution of BiO+ using a current of 25.0 A?
Electrochemistry – ch 11
15. What volume of gas at STP is produced from the electrolysis of
water by a current of 3.5 amps in 15 minutes?
Electrochemistry – ch 11 – Answers
1. Consider the following reaction:
a. What substance is getting reduced? VO2+
b. How many moles of electrons are transferred? 2
2 VO2+ + 4 H+ + Cd → 2 VO2+ + 2 H2O + Cd2+
+5
+4
Oxidation number for V is decreasing
⇒ Getting reduced
Electrochemistry – ch 11 – Answers
2. Balance the following reaction under basic conditions:
Fe(CN)64- + Ce4+ → Ce(OH)3 + Fe(OH)3 + CO32- + NO3Red ½:
Ce4+ + 3H2O  Ce(OH)3 + 3H+
Ox ½: Fe(CN)64– + 39H2O  Fe(OH)3 + 6CO32– +6NO3– + 75H+ + 61e–
Acidic: Fe(CN)64– + 61 Ce4+ + 222H2O  Fe(OH)3 + 6CO32– +6NO3– + 258H+ + 61Ce(OH)3
Basic: Fe(CN)64– + 61 Ce4+ + 258 OH– Fe(OH)3 + 6CO32– +6NO3– + 36H2O + 61Ce(OH)3
Electrochemistry – ch 11 – Answers
3. Which is the strongest oxidizing agent?
a. Mn2+
b. Brc. Br2
d. Ag+
4. Which is the strongest reducing agent?
a. Na+
b. Al
c. Zn2+
d. Fe. Mn
Electrochemistry – ch 11 – Answers
5. Which of the following are true about galvanic cells (aka. voltaic
cells):
a. Spontaneously produce a current
b. A current must be provided in order to run
c. Oxidation occurs at the cathode
d. Have possible plating out of metals at the cathode
e. The current flows from anode to cathode
f. The concentrations at the electrodes are 1M
Electrochemistry – ch 11 – Answers
6. You want to design a galvanic cell with silver and gold electrodes. Show or
describe how you would set it up? Assign the electrodes, write the overall
reaction, and calculate the standard cell potential. As the cell operates what
happens to the masses of the silver and gold electrodes?
Au
Ag
Cathode ½: Au3+ + 3e–  Au
Anode ½: Ag  Ag+ + e–
E° = 1.5 V
E° = 0.8 V
Overall Rxn: Au3+(aq) + 3Ag(s)  Au(s) + 3Ag+(aq)
Ecell° = 0.7 V
Au3+(aq)
Cathode
Ag+(aq)
Anode
The cathode is getting plated with Au so the mass is
getting heavier and at the anode Ag is corroding so the
mass is going down.
Electrochemistry – ch 11 – Answers
7. Using reduction potentials answer the following:
a. Is Cl2 able to reduce Cr3+? No Cl2 is an oxidizing reagent
b. Is Pb2+ able to oxidize Ni? Yes
c. Will Au dissolve in an HCl solution? No
d. Will Zn dissolve in an HCl solution? Yes
e. What can oxidize Al but not Fe? Cr3+, Zn2+, H2O and Mn2+
f. What can reduce Ag+ but not I2? Hg, Fe2+, H2O2 and MnO42-
Electrochemistry – ch 11 – Answers
8. Consider the following reaction at 75 °C.
Pb2+(aq) + 2 Cr2+(aq)  Pb(s) + 2 Cr3+(aq)
a. Calculate the standard potential
Eredº = -0.13V and Eoxº = +0.5 V ⇒ Erxnº = 0.37V
b. Calculate the standard change in free energy ΔGº =-nFEº
ΔGº =-(2mol e-)(96,485C/mol e-)(0.37J/C) = -71.4 kJ
c. Calculate the equilibrium constant ΔGº =-RTlnK
-71.4kJ=(0.008314J/molK)(348K)lnK ⇒ K=5.2x1010
d. If the above cell is allowed to operate spontaneously, will
the voltage increase, decrease or remain the same?
Rxns spontaneously go to equilibrium where E=0
e. If the above cell is allowed to operate spontaneously, what
will happen to the concentrations at the electrodes? The
concentrations of the reactants will decrease and the
products increase
Continue to next slide…
Electrochemistry – ch 11 – Answers
f. Calculate the initial cell voltage for the above reaction if the initial
concentrations are [Pb2+] = 0.25 M, [Cr2+] = 0.20M and [Cr3+]
= 0.005 M. Since the concentrations are not 1M these are
nonstandard conditions ⇒ E = E° E = 0.37V -
(
8.314𝐽
)(348𝐾)
𝑚𝑜𝑙𝐾
−
𝐶
ln
(2 𝑚𝑜𝑙 𝑒 )(96,485𝑚𝑜𝑙𝑒−)
𝑅𝑇
lnQ
𝑛𝐹
0.005 2
0.25 0.2
2
⇒
= 0.46
Electrochemistry – ch 11 – Answers
9. Consider the following cell:
Cu (s) | Cu2+(aq)(0.001 M) || Br2(l) | Br-(aq) (? M), Pt(s)
a. Assign the electrodes. Left side ⇒ anode and right ⇒ cathode
b. Determine the direction of electron flow and the direction of
the current. Electrons flow from anode to cathode and vice versa
for the current
c. Describe the flow of ions in the salt bridge anions flow to the
anode and cations to cathode
d. Calculate the standard potential. Eoxº = -0.34 V and Eredº = 1.09
V ⇒ Ecellº = 0.75 V
Continue to next slide…
Electrochemistry – ch 11 – Answers
e. At 25 °C the measured cell voltage is 0.975 V. Calculate the
concentration of the bromide ion. E = E° 0.975 = 0.75V -
(
8.314𝐽
)(298𝐾)
𝑚𝑜𝑙𝐾
(2 𝑚𝑜𝑙 𝑒
−
𝑅𝑇
lnQ
𝑛𝐹
⇒
ln(0.001)[Br-]2 ⇒ [Br-] = 0.005
𝐶
)(96,485𝑚𝑜𝑙𝑒−)
Electrochemistry – ch 11 – Answers
10. Consider the following cell:
Al(s) | Al3+ (1.0 M) | | Pb2+(1.0 M) | Pb (s)
Calculate the cell potential after the reaction has operated long enough
for the [Al3+] to have changed by 0.6 M at 25 °C.
Anode: Al  Al3+ + 3eE°ox = +1.66V
Cathode: Pb2+ + 2e-  Pb
E°red = -0.13V
Cell: 2Al + 3Pb2+  2Al3+ + 3Pb
E°cell = 1.53V
3Pb2+
2Al3+
1M
1M
-3/2(0.6)
+0.6
0.1
1.6
E = E° -
𝑅𝑇
lnQ
𝑛𝐹
⇒
8.314𝐽
E = 1.53V -
( 𝑚𝑜𝑙𝐾 )(298𝐾)
−
𝐶
ln
(6 𝑚𝑜𝑙 𝑒 )(96,485𝑚𝑜𝑙𝑒−)
1.6
0.1
2
3
= 1.5 V
Electrochemistry – ch 11 – Answers
11. Consider two electrodes connected by a wire. One side has 0.0001 M
Fe2+/Fe (s) and the other side has 10 M Fe2+/Fe(s).
a. Assign the electrodes This is a concentration cell ⇒ the voltage is
strictly due to the difference in concentration ⇒ at equilibrium the
concentrations are equivalent ⇒ the side with lower concentration
will have to increase and vice versa until the concentrations meet in
the middle and there will no longer be a voltage aka equilibrium ⇒
so in order to increase the concentration on the lower side Fe needs
to get oxidized making it the anode ⇒ to decrease the concentration
on the other side the Fe2+ needs to be reduced making it the cathode
b. Determine the direction of electron flow anode to cathode or low
conc to high conc
Continue to next slide…
Electrochemistry – ch 11 – Answers
d. Calculate the cell voltage at 25 °C.
Anode: Fe  Fe2+ + 2eE°ox = -0.44V
Cathode: Fe2+ + 2e-  Fe
E°red = +0.44V
Cell: Fe(anode) + Fe2+(cathode)  Fe(cathode) + Fe2+ (anode)
E°cell = 0
e. What are the concentrations at equilibrium?
at equilibrium E = 0 and Q = K and for a concentration cell [anode]=[cathode]
They will be equal at the midpoint ⇒ [Fe2 ]=
0.0001+10
2
= 5M
Electrochemistry – ch 11 – Answers
12. What mass of Co forms from a solution of Co2+ when a current of
15 amps is applied for 1.15 hours? Co2+ + 2e-  Co
ne- = It/F = (15C/s)(1.15hr)(3600s/hr)/(96,485C/mol e-) = 0.644
mol e(0.644 mol e-)(1mol Co/2mole-)(58.933g Co/mol)= 19 g Co
Electrochemistry – ch 11 – Answers
13. A solution of iron chloride underwent electrolysis for 2 hrs at 10 amps
yielding 20.84 g of iron. What is the oxidation state of the iron in the
iron chloride? FeX+ + Xe-  Fe So we need to figure out how many
moles of electrons per mol of Fe
Fe ⇒ (20.84g)/(55.85g/mol) = 0.373 mol Fe
e- ⇒ ne- =
𝐼𝑡
𝐹
⇒ ne- =(10C/s)(2 hr)(3600s/hr)/(96,485C/mol e-) = 0.746 mol e-
0.373 mol Fe:0.746 mol e- ⇒ 1 mol Fe : 2 mol e- ⇒ Fe2+
Electrochemistry – ch 11 – Answers
14. How long will it take (in min) to plate out 10.0 g of Bi from a
solution of BiO+ using a current of 25.0 A?
BiO+ + 2H+ + 3e-  Bi + H2O
ne- =
𝐼𝑡
𝐹
10.0 g Bi
1 𝑚𝑜𝑙 𝐵𝑖 3 𝑚𝑜𝑙 𝑒−
x
x
209𝑔
1 𝑚𝑜𝑙 𝐵𝑖
=
25𝐶
)(𝑡)
𝑠
96,485𝐶
( 𝑚𝑜𝑙𝑒 )
(
⇒ t = 554 s or 9.23 min
Electrochemistry – ch 11 – Answers
15. What volume of gas at STP is produced from the electrolysis of
water by a current of 3.5 amps in 15 minutes?
2H2O  2H2 + O2
ne- =
𝐼𝑡
=
𝐹
(3.5C/s)(15min)(60s/min)/(96,485C/mol e-)
ne- = 0.0326 mol e-
(0.0326 mol e-)(3mol gas/4mol e-)(22.4 L/mol) = 0.55 L