ELECTROCHEMISTRY STUDY PREPARATION WITH COMPLETE
SOLUTIONS 100% VERIFIED!!
Calculate the oxidation number of sulfur in Na2S2O3.
+6
+2
+3
+4 - ANSWER>>The sum of the oxidation numbers in all neutral compound must equal
0.
2(+1) + 2x + 3(-2) = 0
2 + 2x -6 = 0
x = +2
A galvanic voltaic cell has the generic metals X and Y as electrodes. X is more reactive
than Y, that is, X more readily reacts to form a cation than Y does. Classify the
descriptions by whether they apply to the X or Y electrode. - ANSWER>>X electrode:
anode
anions from the salt bridge flow towards it
loses mass
electrons in the wire flow away from it
Y electrode:
cathode
electrons in the wire flow towards it
cations from the salt bridge flow towards it
gains masses
Arrange these species by their ability to act as an oxidizing agent.
,Na+, Cl2, Pb2+, Zn2+ - ANSWER>>Best oxidizing agent:
Cl2
Pb2+
Zn2+
Na+
Poorest oxidizing agent
A more positive reduction potential indicates a greater tendency to be reduced and thus
to be a better oxidizing agent.
Which of the reagents would oxidize Cr to Cr2+, but not Pb to Pb2+?
Ca, Co, Br2, Br-, Co2+, Ca2+ - ANSWER>>The relevant reduction half-reactions are:
Cr2+ + 2e- = Cr E = -.91 V
Pb2+ + 2e- = Pb E = -0.13 V
Co, Br-, and Ca would all act as reducing agents, not oxidizing agents, so they can be
eliminated.
Co2+ is the only one that would have a reduction potential value between -0.91 and -0.13
(-0.28 V), so it would only oxidize Cr to Cr2+.
Which of the reagents would oxidize Cu to Cu+, but not Cl- to Cl2?
Br2, Br-, Ca, Ca2+, Co2+, Co - ANSWER>>The two relevant reduction half-reactions are:
Cu+ + e- = Cu E = 0.52 V
Cl2 + e- = 2Cl- E = 1.36 V
Co, Br-, and Ca would all be reducers, not oxidizers, so they can be eliminated.
Br2 is the only reagent whose reduction potential falls between 0.52 and 1.36 V (1.07 V),
and thus it would only oxidize Cu to Cu+:.
A galvanic (voltaic) cell consists of an electrode composed of nickel in a 1.0 M nickel(II)
ion solution and another electrode composed of gold in a 1.0 M gold(III) ion solution,
connected by a salt bridge. Calculate the standard potential for this cell at 25C. -
ANSWER>>Identify the anode and the cathode.
, Ni2+ + 2e- = Ni(s) E = -0.26 V
Au3+ + 3e- = Au(s) E = 1.498 V
The reaction with the higher reduction potential is the cathode, so gold is the cathode.
Subtract the standard reduction potential of the half-reaction at the anode from the
cathode.
1.498 V - (-0.26 V) = 1.76 V
Calculate the standard free-energy change for the reaction at 25C.
2Au3+(aq) + 3Zn(s) = 2Au(s) + 3Zn2+(aq) - ANSWER>>First, write the half-reactions.
Anode: 3 Zn = 3Zn2+ + 6e-
Cathode: 2Au3+ + 6e- = 2Au
Use the table of standard reduction potentials to find the overall cell potential.
(1.498 V) - (-0.76 V) = 2.26 V
The following equation relates free-energy change and cell potential: delta G naught =
-nFEcell
The balanced equation indicates that six moles of electrons were transferred, so n=6.
delta G naught = -(6)(96485)(2.26)(1 kJ/1000 J) = -1310 kJ
Given the half-reactions and their respective standard reduction potentials.
1. Cu3+ + 2e- = Cu+ Ei= +1.28 V
2. Cu2+ + e- = Cu+ Ei= +0.15 V
3. Cu2+ + 2e- = Cu Ei= +0.34 V
4. Cu+ + e- = Cu Ei= +0.52 V
Calculate the standard reduction potential for the reduction half-reaction of Cu(III) to
Cu(II). Cu3+ + e- = Cu2+ - ANSWER>>We can rearrange the equation delta G naught
final = -nFE naught final to solve for E naught final:
E naught final = - delta G naught/nF
E naught final = -0.15 V - 2(1.28 V)/1 = 2.41 V
SOLUTIONS 100% VERIFIED!!
Calculate the oxidation number of sulfur in Na2S2O3.
+6
+2
+3
+4 - ANSWER>>The sum of the oxidation numbers in all neutral compound must equal
0.
2(+1) + 2x + 3(-2) = 0
2 + 2x -6 = 0
x = +2
A galvanic voltaic cell has the generic metals X and Y as electrodes. X is more reactive
than Y, that is, X more readily reacts to form a cation than Y does. Classify the
descriptions by whether they apply to the X or Y electrode. - ANSWER>>X electrode:
anode
anions from the salt bridge flow towards it
loses mass
electrons in the wire flow away from it
Y electrode:
cathode
electrons in the wire flow towards it
cations from the salt bridge flow towards it
gains masses
Arrange these species by their ability to act as an oxidizing agent.
,Na+, Cl2, Pb2+, Zn2+ - ANSWER>>Best oxidizing agent:
Cl2
Pb2+
Zn2+
Na+
Poorest oxidizing agent
A more positive reduction potential indicates a greater tendency to be reduced and thus
to be a better oxidizing agent.
Which of the reagents would oxidize Cr to Cr2+, but not Pb to Pb2+?
Ca, Co, Br2, Br-, Co2+, Ca2+ - ANSWER>>The relevant reduction half-reactions are:
Cr2+ + 2e- = Cr E = -.91 V
Pb2+ + 2e- = Pb E = -0.13 V
Co, Br-, and Ca would all act as reducing agents, not oxidizing agents, so they can be
eliminated.
Co2+ is the only one that would have a reduction potential value between -0.91 and -0.13
(-0.28 V), so it would only oxidize Cr to Cr2+.
Which of the reagents would oxidize Cu to Cu+, but not Cl- to Cl2?
Br2, Br-, Ca, Ca2+, Co2+, Co - ANSWER>>The two relevant reduction half-reactions are:
Cu+ + e- = Cu E = 0.52 V
Cl2 + e- = 2Cl- E = 1.36 V
Co, Br-, and Ca would all be reducers, not oxidizers, so they can be eliminated.
Br2 is the only reagent whose reduction potential falls between 0.52 and 1.36 V (1.07 V),
and thus it would only oxidize Cu to Cu+:.
A galvanic (voltaic) cell consists of an electrode composed of nickel in a 1.0 M nickel(II)
ion solution and another electrode composed of gold in a 1.0 M gold(III) ion solution,
connected by a salt bridge. Calculate the standard potential for this cell at 25C. -
ANSWER>>Identify the anode and the cathode.
, Ni2+ + 2e- = Ni(s) E = -0.26 V
Au3+ + 3e- = Au(s) E = 1.498 V
The reaction with the higher reduction potential is the cathode, so gold is the cathode.
Subtract the standard reduction potential of the half-reaction at the anode from the
cathode.
1.498 V - (-0.26 V) = 1.76 V
Calculate the standard free-energy change for the reaction at 25C.
2Au3+(aq) + 3Zn(s) = 2Au(s) + 3Zn2+(aq) - ANSWER>>First, write the half-reactions.
Anode: 3 Zn = 3Zn2+ + 6e-
Cathode: 2Au3+ + 6e- = 2Au
Use the table of standard reduction potentials to find the overall cell potential.
(1.498 V) - (-0.76 V) = 2.26 V
The following equation relates free-energy change and cell potential: delta G naught =
-nFEcell
The balanced equation indicates that six moles of electrons were transferred, so n=6.
delta G naught = -(6)(96485)(2.26)(1 kJ/1000 J) = -1310 kJ
Given the half-reactions and their respective standard reduction potentials.
1. Cu3+ + 2e- = Cu+ Ei= +1.28 V
2. Cu2+ + e- = Cu+ Ei= +0.15 V
3. Cu2+ + 2e- = Cu Ei= +0.34 V
4. Cu+ + e- = Cu Ei= +0.52 V
Calculate the standard reduction potential for the reduction half-reaction of Cu(III) to
Cu(II). Cu3+ + e- = Cu2+ - ANSWER>>We can rearrange the equation delta G naught
final = -nFE naught final to solve for E naught final:
E naught final = - delta G naught/nF
E naught final = -0.15 V - 2(1.28 V)/1 = 2.41 V
