University of Toronto A Redox Reaction Chemistry Questions

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A.P. Chemistry Practice Test – Ch. 17: Electochemistry
SCH4U Electro Unit 4 test
student name
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
1) The gain of electrons by an element is called __________.
A) oxidation
B) sublimation
C) reduction
D) disproportionation
E) fractionation
2) __________ is reduced in the following reaction:
Cr2O72- + 6S2O32- + 14H+ ¬ 2Cr3+ + 3S4O62- + 7H2O
A) Cr2O72-
B) S2O32-
C) H+
D) S4O62-
E) Cr3+
D) Cr2O72-
E) S4O62-
D) (a) and (c)
E) (b) and (c)
D) PbSO4
E) H2SO4
D) +12
E) +6
3) __________ is the oxidizing agent in the reaction below.
Cr2O72- + 6S2O32- + 14H+ ¬ 2Cr3+ + 3S4O62- + 7H2O
A) Cr3+
B) H+
C) S2O32-
4) Which of the following reactions is a redox reaction?
(a) K2CrO4 + BaCl2 ¬ BaCrO4 + 2KCl
(b) Pb22+ + 2Br- ¬ PbBr
(c) Cu + S ¬ CuS
A) (a) only
B) (b) only
C) (c) only
5) Which substance is the reducing agent in the reaction below?
Pb + PbO2 + 2H2SO4 ¬ 2PbSO4 + 2H2O
A) H2O
B) Pb
C) PbO2
6) What is the oxidation number of chromium in the dichromate ion?
A) +7
B) +3
C) +14
7) What is the oxidation number of potassium in potassium permanganate?
A) 0
B) +3
C) +2
D) +1
E) -1
8) What is the oxidation number of manganese in the permanganate ion, MnO 4- ?
A) +2
B) +5
C) +1
1
D) +7
E) +4
9) What is the coefficient of the bromide ion when the following redox equation is balanced?
BrO- + Fe(OH)2 ¬ Br- + Fe(OH)3
A) 4
B) 3
(basic solution)
C) 2
D) 1
E) 5
10) What is the coefficient of Fe3+ when the following equation is balanced?
CN- + Fe3+ ¬ CNO- + Fe2+
A) 1
B) 2
(basic solution)
C) 3
D) 4
E) 5
11) The balanced half-reaction in which sulfate ion is reduced to sulfite ion is a __________ process.
A) three-electron
B) four-electron
C) two-electron
D) one-electron
E) six-electron
12) The half-reaction occurring at the anode in the balanced reaction shown below is __________.
3MnO4- (aq) + 24H+ (aq) + 5Fe (s) ¬ 3Mn2+ (aq) + 5Fe3+ (aq) + 12H2O (l)
A) Fe2+ (aq) ¬ Fe3+ (aq) + eB) MnO4- (aq) + 8H+ (aq) + 5e- ¬ Mn2+ (aq) + 4H2O (l)
C) Fe (s) ¬ Fe3+ (aq) + 3eD) 2MnO4- (aq) + 12H+ (aq) + 6e- ¬ 2Mn2+ (aq) + 3H2O (l)
E) Fe (s) ¬ Fe2+ (aq) + 2e13) The purpose of the salt bridge in an electrochemical cell is to __________.
A) provide a source of ions to react at the anode and cathode.
B) provide oxygen to facilitate oxidation at the anode.
C) provide a means for electrons to travel from the anode to the cathode.
D) provide a means for electrons to travel from the cathode to the anode.
E) maintain electrical neutrality in the half-cells via migration of ions.
14) In a voltaic cell, electrons flow from the __________ to the __________.
A) salt bride, anode
B) anode, cathode
C) anode, salt bridge
D) salt bridge, cathode
E) cathode, anode
15) 1V = _________.
A) 96485 C
B) 1 J/C
C) 1 C/J
2
D) 1 J/s
E) 1 amp œ s
Table 20.1
Half Reaction
F2 (g) + 2e- ¬ 2F- (aq)
+2.87
Cl2 (g) + 2e- ¬ 2Cl- (aq)
+1.359
Ee (V)
Br2 (l) + 2e- ¬ 2Br – (aq)
+1.065
O2 (g) + 4H+ (aq) + 4e- ¬ 2H2O (l) +1.23
Ag+ + e- ¬ Ag (s)
Fe3+ (aq) + e- ¬ Fe2+ (aq)
+0.799
I2 (s) + 2e- ¬ 2I- (aq)
Cu2+ + 2e- ¬ Cu (s)
+0/536
+0.34
2H+ + 2e- ¬ H2 (g)
0
Pb2+ + 2e- ¬ Pb (s)
Ni2+ + 2e- ¬ Ni (s)
-0.126
Li+ + e- ¬ Li (s)
-3.05
+0.771
-0.28
16) Which of the halogens in Table 20.1 is the strongest oxidizing agent?
A) Br2
B) I2
C) Cl2
D) F2
E) All of the halogens have equal strength as oxidizing agents.
17) Which one of the following types of elements is most likely to be a good oxidizing agent?
A) transition elements
B) alkaline earth elements
C) lanthanides
D) alkali metals
E) halogens
3
Table 20.1
Half Reaction
F2 (g) + 2e- ¬ 2F- (aq)
+2.87
Cl2 (g) + 2e- ¬ 2Cl- (aq)
+1.359
Ee (V)
Br2 (l) + 2e- ¬ 2Br – (aq)
+1.065
O2 (g) + 4H+ (aq) + 4e- ¬ 2H2O (l) +1.23
Ag+ + e- ¬ Ag (s)
Fe3+ (aq) + e- ¬ Fe2+ (aq)
+0.799
I2 (s) + 2e- ¬ 2I- (aq)
Cu2+ + 2e- ¬ Cu (s)
+0/536
+0.34
2H+ + 2e- ¬ H2 (g)
0
Pb2+ + 2e- ¬ Pb (s)
Ni2+ + 2e- ¬ Ni (s)
-0.126
Li+ + e- ¬ Li (s)
-3.05
+0.771
-0.28
18) Using Table 20.1, which substance can be oxidized by O2 (g) in acidic aqueous solution?
A) Ni 2+ (aq)
B) Br2 (l)
Table 20.2
Half-reaction
Cr3+ (aq) + 3e- ¬ Cr (s)
C) Ag (s)
D) Cu2+ (aq)
E) Br- (aq)
Ee (V)
-0.74
Fe2+ (aq) + 2e- ¬ Fe (s)
Fe3+ (aq) + e- ¬ Fe2+ (s)
+0.771
Sn4+ (aq) + 2e- ¬ Sn2+ (aq)
+0.154
-0.440
19) The standard cell potential (Ee cell ) for the voltaic cell based on the reaction below is __________ V.
Sn2+ (aq) + 2Fe3+ (aq) ¬ 2Fe2+ (aq) + Sn4+ (aq)
A) +1.39
B) +0.46
C) +0.617
D) +1.21
E) -0.46
20) The standard cell potential (Ee cell ) for the voltaic cell based on the reaction below is __________ V.
3Sn4+ (aq) + 2Cr (s) ¬ 2Cr3+ (aq) + 3Sn2+ (aq)
A) +0.89
B) +1.94
C) -0.59
D) +2.53
E) -1.02
21) The relationship between the change in Gibbs free energy and the emf of an electrochemical cell is given by
__________.
-nF
-nF
-E
B) DG = -nRTF
C) DG =
D) DG =
E) DG = -nFE
A) DG =
E
ERT
nF
4
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Part B short answers
1- A Mn/Mn2+ and Ag+ /Ag electrochemical cell is set up at standard conditions. Draw the
electrochemical cell for this particular reaction. Label all parts of the cell, including the
voltage produced
2- Identify the Reducing Agent by circling it.
Mn+2 + BiO3- →MnO4 – + Bi+3 (acid solution)
3- Identify the Oxidizing Agent by circling it.
MnO4 – + C2O4 -2 →CO2 + MnO2 (Basic solution)
Communication
4- For each of the following compounds, identify the oxidation number of the atom(s)
indicated. a) Mg2TiO4
Mg=
Ti=
O=
b) K2Cr2O7
c) MnO4 –
K=
Mn=
Cr=
O=
O=
5- How does a Galvanic cell differ from and electrolytic cell?
6- Balance the equation for the following redox reaction in an acidic solution: Show the
half-reactions for oxidation and reduction. Show your work and oxidation numbers.
Application
7- Draw a diagram for the following standard cell
b. Write chemical equations to represent the following (3 mks):
The half reaction for reduction
The half reaction for oxidation
The complete net reaction
c. Calculate the electric potential for the previous cell and determine if the reaction is
spontaneous or not?
9-Balance the following redox equations in basic solution
10- Thinking
During the electrolysis of a tin(II) chloride solution, SnCl2(aq), a silver-colored metal, is
deposited onto the cathode. Bubbles of a gas are observed at the anode. A bleach-like odour is
detected. Use a redox table to predict the anode, the cathode, and the net ionic equation
11- An electrolytic cell is set up using inert electrodes in an aqueous copper(II) sulfate solution.
After an hour of electrolysis, the solution is tested. The concentration of copper(II) ions has
decreased, while the concentration of hydrogen ions has increased. The mass of one electrode
has also increased. Explain these observations using the half-reaction equation for each
electrode.
12- Electroplating involves an electrolytic cell in which the object to be plated is made to be one
of the electrodes. What reactions take place at the cathode and the anode when each of the
following solutions is used in an electrolytic cell? Write the half-reaction equations for the anode
and cathode, and state whether the reaction would result in metal being plated onto the object.
(a) 1.0 mol/L nickel bromide solution, NiBr2(aq)
(b) 1.0 mol/L aluminum fluoride solution, AlF3(aq)
(c) 1.0 mol/L manganese iodide solution, MnI2(aq)
13- Predict the half-reaction that will take place at the cathode and anode when each salt is
melted and electrolyzed.
(a) NiBr2(l)
(b) AlF3(l)
(c) MnI2(l)
9.2 Balancing Redox Reactions
Using Oxidation Numbers
Learning Goals …
… balance complex redox reactions using oxidation
numbers in aqueous solutions
… balance redox reactions in acidic and basic
solutions
• Most simple redox reactions can be balanced by
inspection.
• For more complex reactions, we can use either of
two methods:
1. Oxidation Numbers Method
2. Half Reactions Method
• For redox reactions, the charge must be balanced,
and so must each element.
Oxidation Numbers Method
1. Determine the oxidation numbers for each element in the
equation and identify the elements for which the
oxidation numbers change.
2. Adjust the values of the coefficients to balance the
electrons transferred.
3. Balance the rest of the equation by inspection. If
necessary, balance oxygen by adding H2O.
4. If necessary, balance hydrogen by adding H+ and/or OH-.
5. Check your answer.
6. Write the balanced equation.
• Redox reactions often take place in aqueous
solutions, which can be acidic, basic or
neutral.
when balancing, you may need to include
H2O, H+ or OH-.
Eg 1)
This number becomes the coefficient
required to balance the electrons
in ox #, oxidized, 7e- lost (x4)
4 NH3 + 7 O2
(-3)(+1)
(0)
4 NO2 + 6 H2O
(+4)(-2)
in ox #, reduced, 2e- gained /O
4e- gained /O2 (x7)
• Add H2O to balance the oxygens
Eg 2) A Redox Reaction in Acidic Solution
in ox #, oxidized, 1e- lost (x5)
8 H+ + MnO4- + 5 Fe2+
(+7)(-2)
(+2)
Mn2+ + 5 Fe3+ + 4 H2O
(+2)
(+3)
in ox #, reduced, 5e- gained (x1)
• Add H2O to balance the oxygens
• Add H+ to balance the hydrogens
Note: The charges must balance (+17 on each side)
Eg 3) A Redox Reaction in Basic Solution
in ox #, red., 6e- gained (x1)
I- + 6 CO2 + 3 H2O + 6 OH6 H+ + IO3- + 3C2O42(-1) (+4)(-2)
6 OH- (+5)(-2) (+3)(-2)
6 H2O
3 H2O
in ox #, ox. 1e- lost /C
2e- lost /C2
(x3)
• Add H2O to balance the oxygens
• Add H+ to balance the hydrogens
• Add same # of OH- (as H+ added) to both sides.
Note: H+ + OHH2O (cancel out extra H2O)
The charges must balance (-7 on each side)
Self Check
How prepared am I to start my homework? Can I …
… balance complex redox reactions using oxidation
numbers in aqueous solutions?
… balance redox reactions in acidic and basic
solutions?
HOMEWORK
p613 #1-6
p617 #3,4
10.2 Standard Reduction
Potentials
Learning Goals …
… calculate the cell potential using a table of standard
reduction potentials
… predict the spontaneity of redox reactions based on
overall cell potentials
• The Cell Potential is the electric potential difference or
voltage between the anode and cathode.
• The Standard Reduction Potential is the ability of a halfcell to attract electrons in a cell that is operating under
standard conditions (1.0M, standard pressure and ambient
temperature (101.3 kPa, 25°C)
Calculating Standard Cell Potential:
E(cell) = Er(cathode) – Er(anode)
❖ If E°(cell) > 0 then the reaction is spontaneous
❖ If E°(cell) < 0 then the reaction is not spontaneous Ex 1) Determine the cell potential and write the net ionic equation for the following spontaneous reaction: Mn2+ + 2e-  Mn(s) Er = -1.18 V Fe3+ + e-  Fe2+ Er = 0.77 V Lower  reducing agent  OXIDIZED (anode) Higher  oxidizing agent  REDUCED (cathode) Calculating the Standard Cell Potential: E(cell) = Er(cathode) - Er(anode) E(cell) = 0.77 - (-1.18) = 1.95 V Ex 1) Determine the cell potential and write the net ionic equation for the following spontaneous reaction: Mn2+ + 2e-  Mn(s) Er = -1.18 V Fe3+ + e-  Fe2+ Er = 0.77 V Lower  reducing agent  OXIDIZED (anode) Higher  oxidizing agent  REDUCED (cathode) Net Ionic Equation: • Since the above reactions are both reduction reactions, you must reverse the reaction occurring at the anode • Also multiply the equations in order for the electrons to be equal (Note: Do not multiply the reduction potential (E r) Ex 1) Reverse reaction Mn2+ + 2e-  Mn(s) Er = -1.18 V Fe3+ + e-  Fe2+ Er = 0.77 V Lower  reducing agent  OXIDIZED (anode) Higher  oxidizing agent  REDUCED (cathode) Multiply reaction by 2 Net Ionic Equation: Mn(s) → Mn2+ + 2e2Fe3+ + 2e- → 2Fe2+ Mn(s) + 2Fe3+ → Mn2+ + 2Fe2+ Ex 2) Determine the cell potential and write the net ionic equation for the following: Reaction at the anode is reversed Zn(s)  Zn2+(aq) Sn2+(aq) Sn(s) anode cathode Zn2+(aq) + 2e-  Zn(s) Er = -0.76 V (anode) Sn2+(aq) + 2e-  Sn(s) Er = -0.14 V (cathode) E(cell) = Er(cathode) - Er(anode) E(cell) = -0.14-(-0.76) = 0.62 V Net Ionic Equation: Half reactions: Zn(s) → Zn2+(aq) + 2eSn2+(aq) + 2e- → Sn(s) Zn(s) + Sn2+(aq) → Zn2+(aq) + Sn(s) Ex 3) Using cell potential, determine if the following redox reaction will be spontaneous: Fe3+ + 2Cl- Fe2+ + Cl2 Find the reactions on the standard reduction potential table Fe3+ + e-  Fe2+ Er= 0.77 V (cathode) Cl2 + 2e-  2ClEr= 1.36 V (anode) Note: • The above reactions are both reduction reactions • Since the chlorine is reversed in the net ionic equation, this reaction is occurring at the anode (oxidation) E(cell) = Er(cathode) - Er(anode) E(cell) = 0.77 - 1.36 = -0.59 V  Not a spontaneous reaction Self Check How prepared am I to start my homework? Can I … … calculate the cell potential using a table of standard reduction potentials? … predict the spontaneity of redox reactions based on overall cell potentials? HOMEWORK p647 #1 p648 #2-6, 8a 9.3 Predicting Redox Reactions Learning Goal … … predict the spontaneity of redox reactions based on position in standard reduction potential table • A redox reaction involves a transfer of valence electrons from one substance to another. • Most atoms, molecules, and ions are stable, and do not readily release electrons. • Since two particles must be involved in an electron transfer, this transfer can be explained as a competition for electrons. If one particle is able to pull electrons away from the other, a spontaneous reaction occurs. If not, the reaction is not spontaneous, and it will not occur on its own. Experiments can be run … … and results can be summarized in a table like the one below Standard Reduction Potentials Table high likelihood of being reduced Standard reduction potential is the tendency for a chemical species to be reduced. S.R.P. is measured in volts at standard conditions. The more positive the potential is, the more likely it will be reduced. The table to the right lists the S.R.P.s of common oxidizing agents and reducing agents (in order from strongest to weakest). The Spontaneity Rule spontaneous reaction no spontaneous reaction spontaneous reaction no spontaneous reaction *note – the oxidation half reaction will appear flipped on the reduction table • • When an oxidizing agent (OA) is above the reducing agent (RA) on the redox table, a spontaneous reaction occurs. But when an oxidizing agent is below the reducing agent, no spontaneous reaction occurs. Ex) Use the redox table to predict whether the following reactions will be spontaneous: *again – the oxidation half reaction will appear flipped on the reduction table a) Cr (s) + NiSO4(aq) ? RA Cr (s) + Ni2+(aq) ? Cr (s)  Cr3+(aq) + 3e- LEO, RA Ni2+(aq) + 2e-  Ni (s) GER, OA OA is above the RA  spontaneous reaction b) Ag(s) + Pb(NO3)2(aq) ? OA Ag (s) + Pb2+(aq) ? OA Ag (s)  Ag+(aq) + e- LEO, RA Pb2+(aq) + 2e-  Pb (s) GER, OA OA is below the RA  no spontaneous reaction RA Ex) What will happen if an aluminum spoon is used to stir a solution of iron (II) nitrate? Balanced Chemical Equation: 2Al(s) + 3Fe(NO3)2(aq) → 2Al(NO3)3(aq) + 3Fe(s) Net Ionic Equation: 2Al(s) + 3Fe2+(aq) → 2Al3+(aq) + 3Fe(s) Half Reactions: Al(s) → Al3+ + 3e- LEO, RA OA Fe2+ + 2e- → Fe(s) GER, OA  Since the oxidizing agent is above the reducing agent, the reaction will occur spontaneously & the aluminum spoon will dissolve in the solution RA Self Check How prepared am I to start my homework? Can I … … predict the spontaneity of redox reactions based on position in standard reduction potential table? HOMEWORK P623 #1, 3, 4, 6 p628 #7-9, 28-31, 59 9.1 Electron Transfer Reactions Learning Goals … … identify the reactant being oxidized and reduced in a redox reaction … assign oxidation numbers to determine if a chemical reaction is a redox reaction … identify the reducing and oxidizing agents in a redox reaction OXIDATION • the process by which one or more electrons is lost by a chemical entity REDUCTION • the process by which one or more electrons is gained by a chemical entity OXIDATION-REDUCTION (REDOX) REACTION • the reaction in which one or more electrons are transferred between chemical entities e.g. When copper wire placed in a solution of silver nitrate. The copper atoms in the wire are displaced by silver ions in solution. The blue tint of the solution is due to copper ions, and the fuzzy coating on the wire is the silver metal. Cu(s) + 2 AgNO3(aq) → 2 Ag(s) + Cu(NO3)2(aq) Total ionic equation Cu(s) + 2 Ag+(aq) + 2 NO3- (aq) → 2 Ag(s) + Cu2+(aq) + 2 NO3- (aq) Net ionic equation Cu(s) + 2 Ag+(aq) → 2 Ag(s) + Cu2+(aq) Half-Reactions • redox reactions and can be broken into two half-reactions: Oxidation Cu(s) → Cu2+(aq) + 2e- Reduction Ag+(aq) + e- → Ag(s) e.g. Write the half-reaction equations for the redox reaction that occurs when aluminum is placed in a silver nitrate solution. Balanced chemical equation: Al(s) + 3 AgNO3(aq) → 3 Ag(s) + Al(NO3)3 (aq) Ionic Equation: Al(s) + 3Ag + (aq) + 3NO3- (aq) → 3Ag(s) + Al3+(aq) + 3NO3- (aq) Net Ionic Equation: Al(s) + 3Ag+ (aq) → 3Ag(s) + Al3+(aq) Oxidation half reaction: Al(s) → Al3+ (aq) + 3eReduction half reaction: Ag + (aq) + e- → Ag (s) OXIDATION NUMBERS • a number used to keep track of electrons in redox reactions • also known as oxidation state Definition • the apparent net electric charge that an atom would have if the electron pairs in covalent bonds belonged entirely to the more electronegative atom -2 (2 extra e-) +3 -3 +1 (1 less e-) -2 Rules for Assigning Oxidation Numbers Rule Examples 1. A pure element has an oxidation Na O2 Fe number of zero (0) ( 0) ( 0) 2. The oxidation # of an ion equals the Al3+ O2charge of the ion (+3) (-2 ) 3. The oxidation # of an ion in an ionic compound is its ionic charge 4. The oxidation # of hydrogen is +1, except in metal hydrides, H is –1. Na2S (+1)(-2 ) MgO (+2)( -2) H2S H2O MgH2 (+1)( -2) (+1)( -2) (+2)(-1 ) 5. The oxidation # of oxygen is –2, unless it is a peroxide. (eg. H2O2) 6. The sum of all oxidation #s of all elements in a compound is zero Li2O CO2 H2O2 (+1)( -2) (+4)( -2) (+1)( -1) +4 -4 +2 -2 Na2SO4 KNO3 (+1)(+6)(-2 ) (+1)(+5)( -2) +2 +6 -8 +1 +5 -6 HClO4 (+1)( +7)(-2 ) +1 +7 -8 PCl3 CS2 N2O5 (+3)( -1) (+4)( -2) (+5)(-2) +3 -3 +4 -4 +10 -10 7. In covalent compounds, the more electronegative element is assigned an oxidation # that equals the negative charge it usually has in its ionic compounds 8. The sum of all oxidation #s of all the NO2- SO42Cr2O72elements in a polyatomic ion equals (+3)( -2) (+6)(-2) (+6)( -2) +3 -4 +6 -8 +12 -14 the charge of the ion Applying Oxidation Numbers to Redox Reactions Oxidation • increase in oxidation number Reduction • decrease in oxidation number * if there is no change in oxidation number then a redox reaction does not take place * e.g. Determine whether the following reactions are redox reactions. If so, identify what is being oxidized and reduced. a) CH4 + Cl2 → CH3Cl + HCl (-4)(+1) (0) (-2)(+1)(-1) (+1)(-1) Oxidation (↑ in Ox #): Carbon: goes from -4 to -2 (↑2) Reduction (↓ in Ox #): Chlorine: goes from 0 to -1 (↓1) b) CaCO3 + 2HCl → CaCl2 + H2O + CO2 (+2)(+4)(-2) (+1)(-1) (+2)(-1) (+1)(-2) (+4)(-2) *no ion undergoes a change in oxidation # therefore not a redox reaction! Oxidizing and Reducing Agents OXIDIZING AGENT • the reactant that is reduced • gains electrons REDUCING AGENT • the reactant that is oxidized • loses electrons e.g. Identify the oxidizing and reducing agents in each of the following. a) 3 H2S + 2 NO3- + 2 H+ → 3 S + 2 NO + 4 H2O +1 -2 +5 -2 +1 0 +2 -2 +1 -2 oxidation reduction reducing agent: H2S oxidizing agent: NO3b) MnO2 + 4 HBr → MnBr2 + Br2 + 2 H2O +4 -2 +1 -1 +2 -1 0 +1 -2 reduction oxidizing agent: MnO2 oxidation reducing agent: HBr Self Check How prepared am I to start my homework? Can I … … identify the reactant being oxidized and reduced in a redox reaction … assign oxidation numbers to determine if a chemical reaction is a redox reaction … identify the reducing and oxidizing agents in a redox reaction HOMEWORK p601 #1a, 4 p604 #1-4 p606 #1b, 3 p607 #1, 5, 6ab, 7-10 10.7 Electrolysis Learning Goals … … explain the difference between a galvanic and an electrolytic cell … predict the products of the electrolysis of molten and aqueous solution There are two types of electrochemical cells: 1. A galvanic cell produces a current when an oxidationreduction reaction occurs spontaneously 2. An electrolytic cell uses electrical energy to produce a chemical reaction that would not occur spontaneously Examples of electrolytic cells: • Recharging a battery • Electroplating metals Galvanic Cell vs Electrolytic Cell Galvanic Cell vs Electrolytic Cell Electrolysis of Water The electrolysis of water produces hydrogen gas at the cathode and oxygen gas at the anode. Note: electrolysis is not a spontaneous reaction and requires a battery Cathode (reduction): 2 H2O(l) + 2e− → H2(g) + 2 OH-(aq) Er= -0.83V Anode (oxidation): 2 H2O(l) → O2(g) + 4 H+(aq) + 4e− Er = 1.23V Overall reaction: 2 H2O(l) → 2 H2(g) + O2(g) Ecell = -2.06V Predicting the Products of Electrolysis of an Aqueous Solution 1. Identify the two possible electrolysis reactions that may occur (electrolysis of salt and electrolysis of water). 2. List the four relevant half reactions and their reduction potentials. 3. Determine which reaction will occur at cathode (the reaction with highest reduction potential will be reduced) 4. Determine which reaction will occur at the anode (the reaction with the lowest reduction potential will be oxidized) e.g. Predict the products of the electrolysis of 1.0 mol/L NiCl2 (aq). Write the net ionic equation and calculate the cell potential. 2 possible reactions (4 half-reactions) Cathode (reduction): NiCl2(aq) → Ni(s) and Cl2(g) 2 H2O(l) → 2H2(g) + O2(g) highest reduction potential E = - 0.23 V Ni2+ + 2e- → Ni(s) 2 H2O(l) + 2e− → H2(g) + 2 OH-(aq) Anode (oxidation): r Er = - 0.83 V lowest reduction 2Cl- → Cl2(g) + 2eEr = 1.36 V potential 2 H2O(l) → O2(g) + 4 H+(aq) + 4 e− Er = 1.23 V Net Ionic Equation: Do not change these values Multiply by 2 Cathode Ni2+ + 2e- → Ni(s) Er = - 0.23 V Anode 2H2O(l) → O2(g) + 4H+(aq) + 4e− Er = 1.23 V 2Ni2+ + 4e- → 2Ni(s) 2H2O(l) → O2(g) + 4H+(aq) + 4e− Er(cathode) = - 0.23 V Er(anode) = 1.23 V 2Ni2+(aq)+2H2O(l) → 2Ni(s) + O2(g) + 4H+ Ecell = - 1.46 V (Cathode - Anode) Self Check How prepared am I to start my homework? Can I … … explain the difference between a galvanic and electrolytic cell? … predict the products of electrolysis of a molten and an aqueous solution? HOMEWORK p670 #3 (molten solutions), 8 p682 #60 (aqueous solutions) include net ionic equations and calculate cell potential for each