Extra Credit 20
- Page ID
- 83414
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In each of the following examples, write the cell diagram and cathode. Indicate the direction of the electron flow; write a balanced equation.
- Al(s)+Cu2+→Cu(s)+Al3+
- Fe3++Ag(s)→Ag++Fe2+
- Cu(s)+2Fe3+→Cu2++2Fe2+
Solutions:
In an electrochemical cell,
- Oxidation occurs at the anode, drawn on the left hand side. Reduction occurs at the cathode, drawn on the right hand side.
- A voltmeter determines the voltage produced by the reaction. Voltage is measured in volts (V) which is J/C. Voltage is the same as electromotive force (emf) and cell potential.
- A solid metal in the half reaction is often used as an electrode in the half cell. If the reaction does not contain a solid, use Pt (s) or C (graphite) as the electrode.
- An aqueous compound is used as the solution in the half cell.
- Half cells are connected with a salt bridge to prevent charge buildup by allowing ions to exchange between the bridge and the half cell.
- A wire is used to allow electrons to transfer from the oxidation to reduction half-reaction. Electrons in the diagram are shown flowing in the spontaneous direction.
- Electrons flows from the left side to the right hand side.
1.
Step 1: we need to identify which half reaction is reduction or oxidation. Recall the reduction is gains of elections and Oxidation is loss electrons. Oxi: Al → Al3+ + 3e- Red: Cu2+ +2e- → Cu
Step 2: when you write the cell diagram, salt bridge symbol as the center and you will always starts from aqueous, gas, liquid, and solid. Cell diagram shown below: Al (s) | Al3+ (aq) || Cu2+ (aq) | Cu (s)
where | is change in states; || is salt bridge. Cathode is Cu2+ +2e- → Cu, and anode is Al → Al3+ + 3e-
Step 3: writing a balance equation, start with write out the half reactions: Oxi: Al → Al3+ Red: Cu2+ → Cu
Step 4: Balance the charge or oxidation number with electrons Oxi: Al → Al3+ + 3e- Red: Cu2+ +2e- → Cu
Step 5: Balance O by adding H2O. In this case, this equation has no Oxygen so we can skip it. Oxi: Al → Al3+ + 3e- Red: Cu2+ +2e- → Cu Step 6: Balance H by adding H+. In this case, this equation has no Hydrogen either.
Step 7: multiply by some integer to make electrons (lost) = electrons (gained). We try to cancels out the electrons on both side Oxi: (Al → Al3+ + 3e- ) x2 Red: (Cu2+ +2e- → Cu) x3
Step 8: check atom balance and charge balance on both sides of the equation! 2 Al + 3Cu2+ → 2 Al3+ + 3Cu
Step 9: At the cathode, Cu2+ (aq) is reduced to Cu (aq). We uses the standard cell equation:
Ecell°=Ecathode°−Eanode° E=0.34 V
Step 10: E = [electrode potential for Al3+ (aq) ] E = -(-1.66 V) = 1.66V Ecell = 0.34 V + 1.66 V = 2.00 V
2.
Step 1: we need to identify which half reaction is reduction or oxidation. Recall the reduction is gains of elections and Oxidation is loss electrons. Oxidation: Fe3+ (aq) +e−→Fe2+ (aq) Reduction: Ag+ (aq) + e- → Ag (s)
Step 2: construct the cell diagram as the order, (aq), (g), (l), and (s) Pt (s) | Fe2+ (aq) | Fe3+ (aq) || Ag+ (aq) | Ag (s)
Step 3: write out the half reactions in order to reach the processing of balancing the equation Oxidation: Fe3+ (aq) →Fe2+ (aq) Reduction: Ag+ (aq) → Ag (s)
Step 4: Balance the charge or oxidation number with electrons Oxidation: Fe3+ (aq) +e−→Fe2+ (aq) Reduction: Ag+ (aq) + e- → Ag (s)
Step 5: multiply by some integer to make electrons (lost) = electrons (gained). We try to cancels out the electrons on both side, and check atom balance and charge balance on both sides of the equation. Fe3++Ag(s)→Ag++Fe2+
Step 6: According to the chart from Q19.59B, we known: Fe3+ (aq) +e−→Fe2+ (aq) E=0.771 V Ag+ (aq) + e- → Ag (s) E=0.80 V
Step 7: use the equation: Ecell°=Ecathode°−Eanode° E=(0.80 V) - (0.771 V) = 0.03 V
3.
Step 1: we need to identify which half reaction is reduction or oxidation. Recall the reduction is gains of elections and Oxidation is loss electrons Oxi: Cu (s)→ Cu2+ (aq) + 2e- (Anode) Red: Fe3+ (aq) +e−→Fe2+ (aq) (Cathode)
Step 2: construct the cell diagram as the order, (aq), (g), (l), and (s) Cu (s) | Cu2+ (aq) || Fe3+ (aq) | Fe2+ (aq) | Pt (s)
Step 3: balance the equation Cu(s)+2Fe3+→Cu2++2Fe2+
Step 4: Find the cell potential 2Fe3+ (aq) +e−→2Fe2+ (aq) E = 0.77 V Cu (s)→ Cu2+ (aq) + 2e- -E= - 0.34 V Ecell°=Ecathode°−Eanode° Ecell°=0.77 V - 0.34 V =0.43 V
Q19.59B
What would happen to the corrosion of iron nail under phenolphtalein indicator, if:
- Lead wire are wrapped around the nail
- Deep scratch occurred at the iron nail
- Galvanized nail is substituted for the iron nail
Solutions:
First of all, corrosion is a process through which metals in fabricated states come back to their regular oxidation states. This process is a reduction-oxidation reaction in which the metal is being oxidized by its environment, frequently the oxygen in air. However, we also need to understand that the phenolphtalein indicator is used for as an indicator in base-acid titration where it usually turn to colorless. Corrosion is essentially an unwanted galvanic cell.
- Based on the chart of the reducing agent, we noticed that Pb, as Lead wire is a weaker reducing agent(weaker reducing agent means lower tendency to get oxidized, mostly by the air, oxygen) than iron, it tend to less reactive with metal. When the Lead wire wrapped around the nail, OH- (aq) would be produced near the lead wire. The following reactions might occur: Anode: 2Fe (s) → 2Fe2+ (aq) +4e- Cathode: O2 (g) +2H2O (l) +4e-→ 4 OH- (aq) The reaction at the cathode is actually more complicated than shown here. Fe2+ (aq) reacts further in a series of reactions to form 2 Fe2O3 . H2O, which is rust.
- Scratch exposes the metal, more susceptible to corrosion. Blue precipitate around the scratch should occur.
- Zinc protect iron from corrosion as it is a strong reducing agent. Thus there should be no blue precipitate. Zinc corrodes instead. Pink color of OH−(aq) would still continue to form.
Q21.3E
- Sketch the geometric isomers of [FeCl2(CO)2(NH3)2].
- Sketch the geometric isomers of [NiCl4(NO)2], Label whether it is cis, trans, mer, fac, or none.
Solutions:
Geometric Isomers: A stereoisomer where the arrangement of the ligands are either neighboring (cis or fac) or across (trans or mer) from each other.
- This cis/trans isomerism occurs only in square planar and octahedral complexes when 2 identical ligands are either neighboring (cis) or across (trans or mer) from each other.
- Trans occurs when 2 identical ligands are bonded across from each other. In other words, these 2 ligands are on the same axis.
- Cis occurs when 2 identical ligands are bonded neighboring each other. In other words, these 2 ligands are on different axes
- The fac/mer isomerism occurs only in octahedral complexes when 3 identical ligands are either neighboring (fac) or across (mer) from each other.
- Fac occurs when 3 identical ligands are bonded neighboring each other.
- Mer occurs with 3 identical ligands occur, and 2 of the 3 are bonded on the same axes.
1.
Step 1: since the formula [FeCl2(CO)2(NH3)2], which we can expected that the geometric isomers will inside of the bracket. In this case, the transition metal is Fe so that we can draw it as the center.
Step 2: Since there is 6 stereoisomers so that we can expected it is octahedral.
2.
Step 1: [NiCl4(NO)2] also a octahedral structure but there is no mer or fac because it only have 4 identical ligands (Cl) and 2 identical ligands (NO) rather than 3 identical ligands.
Q24.11B
The initial rate of the reaction A+B→C+D is determined for different initial conditions, with the results listed in the table:
Expt |
[A], M |
[B], M |
Initial Rate, Ms-1 |
---|---|---|---|
1 |
0.331 |
0.203 |
2.205 X 10-4 |
2 |
0.331 |
0.406 |
8.82 X 10-4 |
3 |
0.662 |
0.203 |
4.41 X 10-4 |
4 |
0.662 |
0.406 |
1.7 X 10-3 |
a. What is the order of reaction with respect to A and B?
b. What is the overall reaction order?
c. What is the value of the rate constant k?
Solutions:
We have noted that in general the rate of a reaction slows down as time passes. The dependence of the rate on concentration for a reaction can generally be stated as an experimentally determined mathematical expression, called the differential rate law. In general, it was written as:
The overall order is supposed to be m+n, k is the rate constant for the reaction at a certain temperature. For this problem, it is important to note factors that change between reactions as the table shown above.
a. When you try to find the order of reaction with respect to A, we need to look at the values of the [A], M and [B],M. However, we have noticed that the Reaction 1 to Reaction 3 has different values of [A],M but the same values of [B],M. So that we can do a simple math as the experiment 3 divide by the experiment 1: Reaction 3/Reaction 1=(4.41 × 10-4)/(2.205 × 10-4)={[0.406]m[0.662]n}/{[0.203]m[0.662]n}=2m, m=1. Thus the reaction is first order with respect to A. We do the same thing for B that we find where A is a constant and B changes (Reaction 2 divide by Reaction 1). Then we should get the value as 4=2n, n=2. Thus the reaction is second order with respect to B.
b. Reaction order = reaction order of A + reaction order of B = 1+2=3. Thus the overall reaction order is third order reaction.
c. Since we already figure out the values of m and n, we can now write the rate equation as: rate=k[A][B]2. In order to find k which the rate constant, we can plug it in into any set of the reactions. We can use the reaction 1: 2.205 x 10-4=k x [0.331][0.203]2, k=0.016, or 1.6 x 10-2.
Q24.51B
The following observations of a reaction’s rate constant have been made: at T=325 K, k=3.2×10-6 M-1s-1; at T=456 K, k=2.8×10-5 M-1s-1. What is the activation energy of this reaction?
Solutions:
Use the equation ln k1 / k2 = Ea / R ( 1/T1 - 1/T2)
ln (2.8 x 10-5)/(3.2 x 10-6) = Ea/8.3145J/mol K (1/325 - 1/456)
-4.21 = Ea (-2.22 x 10-4)/8.3145
-35.004 = Ea(-2.22 x 10-4)
Ea = 2.25 x 104 J/mol or 159 kJ/mol
Q25.27B
A museum has just bought an ancient Greek vase from a collector. Radiocarbon dating of the vase reveals a disintegration rate of 8.0 dis min-1 g-1. The original disintegration rate is 15.0 dis min-1 g-1. If the vase was from the “Golden Age” of Athens around 480 BC-404 BC do you think the object is authentic? Explain
Solutions:
ln Nt / N0 = -λt
t1/2 = ln 2 /λ
λ = ln 2/ t1/2 = ln 2/ 5730 days = 1.21 x 10-4 years-1
ln 8.0 dis min-1 g-1 / 15 dis min-1g-1 = -(1.21 x 10-4 years-1)t
t= ln ( 8.0 dis min-1 g-1 / 15 dis min-1g-1) / -(1.21 x 10-4 years-1) = 5195.11 years.
Q18.7
A voltaic cell has an aluminum electrode in Al2(SO4)3 solution in one compartment and the other compartment has a lead electrode in PbSO4 solution.
- Which has a greater tendency to be oxidized, Al or Pb? Write a balanced equation for the spontaneous reaction.
- Draw a diagram of the voltaic cell, including
- the anode and cathode
- the direction of flow of electrons, positive ions and negative ions
- Which electrode will increase in mass?
Solutions:
1.
Step 1: balance this equation
2 Al (s) + 3 PbSO4 (aq) → 3 Pb (s) + Al2(SO4)3 (aq)
Step 2: Identify the oxidation and the reduction
Since Aluminum loses electrons, it was oxidized. Lead gains electrons, it was reduced. As a result, Al tend to has the greater tendency to be oxidized.
2.
In terms of the voltaic cell appears as NO3-, we use SO4.
3.
The electrode of Pb will increase in mass since Pb2+ gain 2 electrons and that will makes solid Pb.
Q21.4.4
Ethyl-2-nitrobenzoate (NO2C6H4CO2C2H5) hydrolyzes under basic conditions. A plot of [NO2C6H4CO2C2H5] versus t was used to calculate t½, with the following results:
[NO2C6H4CO2C2H5] (M/cm3) | t½ (s) |
0.050 | 240 |
0.040 | 300 |
0.030 | 400 |
Is this a first-order reaction? Explain your reasoning.
Solutions:
The half-life (t1/2) is a timescale on which the initial population is decreased by half of its original value, represented by the following equation.
[A]=1/2[A]o
After a period of one half-life, t=t1/2t=t1/2 and we can write
[A]1/2 / [A]o=1 / 2 = e−kt1/2
Taking logarithms of both sides (remember that lnex=xlnex=x) yields
ln 0.5 = −kt
Solving for the half-life, we obtain the simple relation
t1/2 = ln 2 / k ≈ 0.693k
This indicates that the half-life of a first-order reaction is a constant.
Now look at this problem, the table shows [NO2C6H4CO2C2H5] [M/cm3] is decreasing while the half life is increasing, thus this is a first half-life order.