Extra Credit 20

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 E_cell = 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

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.

1. 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.
2. Scratch exposes the metal, more susceptible to corrosion. Blue precipitate around the scratch should occur.
3. 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.

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.

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}=2^m, 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.

Solutions:

Use the equation ln k₁ / k₂ = E_a / R ( 1/T₁ - 1/T₂)

ln (2.8 x 10^-5)/(3.2 x 10^-6) = E_a/8.3145J/mol K (1/325 - 1/456)

-4.21 = E_a (-2.22 x 10^-4)/8.3145

-35.004 = E_a(-2.22 x 10^-4)

E_a = 2.25 x 10⁴ J/mol or 159 kJ/mol

Solutions:

ln N_t / N₀ = -λt

t_1/2 = ln 2 /λ

λ = ln 2/ t_1/2 = 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.

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.

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=xln⁡ex=x) yields

ln 0.5 = −kt

Solving for the half-life, we obtain the simple relation

t_1/2 = 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.