Extra Credit 22
- Page ID
- 82882
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For each reaction listed, determine its standard cell potential at 25 °C and whether the reaction is spontaneous at standard conditions.
\[a. Mg_{(s)}+Ni^{2+} _{(aq)}⟶Mg^{2+}_{(aq)}+Ni_{(s)}\]
\[b. 2Ag^{+} _{(aq)} + Cu_({s})⟶Cu^{2+}_{(aq)}+2Ag_{(s)}\]
\[c. Mn_{(s)} + Sn(NO_3)_2(aq)⟶Mn(NO_3)_2(aq)+Sn_{(s)}\]
\[d. 3Fe(NO_3)_2(aq) +Au(NO_3)_3(aq)⟶3Fe(NO_3)_3(aq)+Au_{(s)}\]
Strategy
1. In order to find the standard cell potential, first write the equation for each half-reaction and determine which one corresponds to the anode and cathode. Remember that oxidation occurs at the anode and reduction occurs at the cathode.
2. To calculate the standard cell potential use the equation:
\[E°_{cell}=E°_{cathode} − E°_{anode}\]
3. Use the standard reduction potential table below and find the corresponding values for each of your half reactions. Then, substitute the values into the equation in step 2.
4. If the standard cell potential is a positive value, it will be spontaneous. If it is a negative value, it will be non-spontaneous.
Solutions
a.
Step 1:
Anode:
\[Mg_{(s)}→ Mg_{(aq)}^{2+}+ 2e^−\]
Oxidation occurs here because the oxidation state of Mg increases from 0 to +2.
Cathode:
\[Ni^{2+}_{(aq)}+2e−→Ni_{(s)}\]
Reduction occurs here because the oxidation state of Ni decreases from +2 to 0.
Step 2 & 3:
From the standard reduction potential table, the anode's reduction potential is -2.37V and the cathode's reduction potential is -0.25V. Now substitute that into the equation:
\[E°_{cell}= (-0.25V) − (-2.37V)\]
\[E°_{cell}= +2.12V\]
Step 4:
+2.12V is a positive value therefore the reaction is spontaneous.
b.
Step 1:
Anode:
\[Cu_{(s)}→ Cu_{(aq)}^{2+}+ 2e^−\]
Cathode:
\[2Ag^{+}_{(aq)}+2e−→2Ag_{(s)}\]
Step 2 & 3:
From the standard reduction potential table, the anode's reduction potential is +0.34V and the cathode's reduction potential is +0.80V. Now substitute that into the equation:
\[E°_{cell}= (+0.80V) − (+0.34)\]
\[E°_{cell}= +0.46V\]
Step 4:
+0.46V is a positive value therefore the reaction is spontaneous.
c.
Step 1:
Anode:
\[Mn_{(s)} + 6H_{2}O_{(l)} → Mn(NO_{3})_{2(aq)} + 12H^+ + 12e^-\]
Cathode:
\[Sn(NO_{3})_{2(aq)} + 12H^+ + 12e^-→Sn_{(s)} + 6H_{2}O_{(l)}\]
*NO3- (aq) is a spectator ion
Step 2 & 3:
From the standard reduction potential table, the anode's reduction potential is -1.18V and the cathode's reduction potential is -0.14V. Now substitute that into the equation:
\[E°_{cell}= (-0.14) − (-1.18)\]
\[E°_{cell}= +1.04V\]
Step 4:
+1.04V is a positive value therefore the reaction is spontaneous.
d.
Step 1:
Anode:
\[Fe^{2+}_{(aq)} → Fe^{3+}_{(aq)} + e^-\]
Cathode:
\[Au^{3+}_{(aq)} + 3e^-→Au_{(s)}\]
*NO3- (aq) is a spectator ion
Step 2 & 3:
From the standard reduction potential table, the anode's reduction potential is +0.77V and the cathode's reduction potential is +1.50V. Now substitute that into the equation:
\[E°_{cell}= (+1.50) − (+0.77)\]
\[E°_{cell}= +0.73V\]
Step 4:
+0.73V is a positive value therefore the reaction is spontaneous.
Q19.1.20
What is the gas produced when iron(II) sulfide is treated with a nonoxidizing acid?
A nonoxidizing acid is when the anion is a weaker oxidizing agent that H3O+. A common example of a nonoxidizing acid is HCl. Cl- cannot act as an oxidizing agent because it cannot gain any more electrons. It's outermost orbital already has a complete octet. Using HCl as the nonoxidizing acid the reaction yielded is:
\[FeS + 2HCl → H_{2}S + FeCl_2\]
The gas produced is H2S which is hydrogen sulfide which is a colorless gas.
Q19.3.12
Would you expect salts of the gold ion, Au+, to be colored? Explain.
The electronic configuration of Au+ is [Xe]4f145d10. The usual definition of a transition metal is one which forms one or more stable ions which have incompletely filled d orbitals. In this case, Au+ is not a transition metal as it has a full 5d sublevel. Non-transition metals don't have any electron transitions which can absorb wavelengths from visible light, which is the case for Au+. Salts of Au+ would be colorless because no visible light is absorbed. Au+ is colorless because there higher energy orbitals that an excited electron would normally go into is full. The metal cannot absorb any light.
Provided here is a visualization:
Note that this also occurs when there are no electrons in the d sublevel (i.e d0)
Q12.4.12
Some bacteria are resistant to the antibiotic penicillin because they produce penicillinase, an enzyme with a molecular weight of 3 × 104 g/mol that converts penicillin into inactive molecules. Although the kinetics of enzyme-catalyzed reactions can be complex, at low concentrations this reaction can be described by a rate equation that is first order in the catalyst (penicillinase) and that also involves the concentration of penicillin. From the following data: 1.0 L of a solution containing 0.15 µg (0.15 × 10−6 g) of penicillinase, determine the order of the reaction with respect to penicillin and the value of the rate constant.
| [Penicillin] (M) | Rate (mol/L/min) |
|---|---|
| 2.0 × 10−6 | 1.0 × 10−10 |
| 3.0 × 10−6 | 1.5 × 10−10 |
| 4.0 × 10−6 | 2.0 × 10−10 |
Strategy
- Determine the order of the reaction by calculating how change in concentrations of Penicilin affect the rate.
- If it is a zero order reaction, changing the concentration of the reactant will not affect the rate.
- If it is a first order reaction, doubling a particular reactant concentration, makes the reaction rate double, then the
rate is directly proportional to that species. - If it is second order, doubling a particular reactant concentration, makes the reaction rate quadruple, then the rate law is second‐order dependent in that species (i.e 22 =4)
- If it is third order, tripling a particular reactant concentration makes the reaction rate increases by a factor of eight, then the rate law is third‐order in that species (i.e 23=8)
- Form the rate law according to the order determined. The overall order is determined by adding the coefficients of each of the reactants. In this case there is only one reactant so the rate law would be: rate = k[Penicilin]x (Determine x based on the previous steps)
- Choose any row from the table and substitute for rate and concentration to determine k, the rate constant.
Solution
Looking at rows 1 and 3, the concentration 2.0 × 10−6 M doubles to 4.0 x 10-6M and correspondingly, the rates double from 1.0 × 10−10 mol/L/min to 2.0 × 10−10 mol/L/min. This means that it is a first order reaction.
Thus the rate is:
\[rate = k[penicilin]^1\]
To determine k let us substitute the values of row 1:
\[1.0\cdot 10^{-10} = k[2.0\cdot 10^{-6}\]
\[k = \frac{1.0\cdot 10^{-10}}{2.0\cdot 10^{-6}}\]
\[k = 5\cdot 10^{-5} s^{-1}\]
Q21.2.7
What are the two principal differences between nuclear reactions and ordinary chemical changes?
| Nuclear Reactions | Ordinary Chemical Changes |
|
Involves a change in an atom's nucleus, usually producing a different element. Species is changed at a fundamental level. |
Involves only a rearrangement of electrons and do not involve changes in the nuclei. The same number of each kind of atom appear in the reactants and products. |
|
Mass of reactants ≠ Mass of products (We no longer have conservation of mass. This is because some mass is converted into energy.) |
Mass of reactants = Mass of products (Mass is conserved) |
Q21.6.2
Technetium-99m has a half-life of 6.01 hours. If a patient injected with technetium-99 is safe to leave the hospital once 75% of the dose has decayed, when is the patient allowed to leave?
Strategy
- For nuclear reactions, assume first order reaction. Determine the rate constant, k, of the decay by using the equation:
\[t_{\frac{1}{2}} = \frac{0.693}{k}\]
- Use the half life, 6.01 hours and substitute it into the equation, allowing to solve for k.
- Determine the fraction of technetium-99 remaining.
- Use this equation to solve for time:
\[ln\frac{N}{N_o} = -kt\]
Solution
\[k = \frac{0.693}{6.01}\]
\[k = 0.1153\]
Fraction of technium-99 remaining when 75% has decayed is:
\[1 - 0.75 = 0.25\]
(This is the same as 25%)
Thus the patient is allowed to leave when the time passed reaches:
\[ln\frac{25}{100} = -(0.1153)t\]
\[\frac{-1.38629}{-0.1153} = t\]
\[t = 12 hours\]
Q20.4.8
Identify the oxidants and the reductants in each redox reaction.
- \[Br_{{2}(s)}+2I^{-} _{(aq)}⟶2Br^{-}_{(aq)}+I_{{2}(s)}\]
- \[Cu_{(aq)}^{2+} + 2Ag_{(s)} → Cu_{(s)} + 2Ag_{(aq)}^{+}\]
- \[H^+_{(aq)} + 2MnO4^−_{(aq)} + 5H_2SO_{3(aq)} → 2Mn_{2(aq)} + 3H_2O_{(l)} + 5HSO_{4(aq)}^-\]
- \[IO_{3(aq)}^- + 5I^−_{(aq)} + 6H_{(aq)}^+ → 3I_{2(s)} + 3H_2O_{(l)}\]
Strategy
- Determine the change in oxidation numbers (O.N) of each of the respective species.
Refer to the rules for assigning oxidation numbers:
Rule 1: The oxidization number of an element in its free (uncombined) state is zero. (ex: Al(s) or Zn(s))
- Rule 2: The oxidation number of a monatomic ion is the same as the charge on the ion. (ex: Na+ or S2-)
- Rule 3: The sum of all oxidation numbers in a neutral compound is zero. The sum of all oxidation numbers in a polyatomic ion is equal to the charge on the ion.
- Rule 4: The oxidation number of an alkali metal (IA group) in a compound is +1, while the oxidation number of an alkaline earth metal (IIA group) in a compound is +2.
- Rule 5: The oxidation number of oxygen in a compound is usually -2; however, if the oxygen is in a peroxide it has an oxidation number of -1. If oxygen is bounded to fluorine, the oxidation number is +1.
- Rule 6: The oxidation state of hydrogen in a compound is usually +1. If the hydrogen is part of a binary metal hydride (compound of hydrogen and some metal), the oxidation state of hydrogen is -1.
- Rule 7: The oxidation number of fluorine is always -1.
- To determine the oxidant and reductant, remember that oxidants are reduced in the chemical reaction, which means that the O.N decreases. Reductants are oxidized in the chemical reactions, which means that the O.N increases. Think of the acronym OILRIG. Oxidation is loss of elections so O.N. goes up. Reduction is gain of electrons so O.N. goes down.
1. Br2 has an O.N of 0 and decreases to -1 in Br−. The O.N decreases which shows reduction so it is the oxidant. I− has an O.N of -1 and increases to 0 in I. The O.N increases which shows oxidation so it is the reductant.
2. Cu2+ has an O.N of +2 and decreases to 0 in Cu. The O.N decreases which shows reduction so it is the oxidant. Ag has an O.N of 0 and increases to +1 in Ag+. The O.N increases which shows oxidation so it is the reductant.
3. This reaction is more complex, as it has 3 species. First realize that the O.N of H in H+ and H2O remain the same with an O.N of +1. Then, determine the O.N of MnO4− and Mn2+. Mn in MnO4− has an O.N of +7 because (-2 x 4) + 7 = -1. (Refer to rule 2 and 5 for better understanding). This decreases to +2 in Mn2+ which means that reduction occurs and MnO4− is the oxidant. The O.N of S in H2SO3 is +4 because (1 x 2) + (3 x -2) + 4 = 0. (Note that 0 is the O.N of the neutral compound, refer to Rule 3). This increases to +6 in HSO4− because 1 + (4 x -2) + 6 = -1, which is the charge of the ion. (Refer to Rule 3). This means that oxidation occurs and H2SO3 is the reductant.
4. First realize that the O.N of H+ and H2O remain the same with an O.N of +1. Then, determine the O.N of IO3− and I2. I in IO3− has an O.N of +5 because (-2 x 3) + 5 = -1. (Refer to rule 3 for better understanding). This decreases to 0 in I2 which means that reduction occurs and IO3− is the oxidant. The O.N of I in I- is -1 (refer to Rule 2). This increases to 0 in in I2. (Refer to rule 1) This means that oxidation occurs and I- is the reductant.
Q20.7.5
This reaction is characteristic of a lead storage battery:
\[Pb_{(s)} + PbO_{2(s)} + 2H_2SO_{4(aq)} → 2PbSO_{4(s)} + 2H_2O_{(l)}\]
If you have a battery with an electrolyte that has a density of 1.15 g/cm3 and contains 30.0% sulfuric acid by mass, is the potential greater than or less than that of the standard cell?
Firstly, you have to consider the reactions that occur at the anode and the cathode:
Cathode:
\[2e^- + PbO_2 (s) + HSO_4^- (aq) + 3H^+ (aq) \to PbSO_4 (s) + 2H_2O (l) E°=+1.685 V\]
Anode:
\[Pb (s) + HSO_4^- (aq) \to PbSO_4 (s) + H^+ (aq) + 2e^- E°=-0.356 V\]
Then, we can determine the reduction potential of the reaction at standard state.
\[E° = E°_{cathode} - E°_{anode} = 1.685 V - (-0.356 V) = 2.041 V\]
Now that we have the E cell, we can use the Nernst equation:
\(E= 2.041 V - {8.3145 {J \over mol \times K} (298 K)\over 2 mol e^- \times 96485 {C\over mol e^-} } ln{1\over[H_2SO_4]^2}\)
\(E= 2.041 V - 0.01284{ln {1\over[H_2SO_4]^2}}\)
To solve for the unknown molarity of sulfuric acid, we assume that the solution is 100g as it is given that the battery is 30% sulfuric acid. Thus, 30g is sulfuric acid and 70g is water. Now we must determine the number of moles of sulfuric acid and use density as a conversion factor to find the volume of the solution. This will then allow us ot calculate the molarity of the solution:
Firstly, find the moles of sulfuric acid using the information that it is 30g:
\({Moles H_2SO_4\over 100 g solution}={30 g H_2SO_4\over 100 g solution} \times {1 mol H_2SO_4\over 98.1 g H_2SO_4} = {{0.30581 mol H_2SO_4}\over 100 g solution}\)
Now you can find the volume of the 100g of solution by using density as a conversion factor:
Volume = \(100.0 g soln \times {1 mL \over 1.15 g solution}= 86.9565 mL = 0.086957 L\)
\(Molarity of H_2SO_4 = {moles H_2SO_4\over liters of solution}={ 0.30581 mol H_2SO_4\over 0.086957 L}= 3.52 M\)
This enables us to plug in the molarity into the Nernst equation:
\(E= 2.041 V - 0.01284ln {1\over[H_2SO_4]^2} = 2.041 V - 0.01284ln {1\over (3.52 M)^2}\)
=2.073 V
Thus,
\(E°\)= 2.041 V
\(E\)=2.073 V
So \(E> E°\). The potential is greater than that of the standard cell.