Extra Credit 22
- Page ID
- 82729
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Q17.3.1 
For each reaction listed, determine its standard cell potential at 25 °C and whether the reaction is spontaneous at standard conditions.
- Mg(s)+Ni2+(aq)⟶Mg2+(aq)+Ni(s)
- 2Ag+(aq)+Cu(s)⟶Cu2+(aq)+2Ag(s)
- Mn(s)+Sn(NO3)2(aq)⟶Mn(NO3)2(aq)+Sn(s)
- 3Fe(NO3)2(aq)+Au(NO3)3(aq)⟶3Fe(NO3)3(aq)+Au(s)
Q17.3.1 Answer:
The standard cell potential (E°cell) is the voltage/potential difference produced from the oxidation (happens at the anode) and reduction (happens at the cathode) half reactions in the galvanic cell. The first tool that we need to calculate for E°cell is the Standard Reduction Potentials, where negative potentials (top) are more likely to oxidize than the positive potentials (bottom) which are more likely to reduce. The second tool that we need is the formula, E°cell= E°(cathode)-E°(anode). From the result, we can determine the spontaneity of the reaction by looking at the sign of the E°cell, if E°cell is positive then the reaction is spontaneous and non-spontaneous if E°cell is negative.
1. Mg(s)+Ni2+(aq)⟶Mg2+(aq)+Ni(s)
Step 1: Write the cell half reactions and their respective cell potentials:
\[Mg^{2+}(aq)+2e^{-}\rightarrow Mg(s)\] with \(E^o=-2.356\, V\)
\[Ni^{2+}(aq)+2e^{-}\rightarrow Ni(s)\] with \(E^o=-0.25\, V\)
Step 2: Using the Standard Reduction Potentials table, determine where each half reaction happens (anode or cathode) and balance the electrons between the half reactions:
Anode (oxidation half reaction): \[Mg^{2+}(aq)+2e^{-}\rightarrow Mg(s)\] with \(E^o=-2.356\, V\)
Cathode (reduction half reaction): \[Ni^{2+}(aq)+2e^{-}\rightarrow Ni(s)\] with \(E^o=-0.25\, V\)
Step 3: Calculate the standard cell potential, E°cell, and determine the spontaneity of the reaction:
\[E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode} = (-0.25 V)-(-2.356 V) =+2.106 V\]
You can also determine the standard cell potential by adding the half reactions together. This allows us to add the potential of the anode and cathode together. Remember that the charge (sign) of the potential changes when you flip the reaction.
Example:
Anode (oxidation half reaction): \[Mg(s)\rightarrow Mg^{2+}(aq)+2e^{-}\] with \(E^o=2.356\, V\)
Cathode (reduction half reaction): \[Ni^{2+}(aq)+2e^{-}\rightarrow Ni(s)\] with \(E^o=-0.25\, V\)
\[2.356 + (-0.25) = +2.106 V\]
This reaction is spontaneous because the cell potential (E°) is positive, +2.106 V.
2. 2Ag+(aq)+Cu(s)⟶Cu2+(aq)+2Ag(s)
Step 1: Write the cell half reactions and their respective cell potentials:
\[Ag^{+}(aq)+e^{-}\rightarrow Ag(s)\] with \(E^o=+0.800\, V\)
\[Cu^{2+}(aq)+2e^{-}\rightarrow Cu(s)\] with \(E^o=+0.340\, V\)
Step 2: Using the Standard Reduction Potentials table, determine where each half reaction happens (anode or cathode) and balance the electrons between the half reactions:
**One important thing to remember is that when we multiply a coefficient to balance out the electrons between the half reactions, the cell potential of each reaction doesn't change.**
Anode (oxidation half reaction) : \[Cu^{2+}(aq)+2e^{-}\rightarrow Cu(s)\] with \(E^o=+0.340\, V\)
Cathode (reduction half reaction) : \[2(Ag^{+}(aq)+e^{-}\rightarrow Ag(s))\] with \(E^o=+0.800\, V\)
Step 3: Calculate the standard cell potential, E°cell, and determine the spontaneity of the reaction:
\[E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode} = (+0.800 V)-(+0.340 V) =+0.460 V\]
This reaction is spontaneous because the cell potential (E°) is positive, +0.460V.
3. Mn(s)+Sn(NO3)2(aq)⟶Mn(NO3)2(aq)+Sn(s)
Step 1: Rewrite the overall reaction in the net ionic equation form (because it separates in solution) and cancel out the spectator ion, in this case is NO3-, and balance the equation if needed :
\[Mn(s)+Sn^{2+}(aq)+2NO^{-}_3(aq)\rightarrow Mn^{2+}(aq)+2NO^{-}_3(aq)+Sn(s)\]
Which simplify into: \[Mn(s)+Sn^{2+}(aq)\rightarrow Mn^{2+}(aq)+Sn(s)\] where the spectator ion, NO3-, is taken out from both sides.
Step 2: Write the cell half reactions and their respective cell potentials:
\[Mn^{2+}(aq)+2e^{-}\rightarrow Mn(s)\] with \(E^o=-1.19\, V\)
\[Sn^{2+}(aq)+2e^{-}\rightarrow Sn(s)\] with \(E^o=-0.137\, V\)
Step 3: Using the Standard Reduction Potentials table, determine where each half reaction happens (anode or cathode) and balance the electrons between the half reactions:
Anode (oxidation half reaction) : \[Mn^{2+}(aq)+2e^{-}\rightarrow Mn(s)\] with \(E^o=-1.19\, V\)
Cathode (reduction half reaction) : \[Sn^{2+}(aq)+2e^{-}\rightarrow Sn(s)\] with \(E^o=-0.137\, V\)
Step 4: Calculate the standard cell potential, E°cell, and determine the spontaneity of the reaction:
\[E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode} = (-0.137 V)-(-1.19 V) =+1.053 V\]
This reaction is spontaneous because the cell potential (E°) is positive, +1.053 V.
4. 3Fe(NO3)2(aq)+Au(NO3)3(aq)⟶3Fe(NO3)3(aq)+Au(s)
Step 1: Rewrite the overall reaction in the net ionic equation form (because it separates in solution) and cancel out the spectator ion, in this case is NO3-, and balance the equation if needed:
\[3Fe^{2+}(aq)+6NO^{-}_3(aq)+Au^{3+}(aq)+3NO^{-}_3(aq)\rightarrow 3Fe^{3+}(aq)+9NO^{-}_3+Au(s)\]
Which simplify into: \[3Fe^{2+}(aq)+Au^{3+}(aq)\rightarrow 3Fe^{3+}(aq)+Au(s)\]
Step 2: Write the cell half reactions and their respective cell potentials:
\[Fe^{3+}(aq)+e^{-}\rightarrow Fe^{2+}(aq)\] with \(E^o=+0.771\, V\)
\[Au^{3+}(aq)+3e^{-}\rightarrow Au(s)\] with \(E^o=+1.52\, V\)
Step 3: Using the Standard Reduction Potentials table, determine where each half reaction happens (anode or cathode) and balance the electrons between the half reactions:
**One important thing to remember is that when we multiply a coefficient to balance out the electrons between the half reactions, the cell potential of each reaction doesn't change.**
Anode (oxidation half reaction) : \[3(Fe^{3+}(aq)+e^{-}\rightarrow Fe^{2+}(aq))\] with \(E^o=+0.771\, V\)
Cathode (reduction half reaction) : \[Au^{3+}(aq)+3e^{-}\rightarrow Au(s)\] with \(E^o=+1.52\, V\)
Step 4: Calculate the standard cell potential, E°cell, and determine the spontaneity of the reaction:
\[E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode} = (+1.52 V)-(+0.771 V) =+0.749V\]
This reaction is spontaneous because the cell potential (E°) is positive, +0.749 V.
Q19.1.20
What is the gas produced when iron(II) sulfide is treated with a non-oxidizing acid?
Q19.1.20 Answer:
Step 1: Figure out the formula for iron (II) sulfide and determine what a non-oxidizing acid is:
Formula for iron(II) sulfide: \[FeS\]
Definition of non-oxidizing acid: A non-oxidizing acid is an acid that doesn't act the oxidizing agent. Its anion is a weaker oxidizing agent than H+, thus it can't be reduced. Examples of non-oxidizing acids: \[HCl, HI, HBr, H_3PO_4, H_2SO_4\]
Step 2: Choose one of the non-oxidizing acid, in this case HCl, and write the chemical reaction:
\[FeS(s)+2HCl(aq)\rightarrow FeCl_2(s)+H_2S(g)\]
The gas produced when iron (II) sulfide treated with a non-oxidizing acid, HCl, is H2S (dihydrogen sulfide) gas.
Q19.3.12
Would you expect salts of the gold ion, Au+, to be colored? Explain.
Q19.3.12 Answer:
Color is resulted from electron's natural tendency to fall down down an energy level to back to the ground level from an excited level. When the electron falls down, energy is given off in the form of the light, or in transition metals' case, the light energy released is often in the visible region.
Step 1: Determine the electron configuration of Au+ and how many d electrons it has:
Au: \[[Xe]6s^{2}4f^{14}5d^{10}\]
**For transition metals, when it is oxidized, it loses the electrons from s orbital first before it loses its d electrons:
Au+:\[[Xe]6s^{1}4f^{14}5d^{10}\]
Step 2: Determine whether Au+ can be colored looking at the geometry:
The octahedral geometry for Au+ is all filled by the 10 d electrons, thus there is no empty space for visible light to promote electron from one d-orbital to another. Therefore, Au+ cannot be colored.
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.
| (M)[Penicillin] | 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 |
Q12.4.12 Answer:
The reaction order and rate law is expressed in terms of reactant(s) and determined from experimental data.
Step 1: Set up an equation/simplify to compare the ratio between two different rates, in this case rate 1 and 2 are picked, to find the reaction order:
\[(ratio\space of\space concentration)^{x}=(ratio\space of\space rate)\] \[(0.667)^{x}=(0.667)\]
By inspection, it can be assume that the reaction order is x= 1.
Step 2: Using the reaction order and the rate law formula, r=k[M]x, find the rate constant (k) with rate 1's data:
\[r=k[Penicillin]^{x}\] \[1.0 \times 10^{-10}=k(2.0 \times 10^{-6})^{1}\] \[k=5.0 \times 10^{-5} 1/min\]
Q21.2.7
What are the two principal differences between nuclear reactions and ordinary chemical changes?
Q21.2.7 Answer:
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The first principal difference between nuclear reactions and ordinary chemical changes is how the reaction takes place in the atom. Nuclear reaction happens in the atom's nucleus and involves the splitting of the nuclei which makes the element unstable and emitting neutrons/protons. It results in radiation (alpha, beta, or gamma) in order to come back to stability. On the other hand, chemical reaction happens outside the nucleus and involves the two atoms' electrons where they are rearrange (transfer and share of electrons) in order for atoms to gain stability.
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The second principal difference between nuclear and chemical reactions is their energy change. Nuclear reaction involves high energy change, while chemical reaction only involves low energy change. In nuclear reaction, a tiny difference from the product's total mass and reactant's total mass can produce a massive amount of energy release.
Q21.6.2
Technetium-99m has a half-life of 6.01 hours. If a patient injected with technetium-99m is safe to leave the hospital once 75% of the dose has decayed, when is the patient allowed to leave?
Q21.6.2 Answer:
Each element has a specific half-life where the original concentration decayed into half. In this case, the half-life Technetium-99m is 6.01hr. In nuclear kinetic, all radioactive decay is first-order process (ln(N/N0)=-kt), thus its radioactive half-lives follow the first order half-life's formula (t1/2=ln(2)/k).
Step 1: Determine the decay rate constant k using the formula the following formula: \[t_{1/2}=ln(2)/k\]
\[6.01=ln(2)/k\]
\[k=ln(2)/6.01\]
\[k=0.1153 hr\]
Step 2: Determine the time that takes for 75% of Technetium-99m to decay using the formula, ln(N/N0)=-kt :
Note: 75% can be written as 0.75, which means that N(final concentration)=1- 0.75=0.25 since 75% has decayed and N0(original concentration)=1 since it is assume that the patient was injected 100% in the beginning.
\[ln(0.25/1)=-0.1153(t)\]
\[t=[ln(0.25/1)]/-0.1153\]
\[t=12.02 hrs\]
This means that patient injected with technetium-99m is safe to leave the hospital after 12.02 hrs since 75% of the dose has decayed.
Q20.4.8
Identify the oxidants and the reductants in each redox reaction.
- Br2(l) + 2I−(aq) → 2Br−(aq) + I2(s)
- Cu2+(aq) + 2Ag(s) → Cu(s) + 2Ag+(aq)
- H+(aq) + 2MnO4−(aq) + 5H2SO3(aq) → 2Mn2+(aq) + 3H2O(l) + 5HSO4-(aq)
- IO3−(aq) + 5I−(aq) + 6H+(aq) → 3I2(s) + 3H2O(l)
Q20.4.8 Answer:
The oxidants or oxidizing agents are species that have the ability to oxidize other substances and are themselves reduced (gained electrons). On the other hand, reductants or reducing agents are species that have the ability to reduce other substances and are themselves oxidized (donates/loses their electrons).
1. Br2(l) + 2I−(aq) → 2Br−(aq) + I2(s)
Step 1: Find the oxidation number of each species using the "rules to determine oxidation number" :
\[Br_2=0\]
\[Br^{-}=-1\]
\[I^{-}=-1\]
\[I_2=0\]
Step 2: Determine whether the species is the oxidant or reductant by looking at the flow of the electron(s):
- Br2 is the oxidant because it took an electron from I-.
- I- is the reductant because it donated its electron to Br2.
2. Cu2+(aq) + 2Ag(s) → Cu(s) + 2Ag+(aq)
Step 1: Find the oxidation number of each species using the "rules to determine oxidation number" :
\[Cu^{2+}=+2\]
\[Cu=0\]
\[Ag=0\]
\[Ag^{+}=+1\]
Step 2: Determine whether the species is the oxidant or reductant by looking at the flow of the electron(s):
- Cu2+ is the oxidant because it took 2 electron from Ag.
- Ag is the reductant because each Ag donated 1 electron to Cu2+.
3. H+(aq) + 2MnO4−(aq) + 5H2SO3(aq) → 2Mn2+(aq) + 3H2O(l) + 5HSO4-(aq)
Step 1: Find the oxidation number of each species using the "rules to determine oxidation number" :
\[H^{+}=+1\]
\[Mn^2+=+2\]
\[MnO^{-}_4:Mn=+7, O=-2\]
\[H_2O: H=+1, O=-2\]
\[H_2SO_3: H=+1, S=+4, O=-2\]
\[HSO^{-}_4: H=+1, S=+6, O=-2\]
Step 2: Determine whether the species is the oxidant or reductant by looking at the flow of the electron(s):
- MnO4− is the oxidant because Mn is reduced from +7 to +2.
- H2SO3 is the reductant because each S is oxidized from +4 to +6.
4. IO3−(aq) + 5I−(aq) + 6H+(aq) → 3I2(s) + 3H2O(l)
Step 1: Find the oxidation number of each species using the "rules to determine oxidation number":
\[IO^{-}_3: I=+5, O=-2\]
\[I_2=0\]
\[I^-=-1\]
\[H_2O: H=+1, O=-2\]
\[H^+=+1\]
Step 2: Determine whether the species is the oxidant or reductant by looking at the flow of the electron(s):
- IO3− is the oxidant because I is reduced from +5 to 0.
- I- is the reductant because each I is oxidized from -1 to 0.
In this case, I is both oxidized and reduced.
Q20.7.5
This reaction is characteristic of a lead storage battery:
Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(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?
Q20.7.5 Answer:
Step 1: Write the anode and cathode half reactions for lead storage battery and state the cell potential (E°) for each of them :
Anode: \[Pb(s) + HSO^{-}_4(aq)\rightarrow PbSO_4(s)+H^{+}(aq)+2e^{-}\] with \(E^o=-0.356\, V\)
Cathode: \[PbO_2(s)+HSO^{-}_4(aq)+3H^{+}(aq)+2e^{-}\rightarrow PbSO_4(s)+2H_2O(l)\] with \(E^o=+1.685\, V\)
Step 2: Calculate the E° cell for the lead storage equation:
\[E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode} = (+1.685 V)-(-0.356 V) =+2.041V\]
Step 3: Using the Nernst Equation: \[E_{cell}=E^{o}_{cell}-[(RT/nF)lnQ]\]
Compare how the amount of sulfuric acid affect the Ecell. Given: \[R= 8.3145 J/mol\times K\] \[F=96,485 C/mol\times e^{-}\] \[T=298 K\] \[n=2\]
The Nernst Equation can be simplify to: \[E_{cell}=E^{o}_{cell}-(0.0128\times ln([HSO^{-}_4]^{2}/[H^{+}]^{2}))\]
The standard lead storage cell is about 35% sulfuric acid by mass, thus assuming that all the factors in the equation are the same and the only change is the 30% sulfuric acid, the resulting cell potential is greater than that of the standard cell.