Extra Credit 3

Q17.1.3

For each of the following balanced half-reactions, determine whether an oxidation or reduction is occurring.

1. Fe3++3e−⟶FeFe3++3e−⟶Fe
2. Cr⟶Cr3++3e−Cr⟶Cr3++3e−
3. MnO2−4⟶MnO−4+e−MnO42−⟶MnO4−+e−
4. Li++e−⟶LiLi++e−⟶Li

S17.1.3

(a) reduction, in the reaction Fe3++3e−⟶Fe, we can see that Fe3+ is gaining electrons which denotes reduction is taking place.

(b) oxidation, in the reaction Cr⟶Cr3++3e−Cr⟶Cr3++3e−, we can see that Cr is losing electrons which denotes oxidation is taking place.

(c) oxidation, in the reaction MnO2−4⟶MnO−4+e−, we can see that MnO2 is losing electrons which denotes oxidation is taking place.

(d) reduction. in the reaction Li++e−⟶LiLi++e−⟶Li, we can see that Li+ is gaining electrons which denotes reduction is taking place.

NOTE: A reduction is a half reaction where a substance, called the oxidant, gains electrons and is reduced in the corresponding process.

Oxidation is a half reaction where a substance, called the reductant, loses electrons in the corresponding system.

Q19.1.1

Write the electron configurations for each of the following elements:

1. Sc
2. Ti
3. Cr
4. Fe
5. Ru

S19.1.1

Orbitals are filled according to Hund's rule- unpaired electrons first, before putting two electrons in a single orbital.

1. [Ar]4s²3d¹, by looking at the periodic table we can find the element Sc to be located on the fourth period in group 3. If we count the number of electrons in the outer shell of this element, which is found by the group # for elements in groups 1-12 and group #- 10 for elements in groups 13-18, we get 3. The first 2 are used to fill up the s orbital and the third is for the d orbital.
2. [Ar]4s²3d², by looking at the periodic table we can find the element Ti to be located on the fourth period in group 4. If we count the number of electrons in the outer shell of this element, which is found by the group # for elements in groups 1-12 and group #- 10 for elements in groups 13-18, we get 4. The first 2 are used to fill up the s orbital and the rest are used for the d orbital.
3. [Ar]4s¹3d⁵, by looking at the periodic table we can find the element Cr to be located on the fourth period in group 6. If we count the number of electrons in the outer shell of this element, which is found by the group # for elements in groups 1-12 and group #- 10 for elements in groups 13-18, we get 6. The first electron is used in the s orbital and the rest are used to fill up the d orbital, this is due to the fact that the s orbital will lose an electron in order to fill the d orbital to make the element more stable.
4. [Ar]4s²3d⁶, by looking at the periodic table we can find the element Fe to be located on the fourth period in group 8. If we count the number of electrons in the outer shell of this element, which is found by the group # for elements in groups 1-12 and group #- 10 for elements in groups 13-18, we get 8. The first 2 are used to fill up the s orbital and the rest are used for the d orbital.
5. [Kr]5s¹4d⁷, by looking at the periodic table we can find the element Ru to be located on the fifth period in group 8. If we count the number of electrons in the outer shell of this element, which is found by the group # for elements in groups 1-12 and group #- 10 for elements in groups 13-18, we get 8. The first electron is used to in the s orbital and the rest are used for the d orbital, this is due to the fact that the s orbital will lose an electron in order to fill the d orbital to make the element more stable.

Q19.2.3

Give the coordination number for each metal ion in the following compounds:

1. [Co(CO3)3]3− (note that CO32− is bidentate in this complex)
2. [Cu(NH3)4]2+
3. [Co(NH3)4Br2]2(SO4)3
4. [Pt(NH3)4][PtCl4]
5. [Cr(en)3](NO3)3
6. [Pd(NH3)2Br2] (square planar)
7. K3[Cu(Cl)5]
8. [Zn(NH3)2Cl2]

S19.2.3

According to Libretext, "the coordination number represents the number or coordinate covalent bonds." It can range from 1 to 12, but 2,4, and 6 are more common.

1. 6; since ligand CO₃^-2 is bidentate, which means two electron pairs donated, and there are 3 ligands, 3x2=6
2. 4; since ligand NH₃ is monodentate, which means only one electron pair is donated, and there are 4 ligands, 4x1=4
3. 6; there are 4 NH₃ ligands which are mondentate, 4x1=4, and 2 Br^- ligands which are monodentate, 2x1=2, so adding these up 4+2=6
4. 8; there are 4 NH₃ ligands which are mondentate, 4x1=4, and 4 Cl^- ligands which are monodentate, 4x1=4, so adding these up 4+4=8
5. 6; there are 3 en ligands which are bidentate, 3x2=6
6. 4; there are 2 NH₃ ligands and 2 Br^- ligand which are both monodentate as suggested by the structure's geometry (square planar), 4x1=4
7. 5; there are 5 Cl^- ligands which are monodentate, 5x1=5
8. 4;there are 2 NH₃ ligands and 2 Cl^- ligand which are both monodentate

Q12.3.15

Nitrogen(II) oxide reacts with chlorine according to the equation:

2NO(g)+Cl2(g)⟶2NOCl(g)2NO(g)+Cl2(g)⟶2NOCl(g)

The following initial rates of reaction have been observed for certain reactant concentrations:

1.What is the rate equation that describes the rate’s dependence on the concentrations of NO and Cl2?

2.What is the rate constant?

3.What are the orders with respect to each reactant?

S12.3.15

1. rate=k[NO]²[Cl₂]¹

These values can be determined by first looking at the change in concentration and comparing it to the change in the rate.

we start out with our unknowns m and n:

rate=k[NO]^m[Cl₂]ⁿ

We can find m by looking at the first two concentrations of NO, during this time the concentration of Cl₂ does not change so we do not have to worry about it affecting the rate. The concentration of NO doubles while the rate quadruples, this means that 4=2^m (rate=[NO]^m since k and Cl₂ are both constant) so from this relationship m=2. Then, by looking at the second and third concentrations of Cl₂ , during this time NO is constant, we can see that the concentration of Cl₂ doubles while the rate also doubles in size. This means that 2=2ⁿ (rate=[Cl₂]ⁿ since k and NO are both constant) n=1.

2. k=9.12 L²/(mol²*h)

k can be found by plugging in any of the three initial rates and there subsequent concentrations of reactants.

Ex) [NO]=1M, [Cl₂]=1M, and rate=9.12M/h

9.12=k(1)²(1)

M/h=[k](M)²(M)

3. Reaction orders can be found by the superscript of each reactant. The overall order of the reaction is third since m+n=3. The order with respect to NO is second since m=2 and the order with respect to Cl₂ is first since n=1.

Q12.6.7

Write the rate equation for each of the following elementary reactions:

1. O3→sunlightO2+O
2. O3+Cl⟶O2+ClO
3. ClO+O⟶Cl+O2
4. O3+NO⟶NO2+O2
5. NO2+O⟶NO+O2

S12.6.7

For an explanation of elementary reactions, visit the following link:-

https://chem.libretexts.org/Core/Phy...tary_Reactions

The rate equations for these single step reactions can be determined by the reactants and their subscripts.

Ex) aA⟶bB

rate=k[A]^a

1. rate=k[O₃]; O₃ is the reactant and its coefficient is one
2. rate=k[O₃][Cl]; O₃ and Cl are the reactants and there coefficient are both one
3. rate=k[ClO][O]; ClO and O are the reactants and both have a coefficient of one
4. rate=k[O₃][NO]; O₃ and NO are the reactants and both have a coefficient of one
5. rate=k[NO₂][O]; NO₂ and O are the reactants and both have a coefficient of one

Technetium-99 is often used for assessing heart, liver, and lung damage because certain technetium compounds are absorbed by damaged tissues. It has a half-life of 6.0 h. Calculate the rate constant for the decay of Tc4399.

S21.4.19

0.12/h

For these type of problems we use first order equations, so ln(N_t/N₀)=-kt. Since the time we are given is a half-life, which is the time it takes for half of the activity to decay, we can assume that N_t/N₀=0.5 (this value is unit less since the units cancel out). then we can simply plug in the values to find k:

ln(0.5)=-k(6)

k=ln(0.5)/(-6)= 0.115=0.12

Q20.3.7

Copper(II) sulfate forms a bright blue solution in water. If a piece of zinc metal is placed in a beaker of aqueous CuSO4 solution, the blue color fades with time, the zinc strip begins to erode, and a black solid forms around the zinc strip.

1. What is happening?

2. Write half-reactions to show the chemical changes that are occurring.

3. What will happen if a piece of copper metal is placed in a colorless aqueous solution of ZnCl₂ZnCl₂?

S20.3.7

1. If we look at the list of standard reduction potentials we can see from the values of E^o for each reaction taking place that, Zn²⁺:E^o=-0.76V and Cu²⁺:E^o=0.34V. Zn is the anode and is being oxidized and that Cu²⁺ is the cathode and is being reduced. This can be determined because unless otherwise stated in the problem, a reaction can only occur if the E^o of the cathode is bigger than that of the anode. This redox reaction causes the solution to change color and the zinc to erode and form a black solid Cu. The solution changing color represents the Cu²⁺ being reduced into a solid black Cu, while the Zinc eroding represents Zn being oxidized and forming a clear solution in the water.

2. These reaction can be found on the list of standard reduction potentials or derived from the problem.

Anode: Zn(s)⟶Zn²⁺(aq) + 2e^-

Cathode:Cu²⁺(aq)+ 2e^-⟶ Cu(s)

Net reaction: Zn_(s) + Cu²⁺_(aq) -> Cu_(s) + Zn²⁺_(aq)

3. Nothing will occur. Again if we look at the standard reduction potentials we will find that Cu being oxidized into Cu²⁺ is unlikely when paired with Zn²⁺ since then the cathode would have a lower reduction potential than the anode.

Q20.5.18

In acidic solution, permanganate (MnO4−) oxidizes Cl− to chlorine gas, and MnO4− is reduced to Mn2+(aq).

1. Write the balanced chemical equation for this reaction.
2. Determine E°cell.
3. Calculate the equilibrium constant.

S20.5.18

1. 2MnO₄^-(aq)+10Cl^-(aq)+16H⁺(aq)⟶ 5Cl₂(g)+ 2Mn²⁺(aq)+8H₂O(l)

first we start out with the unbalance reaction explained in the problem:

MnO₄^-(aq)+Cl^-(aq)⟶ Cl₂(g)+ Mn²⁺(aq)

then we separate it into two half reactions:

MnO₄^-(aq)⟶ Mn²⁺(aq)