Extra Credit 15

Q17.2.4

Balance the following reactions and write the reactions using cell notation. Ignore any inert electrodes, as they are never part of the half-reactions.

1. Al(s)+Zr⁴⁺(aq)⟶Al³⁺(aq)+Zr(s)
2. Ag⁺(aq)+NO(g)⟶Ag(s)+NO₃⁻(aq)(acidic solution)
3. SiO₃^2-(aq)+Mg(s)⟶Si(s)+Mg(OH)₂(s)(basic solution)
4. ClO₃⁻(aq)+MnO₂(s)⟶Cl⁻(aq)+MnO₄⁻(aq)(basic solution)

Q17.2.4

1. Al(s)+Zr⁴⁺(aq)⟶Al³⁺(aq)+Zr(s)

First we will balance this chemical equation. Looking at the species involved, we might assume that the equation is balanced but it is not because the electrons on either side are unbalanced. We have one Al one each side and one Zr on each side of the equation as well. Taking a look at the charges, we can see that the Zr⁴⁺ turns into Zr, which means that 4 electrons are transferred in that half reaction. We can also see that the Al turns into Al³⁺ which means that 3 electrons are transferred. To equalize the number of electrons transferred, we must multiply the Al→Al³⁺ equations by 4 and the Zr⁴⁺→Zr equation by 3.

(i) 4[Al(s)⟶Al³⁺(aq)+3e^-].....This gives us 4Al(s)⟶4Al³⁺(aq)+12e^-

(ii) 3[4e^-+Zr⁴⁺(aq)⟶Zr(s)].....This gives us 12e^-+3Zr⁴⁺(aq)⟶3Zr(s)

Now we must add the two half-reactions together to get the final balanced equation. The electrons cancel out and the final equation is, 4Al(s)+3Zr⁴⁺(aq)⟶4Al³⁺(aq)+3Zr(s).

Finally, we put this reaction into cell notation, Al(s) | Al³⁺(aq) || Zr⁴⁺(aq) | Zr(s)

PHASE II: Tepanka- I agree with these solutions and simply edited them so that the charge of the molecules was distinguishable by superscripting them.

2. Ag⁺(aq)+NO(g)⟶Ag(s)+NO₃^- (aq)(acidic solution)

First we will balance this chemical equation. Immediately we see that the Ag molecules are balanced. The N atoms are also balanced. However, the O atoms are not. We will divide this equation into two half-reactions to balance the equation.

(i) Ag⁺(aq)⟶Ag(s) To balance this equation we have to add and electron to the left side.

1e^-+Ag⁺(aq)⟶Ag(s) This equation is balanced.

(ii) NO(g)⟶NO₃^- (aq) To balance this equation, we will add 2 H₂O molecules to the left side.

2H₂O(L)+NO(g)⟶NO₃^- (aq) Now we need to add 4 H⁺ atoms to the right side to balance the H atoms.

2H₂O(L)+NO(g)⟶NO₃^- (aq) + 4H⁺ Lastly, we need to balance the charges on each side so we must add 3 electrons to the right side.

2H₂O(L)+NO(g)⟶NO₃^- (aq) + 3e^- + 4H⁺ This equation is balanced.

Now to combine these two half equations, we must make the electrons equal to each other and add them together.

Equation (i) must be multiplied by 3 to make that happen.

3[1e^-+Ag⁺(aq)⟶Ag(s)] give us 3e^- + 3Ag⁺(aq)⟶3Ag(s)...........and we will add that to..............2H₂O(L)+NO(g)⟶NO₃^- (aq) + 3e^- + 4H⁺

So our final balanced equation for the whole reaction is 3Ag⁺(aq) + 2H₂O(L) + NO(g)⟶ 3Ag(s) + NO₃^- (aq) + 4H⁺

Finally, we put this into cell notation, Pt(s) | NO(g) | NO₃^- (aq) || Ag⁺(aq) | Ag(s)

PHASE II: Tepanka- I agree with these solutions, however I edited the equations to be able to distinguish between the number of atoms of an element in a molecule and the charge of the molecule. I also edited it so some of the solutions read that we are balancing elemental atoms, not always molecules. I also added Platinum (solid) to the cell notation because you need to have a solid electrode on both sides of the cell so that the wires and clamps can attach to them.

3. SiO₃²⁻(aq)+Mg(s)⟶Si(s)+Mg(OH)₂(s)(basic solution)

First we will balance this chemical equation. We start by dividing the equation into two half reactions.

(i) SiO₃^2-(aq)⟶Si(s) The Si atoms are balanced so we start by balancing the O atoms by adding 3H₂O molecules on the right side.

SiO₃^2-(aq)⟶Si(s) + 3H₂O(L) Now we balance the H atoms by adding 6 H⁺ atoms to the left side.

6H⁺(aq) +SiO₃^2-(aq)⟶Si(s) + 3H₂O(L) Now we balance the charges on both sides by adding 4 electrons to the left side.

4e^- + 6H⁺(aq) +SiO₃^2-(aq)⟶Si(s) + 3H₂O(L) This equation is balanced.

(ii) Mg(s)⟶Mg(OH)₂(s) The Mg atoms are balanced so we balance the O atoms by adding 2 H₂O molecules to the left side.

2H₂O (L) + Mg(s)⟶Mg(OH)₂(s) Now we balance the H atoms by adding 2 H⁺ atoms to the right side.

2H₂O (L) + Mg(s)⟶Mg(OH)₂(s) + 2H⁺(aq) We need to balance the charges and we do that by adding 2 electrons to the right side.

2H₂O (L) + Mg(s)⟶Mg(OH)₂(s) + 2H⁺(aq) + 2e^- This half equation is now balanced.

We need to add these two equations together now. We multiply equation (ii) by 2 to make the electrons equal, and thus cancel out.

2[2H₂O (L) + Mg(s)⟶Mg(OH)₂(s) + 2H⁺(aq) + 2e^-] gives us 4H₂O (L) + 2Mg(s)⟶2Mg(OH)₂(s) + 4H⁺(aq) + 4e^-

Add that to 4e^- + 6H⁺(aq) +SiO₃^2-(aq)⟶Si(s) + 3H₂O(L)

The new equation is 2H⁺(aq) + H₂O (L) + SiO₃^2-(aq) + 2Mg(s)⟶Si(s) + 2Mg(OH)₂(s)

However this is a basic solution so we must add 2 OH^- to the left and right sides to get rid of the H⁺ atoms.

3H₂O (L) + SiO₃^2-(aq) + 2Mg(s)⟶Si(s) + 2Mg(OH)₂(s) + 2OH^-(aq) is the final balanced equation in basic solution.

The cell notation is Mg(s) | Mg(OH)2(s) || SiO₃^2-(aq) | Si(s)

PHASE II: Tepanka- I agree with these solutions, but again, used the superscript and subscript buttons to distinguish charge from number of elements in a molecule/compound.

4. ClO₃⁻(aq)+MnO₂(s)⟶Cl⁻(aq)+MnO₄⁻(aq)(basic solution)

First we will balance this chemical equation. We start by dividing the equation into two half reactions.

(i) ClO₃⁻(aq)⟶Cl⁻(aq) The Cl atoms are balanced. We need to balance the O atoms by adding 3 H₂O molecules to the right side.

ClO₃⁻(aq)⟶Cl⁻(aq) + 3H₂O(L) Now we balance the H atoms by adding 6 H⁺ atoms to the left side.

6H⁺(aq) + ClO₃⁻(aq)⟶Cl⁻(aq) + 3H₂O(L) We also need to balance the charges by adding 6 electrons to the left side.

6e^- + 6H⁺(aq) + ClO₃⁻(aq)⟶Cl⁻(aq) + 3H₂O(L) This equation is now balanced.

(ii) MnO₂(s)⟶MnO₄⁻(aq) The Mn atoms are balanced. To balance the O atoms, we add 2 H₂O molecules to the left side.

2H₂O(L) + MnO₂(s)⟶MnO₄⁻(aq) Next, we add 4 H⁺ atoms to the right side.

2H₂O(L) + MnO₂(s)⟶MnO₄⁻(aq) + 4H⁺(aq) We need to add 3 electrons to the right side to balance the charges.

2H₂O(L) + MnO₂(s)⟶MnO₄⁻(aq) + 4H⁺(aq) + 3e^- This equation is now balanced.

Multiply equation (ii) by 2 to make electrons equal. Then add the equations together.

2[2H₂O(L) + MnO₂(s)⟶MnO₄⁻(aq) + 4H⁺(aq) + 3e^-] gives us 4H2O(L) + 2MnO₂(s)⟶2MnO₄⁻(aq) + 8H⁺(aq) + 6e^-

Add that to 6e^- + 6H⁺(aq) + ClO₃⁻(aq)⟶Cl⁻(aq) + 3H₂O(L)

This gives us ClO₃⁻(aq) + H₂O(L) + 2MnO₂(s)⟶Cl⁻(aq) + 2MnO₄⁻(aq) + 2H⁺(aq)

This is a reaction occurring in a basic solution so we need to add 2OH^- molecules to get rid of the H⁺ atoms.

2OH-(aq) + ClO₃⁻(aq) + 2MnO₂(s)⟶Cl−(aq) + 2MnO₄⁻(aq) + H2O(L) This is the final balanced equation in a basic solution.

The cell notation is MnO₂(s) | MnO₄⁻ (aq) || ClO₃⁻(aq) | Cl⁻(aq) | Pt(s)

PHASE II: Tepanka- I agree with these solutions. Again, used superscripts and subscripts. Added Platinum to cell notation for the necessity of a solid electrode.

Q19.1.13

Find the potentials of the following electrochemical cell:

Cd | Cd²⁺ (M = 0.10) ‖ Ni^{2+ (}M = 0.50) | Ni

Q19.1.13

To start this question, we need the Nernst Equation

E=E^o−(RT/nF) (lnQ)

n is the number of electrons transferred which will be 2 since both cathode and anode have a transfer of 2.

R is constant 8.3145 J/K x Mol and T is assumed to be 25 degreed Celsius or 298 degrees Kelvin.

F is Faraday's constant of 96485 C/Mol or J/V x Mol

Q is reaction quotient which means we put concentration of the reducing agent (oxidized element) over the oxidizing agent (reduced element). We assume from the cell notation that Cd is oxidized because it loses electrons and Ni is reduced because it gains electrons. Therefor, Q = [Cd²⁺]/[Ni²⁺], then Q = 0.1/0.5.

Q is equal to 0.2 and lnQ is equal to -1.61 rounded.

E^o is the standard potential which can be found by looking at the table of reduction potentials. That table can be found here.

The anode in this cell is Cd--> Cd²⁺ which has a reduction potential of -0.403V

The cathode is Ni²⁺ --> Ni which has a reduction potential of -0.26V

E^ocell=E^ocathode−E^oanode

So E^o will be -0.26V-(-0.403V)

E^o = 0.143V

From our equation, we can conclude that (RT/nF) = [(8.3145J/K x Mol)(295 K)]/[(2)(96485 J/V x Mol)] = (2477.72 J/Mol)/(197970 J/V x Mol) = 0.0128 V rounded.

E = 0.143V - (0.0128V)(-1.61)

E = 0.143V + 0.021V

E = 0.164V (Answer Key says answer should be 0.167V so, due to rounding, the results are slightly different).

PHASE II: Tepanka- I edited this one a bit so that we could see more of the step-by-step calculations and so the numerical answers were more accurate.

Q19.3.5

Is it possible for a complex of a metal in the transition series to have six unpaired electrons? Explain.

Q19.3.5

Transition metals are located in the d orbitals which have 5 orbitals each. The d orbitals can hold 5 maximum unpaired electrons and 10 maximum paired electrons. However, when the metal wants to accept ligands, it can create hybrid orbitals which allows it to accept up to 6 ligands. According to the ligand fields, the geometry of the complex, and whether or not the transition metal is low spin or high spin, the transition metal is still limited in 5 maximum unpaired electrons because of its orbitals: d_z², d_x^2-y², d_xy, d_xz, d_yz. However, a transition metal itself can have 6 unpaired electrons when it borrows an electron from the s orbital to fill all orbitals in the d orbital (making it more stable). For example, Chromium's electron configuration is [Ar] 4s²3d⁴, where there are 4 unpaired electrons and 1 electron pair. To be more stable, it can take on the electron configuration [Ar] 4s¹3d⁵, which gives us 6 unpaired electrons.

PHASE II: Tepanka- I added the last four sentences.

Q12.4.5

From the given data, use a graphical method to determine the order and rate constant of the following reaction:

2X⟶Y+Z

Q12.4.5

The rate law for this reaction would be rate = k[A]² since the equation has 2X as the reactant. Using the graphing approach, we find that the graph of 1/[X] vs. Time (s) gives us a linear graph. This means that the reaction is a second order reaction which is what we predicted.

Now that we know that rate = k[X]² is the correct rate law, we can find the rate constant. The rate can be calculated by looking at the slope of our plot. The slope will be change in concentration/change in time. Using the two points to find the slope, m = (Y₂-Y₁)/(X₂-X₁), we have (80-10.1)/(40-5) which gives us the slope and rate constant of 1.99.

PHASE II: Tepanka- I agree with this answer.

Q12.7.8

1. Based on the diagrams in Question Q12.7.6, which of the reactions has the fastest rate? Which has the slowest rate?
2. Based on the diagrams in Question Q12.7.7, which of the reactions has the fastest rate? Which has the slowest rate?

Q12.7.8

In the diagrams for question 12.7.6 we can see the activation energy decreases from (a) 25kJ to just (b) 10kJ. This means that the reaction rate will be faster for the reaction with the 10kj activation energy, thus graph (b) will have the slowest rate.

In the diagrams for question 12.7.7 we can see the activation energy for both graphs is the same at 10kJ with an additional 5kJ each after the first transition state. This means that the reaction rate will be the same for these reactions.

PHASE II: Tepanka- I agree with these solutions.

Q21.5.3

Both fusion and fission are nuclear reactions. Why is a very high temperature required for fusion, but not for fission?

Q21.5.3

Two nuclei must collide for fusion to occur. High temperatures are required to give the nuclei enough kinetic energy to overcome the very strong repulsion resulting from their positive charges.

For fission a nucleus splits into two nuclei. It is almost the exact opposite of fusion and that means that it creates heat.

PHASE II: Tepanka- Agreed.

Q20.4.1

Is a hydrogen electrode chemically inert? What is the major disadvantage to using a hydrogen electrode?

Q20.4.1

A hydrogen electrode is chemically inert because the hydrogen gas is contained in a glass tube and the hydrogen gas will not change phases during the cell reaction. The electrode potential is zero, so scientists can use it as a reference for other redox half-reactions. The major disadvantage of using a hydrogen electrode is the fact that hydrogen gas is very difficult to get a hold of and it is very difficult to keep the gas under 1 bar of pressure.

PHASE II: Tepanka- I edited this because the previous answer was copied and pasted from another chem.libretexts.org webpage...

Q20.5.30

Hydrogen gas reduces Ni²⁺ according to the following reaction: Ni²⁺(aq) + H₂(g) → Ni(s) + 2H⁺(aq); E°_cell = −0.25 V; ΔH = 54 kJ/mol.

1. What is K for this redox reaction?
2. Is this reaction likely to occur?
3. What conditions can be changed to increase the likelihood that the reaction will occur as written?
4. Is the reaction more likely to occur at higher or lower pH?

Q20.5.30

1. For this question we need to use the equation that relates E°_cell to K the equilibrium constant. That equation is E°_cell=(RT/nF) lnK.

n is the number of electrons transferred which will be 2 since both cathode and anode have a transfer of 2.

R is constant 8.3145 J/K x Mol and T is assumed to be 25 degreed Celsius or 298 degrees Kelvin.

F is Faraday's constant of 96485 C/Mol or J/ V x Mol.

So plugging in the numbers we are given and constants we have -0.25 V = [(8.3145 J/Mol x K)(298 K)/((96485 J/ V x Mol) x 2))] x lnK

-0.25 V = (2477.721 J/Mol)/(192970 J/V x Mol) x lnK

-0.25 V = 0.0128 V x lnK

-19.53 = lnK ... to get rid of the natural logarithm, we set e^x to each of the sides, x being whatever values are on each side, and e^ln cancels out and become 1.

e^-19.53 = K

K = 3.29 x 10^-9

2. This reaction was not likely to occur because the equilibrium constant is very small. Or we can use ΔG^o= -nE°_cell to determine the Gibbs Free Energy standard of the reaction and thus determine whether the reaction is likely to occur. Using our values, we see that

ΔG^o= -(2)(−0.25) = + 0.5, because ΔG^o is positive, we can assume that the reaction is non-spontaneous and thus less likely to occur naturally.

3. Looking at the values given to us, we can see the ΔH is a positive value. The equation of ΔG is ΔG = ΔH -TΔS.

So if we want a negative ΔG to occur, we must increase the temperature. That way we will have a higher (-TΔS) value that will be subtracted from ΔH to make ΔG negative.

4. Looking at the reaction we see that one of the products is H⁺ . This means that the reaction would be more likely to occur if there were less H⁺ atoms present according to Le Chatelier's principle. So, if the pH is higher meaning that less H⁺ molecules are present the reaction would be more likely to occur.

PHASE II: Tepanka- I agree with these solutions, but added the part about ΔG^o in the answer for question 2 and I added units where applicable.