Extra Credit 47

Q17.7.1

Identify the reaction at the anode, reaction at the cathode, the overall reaction, and the approximate potential required for the electrolysis of the following molten salts. Assume standard states and that the standard reduction potentials in Table P1 are the same as those at each of the melting points. Assume the efficiency is 100%.

a. CaCl₂

Look at the SRP table and find the reactions related to Ca and Cl. You will find the following:

2e^-+Ca²⁺ ⇌ Ca -2.84 V

2e^-+Cl₂ ⇌ 2Cl^- +1.358 V

We know that this is an electrolysis reaction so CaCl₂ will become Ca²⁺ and 2Cl^-. Knowing this, we can break the reaction to:

Ca ⇌ Ca²⁺+2e^- -2.84 V

2e^-+Cl₂ ⇌ 2Cl^- +1.358 V

When you lose electrons, the reaction is an oxidation reaction. Gain is reduction. By this we know that Ca undergoes oxidation and Cl₂ is a reduction reaction. We also know that anodes are the elements that get oxidized. Cathode, vice versa so Ca is the anode and Cl₂ is the cathode.

The approximate potential required for the electrolysis is calculated by:

E°_cell=E°_cathode-E°_anode =1.358-(-2.84)=4.198 V

b. LiH

We follow the same steps as above and find:

2H⁺+2e^- ⇌ H₂ 0.0000V

Li⁺ + e^- ⇌ Li -3.040V

In electrolysis, LiH becomes Li⁺ and H⁺ so the reactions are:

H₂ ⇌ 2H⁺+2e^- 0.0000V (oxidation, anode)

2(Li⁺ + e^- ⇌ Li) -3.040V (reduction, cathode)

E°_cell=E°_cathode-E°_anode =-3.040V

c. AlCl₃

We find:

2e^-+Cl₂ ⇌ 2Cl^- +1.358 V

Al³⁺ + 3e^- ⇌ Al -1.676V

Electrolysis breaks AlCl₃ to Al³⁺ and Cl^- so we have:

3(2e^-+Cl₂-->2Cl- ) +1.358 V (reduction, cathode)

2(Al --> Al³⁺ + 3e^-) -1.676V (oxidation, anode)

E°_cell=E°_cathode-E°_anode = 3.034V

d. CrBr₃

Cr3⁺ + e⁻ ⇌ Cr²⁺ -0.424V

Cr2⁺ + 2e⁻ ⇌ Cr -0.90V

Br₂ + 2e⁻ ⇌ 2Br⁻ 1.087V

Electrolysis breaks CrBr₃ Cr2⁺ and Br^-

2(Cr3⁺ + e⁻ ⇌ Cr²⁺ ) -0.424V (reduction, cathode)

Cr2⁺ + 2e⁻ ⇌ Cr -0.90V(reduction, cathode)

2(2Br− ⇌ Br₂ + 2e⁻) 1.087V (oxidation, anode)

E°_cell=E°_cathode-E°_anode =-2.411V

Q12.3.10

The decomposition of acetaldehyde is a second order reaction with a rate constant of 4.71 × 10⁻⁸ L/mol/s. What is the instantaneous rate of decomposition of acetaldehyde in a solution with a concentration of 5.55 × 10⁻⁴ M?

The instantaneous rate of the concentration is the rate where that concentration is changing at a specific point. Because this is a second order reaction, we can find concentrations using:

1/[A]_t=kt+1/[A₀]

the instantaneous rate can be found by taking the derivative of the equation which would give us k, 4.71 × 10⁻⁸ L/mol/s.

Q12.6.2

In general, can we predict the effect of doubling the concentration of A on the rate of the overall reaction A+B⟶C? Can we predict the effect if the reaction is known to be an elementary reaction?

No, we do not know what order the reaction is, which tells us how concentrations affect the reaction rates. Changing the concentration of A would lead to an unknown change towards the reaction.

Even if the reaction is known to be an elementary reaction, we do not know the order of the reaction, which is needed to determine how concentrations change the reaction rates

Q21.4.14

A 1.00 × 10^–6-g sample of nobelium, No102254, has a half-life of 55 seconds after it is formed. What is the percentage of No102254 remaining at the following times?

1. 5.0 min after it forms
2. 1.0 h after it forms

Since the half life is 55 seconds, we know that t= 55 where half of the original sample is remaining so, 0.5(1.00 x 10^-6)=1.00 x 10^-6e^k55.

We solve for k using algebra and we get k=-0.0126

1. 5 min = 5(60) sec = 300 sec

N = 1.00 x 10^-6e^k300= 2.28 x 10^-8 g sample of nobelium remaining

2. 1 hr = 1(60) min = 60(60) sec = 3600 sec

N = 1.00 x 10^-6e^k3600= 1.97996 x 10^-26 g sample of nobelium remaining

Q20.3.2

If two half-reactions are physically separated, how is it possible for a redox reaction to occur? What is the name of the apparatus in which two half-reactions are carried out simultaneously?

We connect both reactions together using a conductive wire (this allows the transfer of electrons), making a redox reaction. This makes a galvanic cell.

Q20.5.13

For the cell represented as Al(s)∣Al³⁺(aq)∥Sn²⁺(aq), Sn⁴⁺(aq)∣Pt(s), how many electrons are transferred in the redox reaction? What is the standard cell potential? Is this a spontaneous process? What is ΔG°?

Anode is on the left and cathode is on the right so we know that our half reactions are:

2(Al(s) ⇌ Al³⁺(aq) + 3e^-) -1.676V (oxidation)

3(Sn⁴⁺(aq) + 2e ^- ⇌ Sn²⁺(aq) ) 0.154V (reduction)

We balance the electrons and find that a total of 6 electrons are transferred in the redox reaction. The standard cell potential is:

E°_cell=E°_cathode-E°_anode = 1.83V > 0 spontaneous

ΔG°=-nFE°_cell =-(6)( 96485JV/mol)(1.83V)=-10594.053kJ

Q24.6.8

For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present.

1. [Cu(NH₃)₄]²⁺ When we look at the spectrochemical series, we find that NH₃ is a strong field ligand, meaning it is low spin and has a high splitting energy. Next, we find the charge of Cu, which is 2+ since NH₃ is neutral. The electron configuration for Cu²⁺ is 3d⁹ so when we fill the crystal field diagram, we find that there is 1 unpaired electron.
2. [Ni(CN)₄]²⁻ CN^- is a strong field ligand, also meaning low spin and has a high splitting energy. Since CN^- has a -1 charge, we find that Ni has a +2 charge. The electron configuration for Ni²⁺ is 3d⁷ so when we fill the crystal field diagram, there is 1 unpaired electron.

Q14.4.6

Iodide reduces Fe(III) according to the following reaction:

2Fe3+(soln)+2I−(soln)→2Fe2+(soln)+I2(soln)

Experimentally, it was found that doubling the concentration of Fe(III) doubled the reaction rate, and doubling the iodide concentration increased the reaction rate by a factor of 4. What is the reaction order with respect to each species? What is the overall rate law? What is the overall reaction order?

Since doubling the concentration of Fe(III) doubled the reaction rate, this tells us that the order of the reaction with respect to Fe(III) is 1. Doubling iodide concentration increased the reaction rate by a factor of 4 which tells us that the order of the reaction with respect to iodide is 2. The overall rate law is given by:

rate=k[A]^a[B]^b In this case, rate = k[Fe³⁺][I^-]²