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%.

1. CaCl₂
2. LiH
3. AlCl₃
4. CrBr₃

Q17.7.1

1. CaCl2

The first step is to determine the balanced equation for the electrolysis (decomposition) of the given molten salt. This is considered the overall reaction.

CaCl2(aq) -> Ca(s)+Cl₂ (g)

Then, determine the balanced half reactions so the anode and cathode can be determined

\[ \text{ Ca}^\text{2+}(\mathit{ aq})+\text{ 2 e}^- \rightarrow \text{ Ca}(\mathit{ s}) \]

\[ \text{ Cl}_2^\text{-}(\mathit{ aq}) \rightarrow \text{ Cl}_2(\mathit{ g})+\text{ 2 e}^- \]

The second step is to determine what element is being oxidized and which is being reduced. In the compound, CaCl2, Ca has a charge of +2 and each Cl has a charge of -1, making an overall neutral compound. Since the charge of Ca changes from +2 to 0, it gets reduced, which takes place at the cathode. Since the charge of Cl changes from -1 to 0, it gets oxidized, which takes place at the anode.

The third step is to determine the standard reduction potential (SRP) of each half-reaction (These values can be found in table 1P). You will use these values to calculate the approximate Eocell for the reaction using the equation

\[ \text{E}^°_\text{cell} = \text{E}^°_\text{cathode} - \text{E}^°_\text{anode} \]

Using Table P1 we can get the SRPs.

For calcium: -2.84 V

For chlorine: 1.396 V

NOTE: Keep in mind that although chlorine gets oxidized, the equation you will use to calculate Eocell uses the SRP values, so you will want to keep the value from the table that describes the reduction of chlorine.

\[ \text{E}^°_\text{cell} = -2.84 \text{ V} - 1.396\text{ V}=-4.236\text{ V} \]

FOLLOW THE SAME STEPS FOR THE OTHER MOLTEN SALTS.

2. LiH

First step:

2LiH(aq) -> 2Li(s)+H2(g)

\[ \text{ Li}^\text{1+}(\mathit{ aq})+\text{ 1e}^- \rightarrow \text{ Li}(\mathit{ s}) \]

\[ \text{2H}^\text{-}(\mathit{ aq}) \rightarrow \text{ H}_2(\mathit{ g})+\text{ 2 e}^- \]

Second step:

lithium: +1 to 0

reduction (cathode)

hydrogen: -1 to 0

oxidation (anode)

Third step:

lithium: -3.04 V

hydrogen: -2.25 V

\[ \text{E}^°_\text{cell} = \text{E}^°_\text{cathode} - \text{E}^°_\text{anode} \]

\[ \text{E}^°_\text{cell} = -3.04 \text{ V} - -2.25\text{ V}=-0.79\text{ V} \]

3. AlCl3

First step:

2AlCl3(aq) -> 2Al(s)+3Cl2(g)

\[ \text{ Al}^\text{3+}(\mathit{ aq})+\text{ 3 e}^- \rightarrow \text{ Al}(\mathit{ s}) \]

\[ \text{ 2Cl}_3^\text{1-}(\mathit{ aq})\rightarrow \text{ 3Cl}_2(\mathit{ g})+\text{ 6 e}^- \]

Second step:

aluminum: +3 to 0

reduction (cathode)

chlorine: -1 to 0

oxidation (anode)

Third step:

aluminum: -1.676 V

chlorine: 1.396 V

\[ \text{E}^°_\text{cell} = \text{E}^°_\text{cathode} - \text{E}^°_\text{anode} \]

\[ \text{E}^°_\text{cell} = -1.676 \text{ V} - 1.396\text{ V}=-3.072\text{ V} \]

4. CrBr3

First step:

2CrBr3(aq) -> 2Cr (aq)+3Br2(g)

\[ \text{ Cr}^\text{3+}(\mathit{ aq})+\text{ 3 e}^- \rightarrow \text{ Cr}(\mathit{ aq}) \]

\[ \text{ 2Br}_3^\text{1-}(\mathit{ aq}) \rightarrow \text{ 3Br}_2(\mathit{ g})+\text{ 6 e}^- \]

Second step:

chromium: +3 to 0

reduction (cathode)

bromine: -1 to 0

oxidation (anode)

Third step:

chromium: -0.424 V

bromine: 1.087 V

\[ \text{E}^°_\text{cell} = \text{E}^°_\text{cathode} - \text{E}^°_\text{anode} \]

\[ \text{E}^°_\text{cell} = -0.424 \text{ V} - 1.087\text{ V}=-0.79\text{ V} \]

Eocell = Eocathode - Eoanode

Eocell = (-0.424 V) - (1.087 V) = -1.511 V

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?

Q12.3.10

The rate law for the decomposition is

rate = k[C2H4O]^2

because this describes the rate at which the concentration (M) of acetaldehyde is changing.

We are told that the reaction is second order and are told the needed information to set up the rate law

We know that the rate constant, k, is equal to 4.71 × 10−8 L/mol/s. We also know that the concentration of acetaldehyde is 5.55 × 10−4 M. Using this information, we can plug these two values directly into our equation to get the instantaneous rate at which acetaldehyde changes (decomposes) at this particular concentration.

rate = (4.71 × 10−8 L/mol/s)(5.55 × 10−4 M)^2 = 1.45 × 10−14 M/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?

Q12.6.2

No, for the overall reaction you would need the rate law equation specific to the reaction in order to know how altering the concentrations of the reaction affects the reaction rate. This can only be determined experimentally in the lab.

Yes, for an elementary reaction, although the rate constant would still be unknown, we can write the rate law equation of the given reaction (assuming it is balanced) based on its stoichiometry coefficients. In an elementary reaction, the coefficients correspond to the exponents for the given reactant. Since both A and B have a coefficient of 1, the exponent used in the rate law for A and B will be 1. This is enough information to formulate the rate law equation.

A+B⟶C

\[ \text{ A}+\text{B} \rightarrow \text{ C}\]

rate = k[A][B]

We can now conclude that doubling the concentration of A would also double the rate of the reaction.

Q21.4.14

A 1.00 × 10–6-g sample of nobelium, No102;254, has a half-life of 55 seconds after it is formed. What is the percentage of No102;254 remaining at the following times?

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

Q21.4.14

To start with, we need to form a basic equation to relate the initial amount of sample, the sample's half-life, time, and the new amount of sample. Speaking about half-life refers to a process of decay, so we know that the original amount of sample will be decaying as time passes. If we denote initial amount of the sample as "C", time that passes as "t", the half-life as "h", and "X" as the new sample amount, we can formulate the appropriate relationship as

X = C*(0.5)^(t/h)

\[ \text{X}= \text{C}* \text{ 0.5}^{t/h}\]

Solving for X will yield the new amount of sample.

In this case,

C = 1.00 × 10–6 g

t = 55 seconds

We can now plug into our general formula to give

X = (1.00 × 10–6)*(0.5)^(t/55)

\[ \text{X}= \text{1E-6}* \text{ 0.5}^{t/55}\]

which we will use to solve for the amount of sample left after decay.

1. 5.0 min after it forms

First convert minutes to seconds to keep the units consistent.

(5 mins)*(60 sec/min) = 300 seconds

X = (1.00 × 10–6)*(0.5)^(300/55) = 2.28 x 10–8 g

\[ \text{X}= \text{1E-6}* \text{ 0.5}^{300/55}=\text{2.28E-8 g}\]

2. 1.0 h after it forms

(1 hr)*(60 min/hr)*(60 sec/min) = 3600 seconds

X = (1.00 × 10–6)*(0.5)^(3600/55) = 1.98 x 10–26 g

\[ \text{X}= \text{1E-6}* \text{ 0.5}^{3600/55}=\text{1.98E-26 g}\]

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?

Q20.3.2

If the two half-reactions are separated, the redox reaction can still occur if electrons have a way to move from the anode to the cathode (from oxidation to reduction). This requires a complete circuit, or a means to connect the two together to allow electron flow. An external connection of wire can be made to connect the anode to the cathode, allowing the flow of electrons from one to the other (electricity) and enable the redox reaction to occur. The name of this apparatus is an electrochemical cell. In addition to this electrical component, a salt bridge is needed in between the anode and the cathode to replace the charges being lost by the flow of electrons so the electrolysis can continue.

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°?

Q20.5.13

We must first balance the two half-reactions separately in order to see how many electrons are transferred. Since no hydrogen or oxygen is present, only the charge needs to be balanced with electrons.

anode (oxidation): Al(s) -> Al3+(aq) + 3e-

cathode (reduction): Sn4+(aq) + 2e- -> Sn2+(aq)

\[ \text{oxidation:}\text{ Al} \rightarrow \text{ Al}^\text{ 3+}+\text{ e}^- \]

\[ \text{reduction:}\text{ Sn}^\text{ 4+}+\text{2e}^- \rightarrow \text{ Sn}^\text{ 2+}\]

To be added together, the number of electrons on both sides of the equation must be the same, so we need to find the least common multiple of 3 and 2. The least common multiple of 3 and 2 is 6, so that means 6 electrons will be transferred in this redox reaction.

We can find standard cell potential by using the SRP values from Table P1 and plugging into the equation

E˚cell = E˚cathode - E˚anode

\[ \text{E}^°_\text{cell} = \text{E}^°_\text{cathode} - \text{E}^°_\text{anode} \]

aluminum: -1.676 V

tin: 0.154 V

E˚cell = (0.154 V) - (-1.676 V) = 1.83 V

\[ \text{E}^°_\text{cell} = 0.154\text{ V} - -1.676\text{ V}=1.83\text{ V} \]

If the standard cell potential is positive, it is spontaneous, and a negative cell potential indicates a non-spontaneous reaction. Since the result was positive, it is spontaneous.

To find delta(G), we will use the equation

delta(G) = -n*F*E˚cell

\[ \text{ΔG} = -\text{nF}\text{E}^°_\text{cell} \]

where n=number of moles transferred and F=9.65x10^4 C/mol (Faraday's constant).

delta(G) = -(6)*(9.65x10^4)*(1.83) = -1.06x10^6 J

\[ \text{ΔG} = -(6\text{ mol}\text{ of}\text{ e}^-)\times\frac{96486\text{ J}}{\text{V}\times\text{mol}\text{ of}\text{ e}^-}\times1.83 \text{ V} \]

\[ \text{ΔG} = -1.06E6\text{ J} \]

Since delta(G) is negative, this also reveals that the reaction is spontaneous.

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+
2. [Ni(CN)₄]2−

Q24.6.8

1. [Cu(NH₃)₄]2+

The number of places that a ligand is bounded to copper (the transition metal) is 4. This is also known as the coordination number. We will assume that this results in tetrahedral structure (although square planar could be possible). That means there are 3 t_2g higher energy levels and 2 e_g lower energy levels for electrons to be placed.

NH₃ is a strong field ligand, meaning that the pairing energy of electrons will be less that placing lone electrons in the next energy level, resulting in low spin.

P < delta(t)

Copper has an oxidation state of +2, so it has 9 d-electrons. These will be paired first in the lower levels, then be placed in the t_2g level. This leaves one unpaired electron.

2. [Ni(CN)₄]2-

The coordination number is 4, resulting in tetrahedral. Since CN- is a strong field ligand, this means low spin.

P < delta(t)

Nickel has an oxidation state of +2, so it has 8 d-electrons. They will be paired in the first lower energy levels first, and leave two unpaired electrons.

It will look like this:

Q14.4.6

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

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

Q14.4.6

First we must determine the exponents in the rate law. Based on the reactants, all we know is

rate = k[Fe3+]^x[I-]^y

Since doubling Fe3+ doubled the reaction rate, we can solve [2]^x=2 for x and find that x=1. This means that the reaction order with respect to Fe3+ is first order.

Since doubling I- quadrupled the reaction rate, we can solve [2]^y=4 for y and find that y=2. This means that the reaction order with respect to I- is second order.

Using this information yields the overall rate law:

rate = k[Fe3+][I-]^2

The overall reaction order is the sum of the exponents. Since 1+2=3, the overall reaction is third order.