Extra Credit 48

Q17.7.2

What mass of each product is produced in each of the electrolytic cells of Q17.7.1 if a total charge of 3.33 × 10⁵ C passes through each cell? Assume the voltage is sufficient to perform the reduction.

Products from Q17.7.1

1. CaCl₂
2. LiH
3. AlCl₃
4. CrBr₃

S17.7.2

Product 1. \(CaCl_2\)

Step 1. First determine the decompositions reactions of \(CaCl_2\)

\(CaCl_2\rightarrow\ Cl_2\)

\(CaCl_2\rightarrow\ Ca\)

Step 2. Then determine the redox reactions

\(2e^-+Ca^{2+}Cl_2\rightarrow Ca^0\) \(\text{on the right side of the equation, Ca has an oxidation state of 2+ while on the right side it has a oxidation state of 0}\)

\(CaCl_2^{2(-)}\rightarrow\ Cl_2^0+2e^-\) \(\text{on the right side of the equation, } Cl_2 \text{ has an oxidation state of 2- while on the right side it has a oxidation state of 0}\)

So we have found the number of electrons used to convert from moles of electron to moles of a element. (we will use this later)

Step 3. Now use \(n_e=\dfrac{It}{F}\ \) to find mol of electrons

\(n_e=\) mol of electrons

\(I=ampere\) (current)

\(t=time\)

\(F=96485 C mol^{-1}\) (Faraday's constant)

\(I\times t\) is given as \(3.33\times10^5\)C

\(n_e=\dfrac{3.33\times10^5}{96485}\)

\(n_e=3.451\) mol \(e^-\)

Step 4. Now do basic stoichiometry to determine mass of each product

\(n_e\times \dfrac{\text{mol of product}}{\text{mol of e^- transferred}}\times \dfrac{\text{molar mass of product}}{\text{1 mol of product}}=grams\)

\(3.451 mol e^-1\times \dfrac{1molCa}{2mole^-}\times \dfrac{40.08g}{1molCa}=69.16g \text{ of } Ca\)

\(3.451 mol e^-1\times \dfrac{1molCl_2}{2mole^-}\times \dfrac{70.9g}{1molCl_2}=123.35g \text{ of } Cl_2\)

\(\text{mass of } Ca=69.16g\text{ mass of } Cl_2=122g\)

Product 2. \(LiH\)

Step 1. First determine the decompositions reactions of \(LiH\)

\(2LiH\rightarrow\ 2Li\)

\(2LiH\rightarrow\ H_2\)

Step 2. Then determine the redox reactions

\(2e^-+2Li^{1+}H\rightarrow Li^0\) \(\text{on the right side of the equation, Li has an oxidation state of 1+ while on the right side it has a oxidation state of 0}\)

\(2LiH^-\rightarrow\ H_2^0+2e^-\) \(\text{on the right side of the equation, } H_2 \text{ has an oxidation state of - while on the right side it has a oxidation state of 0}\)

So we have found the number of electrons used to convert from moles of electron to moles of a element. (we will use this later)

Step 3. Now use \(n_e=\dfrac{It}{F}\ \) to find mol of electrons

\(n_e=\) mol of electrons

\(I=ampere\) (current)

\(t=time\)

\(F=96485 C mol^{-1}\) (Faraday's constant)

\(I\times t\) is given as \(3.33\times10^5\)C

\(n_e=\dfrac{3.33\times10^5}{96485}\)

\(n_e=3.451\)mol \(e^-\)

Step 4. Now do basic stoichiometry to determine mass of each product

\(n_e\times \dfrac{\text{mol of product}}{\text{mol of e^- transferred}}\times \dfrac{\text{molar mass of product}}{\text{1 mol of product}}=grams\)

\(3.451 mol e^-1\times \dfrac{2molLi}{2mole^-}\times \dfrac{6.94g}{1molLi}=23.9g \text{ of } Li\)

\(3.451 mol e^-1\times \dfrac{1molH_2}{2mole^-}\times \dfrac{2.0158g}{1molH_2}=3.48g \text{ of } H_2\)

\(\text{mass of } Li=23.9g\text{ mass of } H_2=3.48g\)

Product 3. \(AlCl_3\)

Step 1. First determine the decompositions reactions of \(AlCl_3\)

\(AlCl_3\rightarrow\ Al\)

\(AlCl_3\rightarrow\ Cl_3\)

Step 2. Then determine the redox reactions

\(3e^-+Al^{3+}Cl_3\rightarrow Al^0\) \(\text{on the right side of the equation, Al has an oxidation state of 3+ while on the right side it has a oxidation state of 0}\)

\(AlCl_3^{3(-)}\rightarrow\ Cl_3^0+3e^-\) \(\text{on the right side of the equation, } Cl_3 \text{ has an oxidation state of 3- while on the right side it has a oxidation state of 0}\)

So we have found the number of electrons used to convert from moles of electron to moles of a element. (we will use this later)

Step 3. Now use \(n_e=\dfrac{It}{F}\ \) to find mol of electrons

\(n_e=\) mol of electrons

\(I=ampere (current)\)

\(t=time\)

\(F=96485 C mol^{-1}\) (Faraday's constant)

\(I\times t\) is given as \(3.33\times10^5\)C

\(n_e=\dfrac{3.33\times10^5}{96485}\)

\(n_e=3.451\)mol \(e^-\)

Step 4. Now do basic stoichiometry to determine mass of each product

\(n_e\times \dfrac{\text{mol of product}}{\text{mol of e^- transferred}}\times \dfrac{\text{molar mass of product}}{\text{1 mol of product}}=grams\)

\(3.451 mol e^-1\times \dfrac{1molAl}{3mole^-}\times \dfrac{26.982g}{1molAl}=31.04g \text{ of } Al\)

\(3.451 mol e^-1\times \dfrac{1molCl_3}{3mole^-}\times \dfrac{106.359g}{1molCl_3}=122.35g \text{ of } Cl_3\)

\(\text{mass of } Al=31.04g\text{ mass of } Cl_3=122.35g\)

Product 4. \(CrBr_3)\)

Step 1. First determine the decompositions reactions of \(CrBr_3\)

\(CrBr_3\rightarrow\ Cr\)

\(CrBr_3\rightarrow\ Br_3\)

Step 2. Then determine the redox reactions

\(3e^-+Cr^{3+}Br_3\rightarrow\ Crl^0\) \(\text{on the right side of the equation, Cr has an oxidation state of 3+ while on the right side it has a oxidation state of 0}\)

\(AlCl_3^{3(-)}\rightarrow\ Br_3^0+3e^-\) \(\text{on the right side of the equation, } Br_3 \text{ has an oxidation state of 3- while on the right side it has a oxidation state of 0}\)

So we have found the number of electrons used to convert from moles of electron to moles of a element. (we will use this later)

Step 3. Now use \(n_e=\dfrac{It}{F}\ \) to find mol of electrons

\(n_e=\) mol of electrons

\(I=ampere\) (current)

\(t=time\)

\(F=96485 C mol^{-1}\) (Faraday's constant)

\(I\times t\) is given as \(3.33\times10^5\)C

\(n_e=\dfrac{3.33\times10^5}{96485}\)

\(n_e=3.451\)mol \(e^-\)

Step 4. Now do basic stoichiometry to determine mass of each product

\(n_e\times \dfrac{\text{mol of product}}{\text{mol of e^- transferred}}\times \dfrac{\text{molar mass of product}}{\text{1 mol of product}}=grams\)

\(3.451 mol e^-1\times \dfrac{1molCr}{3mole^-}\times \dfrac{51.996g}{1molCr}=59.8g \text{ of } Cr\)

\(3.451 mol e^-1\times \dfrac{1molBr_3}{3mole^-}\times \dfrac{239.7g}{1molBr_3}=276g \text{ of } Br_3\)

\(\text{mass of } Cr=59.8g\text{ mass of } Br_3=276g\)

Q12.3.11

Alcohol is removed from the bloodstream by a series of metabolic reactions. The first reaction produces acetaldehyde; then other products are formed. The following data have been determined for the rate at which alcohol is removed from the blood of an average male, although individual rates can vary by 25–30%. Women metabolize alcohol a little more slowly than men:

Determine the rate equation, the rate constant, and the overall order for this reaction.

S12.3.11

This is a zero order reaction, which we can tell because the rate stays constant throughout the reaction regardless of the concentration.

Step 1. Determine the rate equation

The rate equation for zero order is \(rate=k\times[M]^0\).

So in our case our rate equation is \(rate=k\times[C_2H_5OH]^0\)

Step 2. Find rate constant

Plug in a concentration and its corresponding rate.

\(rate=2.0\times10^{-2}\) \(M=[C_2H_5OH] \)

\(2.0\times10^{-2}=k\times[C_2H_5OH]^0\)

\(2.0\times10^{-2}=k\times1\)

\(k=2.0\times10^{-2}M/h\)

Q12.6.3

Phosgene, COCl₂, one of the poison gases used during World War I, is formed from chlorine and carbon monoxide. The mechanism is thought to proceed by:

1. Write the overall reaction equation.
2. Identify any reaction intermediates.
3. Identify any catalyst. (Question asked for intermediates again)

S12.6.3

1. Overall reaction equation:

Step 1: write the steps involved

Step 2: cancel out anything that appears on both the reaction side of step 1 and product side of step 2 and anything that appears on both the product side of Step 1 and reactant side of step 2.

Step 3: Then rewrite the reaction

2. Identify any reaction intermediates.

An intermediate is anything that appears on the product side of step 1 and the reaction side of step 2.

\(CoCl\) is an intermediate

3. Identify any catalyst.

A catalyst is a substance that increases the rate of a chemical reaction without itself undergoing a permanent chemical change. In other words, something that appears on the reaction side of step 1 and product side of step 2.

As evinced our example, the catalyst is Cl.

Q21.4.15

²³⁹Pu is a nuclear waste byproduct with a half-life of 24,000 y. What fraction of the ²³⁹Pu present today will be present in 1000 y?

S21.4.15

In this problem you will need to use the first order decay equation \(ln(N/N_0)=-kt\) where

N= Percent left in 1000 years N=?

\(N_0\)= initial percent \(N_0\)=100

t=amount of time passed t=1000

\(t_{1/2}=24000\)

k= rate constant K can be found using \(t_{1/2}=ln(2)/k\) because all half-life reactions are first order

Step 1. Find k

\(t_{1/2}=ln(2)/k\)

\(24000=.693/k\)

\(24000\times k=.693\)

\(k=.693/24000\)

\(k=2.8875e^{-5}\)

Step 2. Insert k and find N

\(ln(N/100)=-(2.8875e^{-5})\times(1000)\)

\(ln(N/100)=-.028875\)

\(e^{ln(N/100)}=e^{-.028875}\)

N/100=.9715

N=.9715\times100

N=97.15

97.15% ²⁹³Pu will be present in 1000 years.

Q20.3.3

What is the difference between a galvanic cell and an electrolytic cell? Which would you use to generate electricity?

S20.3.3

• Galvanic cells, (Electrochemical cells) convert chemical energy into electrical energy. However, Electrolytic cells convert electrical energy into chemical energy.
• The redox reaction in Galvanic cells are spontaneous and are responsible for the production of electrical energy. However, the redox reactions in Electrolytic cells are not spontaneous and require electrical energy to be supplied in order to initiate the reaction.
• The two half-cells in a Galvanic cell are set up in different containers and are connected through the salt bridge or porous partition. However, in Electrolytic Cells, both electrodes are placed in the same container in the solution of molten electrolyte.
• In Galvanic cells, the anode is negative and the cathode is the positive electrode. The reaction at the anode is oxidation and the reaction in the cathode is reduction. However, in Electrolytic cells, the anode is positive, and the cathode is negative. The reaction in the anode is oxidation and the reaction in the cathode is reduction.
• In a Galvanic cell, the electrons are supplied by the species that is getting oxidized. They move from the anode to the cathode in the external circuit. However, in an Electrolytic cell, the external battery supplies the electrons. They enter through the cathode and come out through the anode.

You would use a Galvanic cell to generate electricity because galvanic cells convert chemical energy into electrical energy while Electrolytic cells convert electrical energy into chemical energy and they also require an external power source. (edited)

Q20.5.14

Calculate the pH of this cell constructed with the following half reactions when the potential is 0 at 25 °C

\[MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O\]

\[Au^{3+} + 3e^- \rightarrow Au_{(s)}\]

Under this condition, the concentrations of other species in the cell are:

• 0.36 M: \(MnO_4^-\)
• 0.004 M: \(Au^{3+}\)
• 0.001 M: \(Mn^{2+}\)

S20.5.14

Step 1. You first need to determine which redox reaction happens in the cathode, and which happens in the anode

You will do this by looking at which reaction has the highest reduction potential

\(MnO_4^- + 8H^++ 5e^- \rightarrow Mn^{2+} + 4H_2O\) reduction potential=+1.51

\(Au^{3+} + 3e^- \rightarrow Au_{(s)}\) reduction potential=+1.40

Since the \(MnO_4^-\) equation has the higher reduction potential, it happens in the cathode while the \(Au^{3+}\) happens in the anode

Step 2. You then need to balance the electrons in the two equations

\(3\times (MnO_4^-+8H^+5e^-\rightarrow\ Mn^{2+}+4H_2O)\)

\(5\times (Au_{(s)}\rightarrow\ Au^{3+}+3e^-)\)

Step 3. Which gets rid of the electrons (because there is \(15e^-\)on both sides

Cathode \(3MnO_4^-+24H^+\rightarrow\ 3Mn^{2+}+12H_2O\)

Anode \(5Au_{(s)}\rightarrow\ 5Au^{3+}\)

Step 4. We will now use the Nernst equation

\(E_{cell}=E_{cell}^o-\dfrac{RT}{nF}lnQ\)

we have:

\(T=25^oc\) or \(T=298 K\)

\(R=8.3145J\times mol^-\times k^-\)

\(n=\text{ number of electrons transferred}\) \(n=15\)

\(F=96485 C mol^-1\)

\(E_{cell}=0\)

\(E_{cell}^o=E_{cathode}-E_{anode}\) \(1.51V-1.40V=0.11V\)

\(Q=\dfrac{\text concentrations of products}{\text concentration of reactants}\)

Step 5. Now solve for concentration of\(H^+\)

\(0=0.11V-\dfrac{298\times 8.3145}{15\times 96485}ln(Q)\)

\(64.25=ln(Q)\)

\(e^{64.25}=e^{ln(Q)}\)

\(8.03\times 10^{27}=Q\)

\(8.03\times 10^{27}=\dfrac{[0.001]^3\times [0.004]^5}{[0.36]^3\times [H^+]^{24}}\)

\(3.75\times 10^{26}\times[H^+]^{24}=[0.001]^3\times [0.004]^5\)

\(3.63\times 10^{26}\times[H^+]^{24}=1.024\times 10^{-21}\)

\([H^+]^{24}=2.73\times 10^{-48}\)

\(\sqrt[24]{[H^+]^{24}}=\sqrt[24]{2.73\times 10^{-48}}\)

\(H^+=0.01043\)

Step 6. Now use the concentration of \(H^+\) to determine the pH level

\(-log(H^+)=pH\)

\(-log(0.01043)=1.98\)

Consequently, the pH of this cell is \(1.98\); it is very acidic.

Q24.6.9

The ionic radii of Mn²⁺, Fe²⁺, and Zn²⁺ are all roughly the same (approximately 76 pm). Given their positions in the periodic table, explain why their ionic radii are so similar.

S24.6.9