Extra Credit 35

Q17.5.3

Consider a battery made from one half-cell that consists of a copper electrode in 1 M CuSO₄ solution and another half-cell that consists of a lead electrode in 1 M Pb(NO₃)₂ solution.

1. What are the reactions at the anode, cathode, and the overall reaction?

2. What is the standard cell potential for the battery?

3. Most devices designed to use dry-cell batteries can operate between 1.0 and 1.5 V. Could this cell be used to make a battery that could replace a dry-cell battery? Why or why not.

4. Suppose sulfuric acid is added to the half-cell with the lead electrode and some PbSO₄(s) forms. Would the cell potential increase, decrease, or remain the same?

Q17.5.3

1. The anode is the site where oxidation occurs while the cathode is where reduction occurs. The first step to solving this problem is to decide which of the two elements (Pb vs Cu) is the one being reduced/oxidized. By referring to the Standard Reduction Potential (SRP) table (https://sites.google.com/site/chempendix/potentials) the element with the higher reduction potential will be higher on the chart, with a more positive Ecell value. The elements closer to the bottom with more negative Ecell values have a low reduction potential, more likely to be oxidized. Looking at the chart, the reaction \(Cu^{2+}+2e^{-}\rightarrow\) \(Cu\) has an Ecell value of 0.34 V while the reaction \(Pb^{2+}+2e^{-}\rightarrow\) \(Pb\) has an Ecell value of -0.126. Since the Cu²⁺ reaction has a more positive Ecell value and is found higher on the table, it is the cathode being reduced. The Pb²⁺ reaction is lower on the table, therefore is the anode being oxidized. In the overall reaction, since the table only refers to the reactions in terms of reduction potential, the anode would need to be flipped so the electrons would be equal on both sides to cancel each other out.

cathode: \(Cu^{2+}+2e^{-}\rightarrow\) \(Cu\) E°=0.34V

anode: \(Pb^{2+}+2e^{-}\rightarrow\) \(Pb\) E°=-0.126

overall reaction: \(Cu^{2+}+2e^{-}\rightarrow\) \(Cu\)

+ \(Pb\rightarrow\) \(Pb^{2+}+2e^{-}\)

= \(Cu^{2+}+Pb\rightarrow\) \(Pb^{2+}+Cu\)

2. The standard potential for the battery is based on the equation \(E_{cell}=E_{cathode}-E_{anode}\)

Therefore, using the values determined in question 1, the E_cell would be

\(E_{cell}=0.34-(-0.126)\)

\(E_{cell}=0.466\)

3. This battery could not replace a dry-cell standing by itself. Since a dry cell operates between 1V and 1.5V, the reaction that was calculated above would need to be connected via a series of voltage to increase 0.466V. Batteries can be connected via a series of voltages to increase the voltage. When three cells are connected in a series, the potential becomes three times more than the 0.47 V, meaning that 0.466 x 3 = 1.41 V which is within the 1.0-1.5V standard.

4. If lead sulfate forms this would cause Q to become less than one and the log of any number less than one is negative so the Ecell will increase.

Q12.2.2

Go to the PhET Reactions & Rates interactive. Use the Single Collision tab to represent how the collision between monatomic oxygen (O) and carbon monoxide (CO) results in the breaking of one bond and the formation of another. Pull back on the red plunger to release the atom and observe the results. Then, click on “Reload Launcher” and change to “Angled shot” to see the difference.

1. What happens when the angle of the collision is changed?

2. Explain how this is relevant to rate of reaction.

Q12.2.2

1. The atom may take a long time to collide with the molecule, depending on the angle it was released. The atom may not have the necessary force to break the bond. Remember that a molecule must be hit at the correct angle and force to break the bond.

2. Particles must come in contact with one another before they can react, however, the must connect in the right way with a certain amount of force to break apart and recombine.

Q12.5.7

The rate of a certain reaction doubles for every 10 °C rise in temperature.

1. How much faster does the reaction proceed at 45 °C than at 25 °C?

2. How much faster does the reaction proceed at 95 °C than at 25 °C?

Q12.5.7

1. Since the reaction doubles every 10°C rise in temperature, the difference of temperature is

45 °C - 25 °C = 20°C

If the rate doubles every 10°C increase and the temperature rose by 20°C

22 = 4

The rate increases by a factor of 4.

2. Difference of temperature

95°C - 25°C = 70°C

If the rate doubles every 10°C increase and temperature rose by 70°C

27 = 128

The rate increases by a factor of 128.

Q21.4.2

What changes occur to the atomic number and mass of a nucleus during each of the following decay scenarios?

1. an α particle is emitted

2. a β particle is emitted

3. γ radiation is emitted

4. a positron is emitted

5. an electron is captured

Q21.4.2

Nuclear nomenclature is denoted in terms of A and Z, where A is the mass number and Z is atomic number, which is also the number of protons.

The total charge of the reactants must equal the total charge of the products. Z1 = Z2

The total mass number of reactants and the total mass number of the products must be equal. A_¹ = A_²

1. An α particle is a helium nucleus 4He, the mass number of the daughter nucleus is reduced by 4 and the atomic number is reduced by 2

\(^{A}_{Z}X\rightarrow^{A-4}_{Z-2}Y\)+\(^{4}_{2}\alpha\)

example:

\(^{238}_{92}U\rightarrow^{234}_{90}Th+^{4}_{2}He\)

2. A β particle is the emission of an electron. There is no change in atomic number because for a β particle, A=0. The mass would increase by a factor of 1 because Z = -1. To have equal total masses for both sides of the reaction, the mass would need to increase to compensate for the -1.

\(^{14}_{6}C\rightarrow^{14}_{7}N+^{0}_{-1}\beta\)

3. γ radiation has an atomic number of 0 and a mass number of 0. A γ particle has no affect on either the mass or atomic number.

\(^{60}_{27}Co\rightarrow^{60}_{28}Ni+^{0}_{0}\gamma\)

4. A positron (also known as β+) has a mass number of +1 and an atomic number of 0. The positron would have no affect on the atomic number, the mass would decrease by 1.

\(^{11}_{6}C\rightarrow^{11}_{5}B+^{0}_{1}\beta\)

5. An electron captured is the same as beta decay ( β-). The atomic number would have no change, the mass would increase by 1.

\(^{14}_{6}C\rightarrow^{14}_{7}N+^{0}_{-1}\beta\)

Q20.2.6

Of these elements, which would you expect to be easiest to reduce: Se, Sr, or Ni? Explain your reasoning.

Q20.2.6

Remember the pneumonic, OIL RIG. "Oxidation is loss, Reduction is gain." When an element is reduced, it gains electrons.

Elements want to have a full valence shell (outermost shell that accounts for the reactivity of the element). Since Se is a Group 6 element, it has 6 valence electrons. The most stable state has 8 valance electrons (a full shell). Se would be the easiest to reduce because it readily accepts electrons to complete its valence shell.

Sr is in Group 2, meaning it has 2 valence electrons. Elements in Group 1 or Group 2 are more likely to give up electrons to empty their outermost shell. In this case, those elements would be oxidized because they are giving up electrons.

Ni is part of the transition metals. Most transition metals have a lower number of valence electrons. In this case, Nickel has 2 valence electrons, putting it more likely to be oxidized like Sr.

Q20.5.1

State whether you agree or disagree with this reasoning and explain your answer: Standard electrode potentials arise from the number of electrons transferred. The greater the number of electrons transferred, the greater the measured potential difference. If 1 mol of a substance produces 0.76 V when 2 mol of electrons are transferred—as in Zn(s) → Zn²⁺(aq) + 2e, then 0.5 mol of the substance will produce 0.76/2 V because only 1 mol of electrons is transferred.

Q20.5.1

Standard electrode potential is the measure of energy/unit charge available from oxidation/reductions reactions to drive the reaction. Therefore, the standard electron potential is the change of electrons in a REDOX reaction. It depends on concentration of substance (1 M), temperature (298°C), and pressure (1 atm).

This statement is false because at standard conditions, the temperature, concentration, and pressure remains the same so there is no change in the value of the system. If you were use 2 mol of substance in this reaction, it would have the same value as 1 mol of substance. Remember, in standard potentials, we ignore the coefficients because they don't have correlation to the value of the Ecell.

Q20.9.10

Electrolysis of Cr³⁺(aq) produces Cr²⁺(aq). If you had 500 mL of a 0.15 M solution of Cr³⁺(aq), how long would it take to reduce the Cr³⁺ to Cr²⁺ using a 0.158 A current?

Q20.9.10

\(Cr^{3+}+e^{-}\rightarrow\) \(Cr^{2+}\)

\(500mL\) \(\times\) \((\frac{1L}{10^{3}mL})\) \(\times\) \((\frac{0.15molCr^{3+}}{1L})\) \(\times\) \((\frac{1mole^{-}}{1molCr^{2+}})\) \(\times\) \((\frac{1F}{1mol(e^{-})})\) \(\times\) \((\frac{96485^{\circ}C}{1F})\) \(=0.075molCr^{3+}\)

The formula for electrolysis is:

\(Q=It=nF\) where:

I is current (in amperage)

t is time (seconds)

F is Faraday's constant \((\frac{96485^{\circ}C}{mol})\)

N is number of moles of electrons

Q14.6.8

Nitramide (O₂NNH₂) decomposes in aqueous solution to N₂O and H₂O. What is the experimental rate law (Δ[N₂O]/Δt) for the decomposition of nitramide if the mechanism for the decomposition is as follows?

Assume that the rates of the forward and reverse reactions in the first equation are equal.

Q14.6.8

The first step in determining the rate law using a steady-state approximation is identifying the intermediate(s). The species O₂NNH^- is an intermediate because it is produced in step one of the mechanism and consumed in step two, while not appearing in the overall reaction.

Constructing a steady-state expression for the intermediate O₂NNH^-

\((\frac{dO_2NNH^{-}}{dt})\) \(=0\)

First we need to derive the rate of formation and consumption.

From step (1) of the mechanism

rate of formation of O₂NNH^- =K₁[O₂NNH^-]

rate of consumption of O₂NNH^- = k_-1[O₂NNH^-][H⁺]

From step (2) of the mechanism

rate of consumption of O₂NNH^- = k₂[O₂NNH^-]

For the O₂NNH^- to remain constant, the rate of consumption must equal the rate of formation

\((\frac{k_1O_2NNH_2}{k_-1O_2NNH^{-}})\)

Solving for the concentration of the intermediate O2NNH-

\(O_2NNH^{-}\) \(=\) \((\frac{k_1O_2NNH_2}{k_-1H^{+}})\)

You now need to substitute the stead-state expression of [O₂NNH^-] for the above rate equation

rate = k₂\((\frac{k_1O_2NNH_2}{k_-1H^{+}}\))

simplify

rate = k\((\frac{O_2NNH_2}{H^{+}}\))