Extra Credit 35

Q 17.5.3

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?

Solution:

For 1.

Reactions at the anode: Oxidation of lead electrode: \[Pb\rightarrow Pb^{2+}+2e^{-}\]

Reactions at the cathode: Reduction of \(Cu^{2+}\) ions in the \(CuSO_{4}\) solution: \[Cu^{2+}+2e^{-}\rightarrow Cu\]

Overall reaction: \[Pb^{2+}+CuSO_{4}\rightarrow PbSO_{4}+Cu\]

For 2.

To find the standard cell potential, you would use the equation: E(cathode) - E(anode)= 0.34V - (-0.13V)= 0.47V. The cell potentials can be found on the table above.

For 3.

This cell can't be used to make a battery to replace a dry-cell battery because the cell potential of this battery is only 0.47V. The voltage is not in the range.

For 4.

The cell potential will increase because the sulfuric acid can act as the "salt bridge" to make sure the oxidation in the anode keeps going.

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.

Solution:

For 1.

There will be a decrease in collisions frequency or decrease the successful rate in collision.

For 2.

The angle of the collisions may affect the successful rate of forming bonding between the reactants and thus affect the rate of reaction.

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?

Solution:

For 1.

The reaction rate at 45 degrees Celsius will be 4 times of the reaction rate at 25 degrees Celsius.

For 2.

The reaction rate at 95 degrees Celsius will be 128 times of the reaction rate at 25 degrees Celsius.

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

Solution:

The atomic number and mass of nucleus have to be equal on both sides of the equation.

1. Atomic number: -2; Mass of Nucleus: -4

An α particle \(_{2}^{4}\textrm{He}\) is emitted so the nucleus loses two protons and loses a mass of 4 amu.

For example: \[_{5}^{10}\textrm{B}\rightarrow _{2}^{4}\textrm{He}+_{3}^{6}\textrm{Li}\]

2. Atomic number: +1; Mass of Nucleus: 0

A β particle \(_{-1}^{0}\textrm{e}\) is emitted so the mass doesn't change but the atomic number increases by one.

For example: \[_{5}^{10}\textrm{B}\rightarrow _{-1}^{0}\textrm{e}+_{6}^{10}\textrm{Li}\]

3. Atomic number: 0; Mass of Nucleus: 0

A γ particle is \_{0}^{0}\textrm{γ}\). Since the atomic number and mass number are both zero, there is no change in the nucleus.

4. Atomic number: -1; Mass of Nucleus: 0

A positron is \(_{1}^{0}\textrm{e}\) emitted so the atomic number decreases by one but the mass number doesn't change.

For example: \[_{5}^{10}\textrm{B}\rightarrow _{1}^{0}\textrm{e}+_{4}^{10}\textrm{Be}\]

5.Atomic number: -1; Mass of Nucleus: 0

An electron \(_{-1}^{0}\textrm{e}\) is captured so the atomic number is decreased by one but the mass number stays the same.

For example: \[_{5}^{10}\textrm{B}+_{-1}^{0}\textrm{e}\rightarrow _{4}^{10}\textrm{Be}\]

Q20.2.6

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

Solution:

Se will be the easiest one to reduce for two reasons. First of all, it is the most electronegative among the three elements and it will easy to attract positive ions. Secondly, it has the strongest ionization energy among these three. Ionization energy and electronegativity increases in strentgh from left to right on the periodic table and decreases in strength from top to bottom on the periodic table.

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.

Solution:

I disagree with this reasoning because the molarity of the solution should not effect the moles of ions that participated in the reaction. The coefficient of the chemical reaction is not the same as the actual moles of reactants in the reaction.

Q20.9.10

Electrolysis of \(Cr^{3+}(aq)\) produces \(Cr^{2+}(aq)\). If you had 500 mL of a 0.15 M solution of \(Cr^{3+}(aq)\),how long would it take to reduce the \(Cr^{3+}\) to \(Cr^{2+}\) using a 0.158 A current?

Solution:

\[It=nF\]

\[t=\frac{nF}{I}\]

\[t=\frac{1*96485}{0.158A}\]

\[t=169.629hr\]

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.

Solution:

The rate law of first reaction=k1 [O2NNH2]=k-1[O2NNH-]*[H+]

[O2NNH-]=k1[O2NNH2]/k-1[H+]

The rate law of second reaction=k2[O2NNH-]=k2 *{k1[O2NNH2]/k-1[H+]}, this is the rate determined reaction.

The overall rate law:k= k2*k1/k-1; Rate law={k*[O2NNH2]}/[H+]