Extra Credit 9

Q19.2B

Use the table of Standard Reduction Potential to predict whether these reactions will happen spontaneously.

1. \[Ag^+(aq)+Cu(s)→2Ag(s)+Cu^{2+}(aq)\]
2. \[Fe^{3+}(aq)+Na(s)→Fe^{2+}(aq)+Na^{+}(aq)\]
3. \[Zn^{2+}(aq)+2I^{-}(aq)→Zn(s)+I_{2}(l)\]

Tutorial

The first thing to do in this problem is to separate the reaction into two reduction reactions. The reduction reaction is a half reaction that shows the proper gain of electrons on the reactant side. Once the reduction reactions are created for each of the compounds in the reaction, the Standard Reduction Potential table can be used. It gives the \(E^°{(V)}\) that corresponds to the reduction reaction. For the half reaction with oxidation, you must make the \(E^°{(V)}\) negative because it needs to be the oxidation potential rather than the reduction potential. The equation that can be used to find whether or not the reaction is spontaneous is \[E^°_{cell}=E^°_{cathode}- E^°_{anode}\] where the cathode is where reduction takes place and the anode is where oxidation takes place. If the \(E^°_{cell}\) is positive then the reaction will be spontaneous because it will produce a negative ΔG.

Solution

1. \[2(Ag^{+}(aq)+e^{-}→Ag(s)) \qquad E^°=0.80V{(cathode)}\]

\[Cu^{2+}(aq) + 2e^{-} → Cu(s) \qquad E^°=-0.34V{(anode)}\]

\[E^°_{cell}=0.80V-(-0.34V)=1.14 V\] so the reaction would be spontaneous since it is positive.

2. \[Fe^{3+}(aq)+e^{-}→Fe^{2+}(aq) \qquad E^°=0.77V {(cathode)}\]

\[Na^{+}(aq) + e^{-}→Na(s) \qquad E^°=2.71V{(anode)}\]

\[E^°_{cell}=0.77-2.71V=-1.94V\] so the reaction would be non-spontaneous.

3. \[Zn^{2+}(aq) + 2e^{-}→ Zn(s) \qquad E^°=-0.76V{(cathode)}\]

\[I_{2}(l) + 2e^- →2I^-(aq) \qquad E^°=-0.54V{(anode)}\]

\[E^°_{cell}=-0.76V-(-0.54V)=-0.223V\] so the reaction would be non-spontaneous

Original solution was incorrect (forgot to make oxidation reduction potentials opposite sign as SRPs).

Q19.35C

The voltaic cell in the following diagram has an \(E^°_{cell}=0.5464V\). Solve for the \([Cl−]\) in the cell

\[Ag(s)|Ag^+(0.40M)||Cl^-(xM),Cl_{2}(0.60atm)|Pt(s)\]

Tutorial

For this problem, we must start by writing the two half reduction reactions and deciding which is the anode and the cathode. The anode is where oxidation occurs and the cathode is where reduction occurs. The reduction reactions are the half reactions with the gain of electrons written on the left side. Once these are written, we can look at the Standard Reduction Potential table to find the corresponding \(E^°\). Then we can use the equation \(E^°_{cell}=E^°_{cathode}- E^°_{anode}\) to find the overall voltage. Once we find this, we have to use Nernst equation \[E_{cell}=E^°_{cell}-\frac{(RT)}{(nF)}ln(\frac{[reduction]}{[oxidation]})\] where R is the universal gas constant, T is temperature \((298K)\), n is the number of electrons transferred, and F is Faraday's constant. Plug in all information and solve for the missing concentration.

Solution

The half reactions are: \[2(Ag(s) → Ag^+(aq) + e^-) \qquad E^°=-0.80V{(anode)}\]

\[Cl_{2}(g)+2e^-→ 2Cl^-(aq) \qquad E^°=1.36V{(cathode)}\]

\[E^°_{cell}=1.36V-(-0.80V)=2.16V\]

Then we use Nernst Equation:

\[0.5464V=2.16V-(R)(T)/(n)(F)ln([xM]^{2}/[0.40]^{2})\]

\[-1.6136V=-(0.0591V)/(n)log([xM]^{2}/[0.40]^{2})\]

\[-1.6136V=-(0.02955)log([xM]^{2}/[0.40]^{2})\]

\[54.6=log([xM]^{2}/[0.40]^{2})\]

\[[xM]^{2}/[0.40M]^{2}=10^{54.6}=4.03 x 10^54\]

\[[xM]^{2}=4.03 x 10^54(0.16)=6.45x10^53\]

\[[xM]=6.45x10^53 M\]

Q20.21C

Write an equation using \(Mn^{2+}\) ion as an oxidizing agent in basic solution. Write an equation using \(Fe(s)\) as a reducing agent.

Tutorial

First, we must know the difference between oxidizing and reducing agents. An oxidizing agent is a compound that takes electrons from another compound and induces oxidation, thus being reduced itself. The reducing agent is a compound that gives electrons to the other, causing a reduction reaction, and oxidizing. To write the equation, we start with the two half reactions. The number of electrons being transferred must be equal on both reactions so one equation may need to be multiplied by an integer to accomplish this. Once the equations are balanced, they can be summed together to create the full reaction, where the electrons and any other repeating species will cancel out.

Solution

Half reactions: \[Mn^{2+}(aq)+2e^-→ Mn(s)\]

\[Fe(s)→ Fe^{2+}(aq) + 2e^-\]

Full reaction: \[Mn^{2+}(aq) + Fe(s) → Mn(s) + Fe^{2+}(aq)\]

Q24.1A

In the reaction \(4A+3B→2C+3D\) reaction \(A\) is found to disappear at a rate of 5.1 X 10-5 Ms-1

1. What is the rate of reaction?
2. What is the rate of disappearance of B?
3. What is the rate of formation of C?

Tutorial

The rate of the reaction can be found by the differential rate law that is based on the stoichiometry of the reaction. The differential rate law shows how the rate changes based on concentration and can be defined in terms of the reactants or the products. If defined in terms of the reactants then the derivative would be negative so the overall rate of disappearance (-) could be positive. The rate law for the products would be positive since the rate of appearance would be positive. For the reaction \(aA+bB→cC+dD\) the rate can be determined by the equation \[\frac{-Δ[A]}{aΔt}=\frac{-Δ[B]}{bΔt}=\frac{Δ[C]}{cΔt}=\frac{Δ[D]}{dΔt}\]

Solution

1. The rate of the reaction could be defined in products or reactants but the rate of the reaction would be \(\frac{-Δ[A]}{4Δt}\) where \(\frac{Δ[A]}{Δt}\) is the rate of disappearance. The rate would be equal to \[\frac{-1}{4(-5.1x10^{-5}}M/s)=1.275x10^{-5}M/s.\]

2. The rate of disappearance of B is \(\frac{Δ[B]}{Δt}\). We know that \(\frac{-Δ[A]}{aΔt}=\frac{-Δ[B]}{bΔt}\) so in this case \(\frac{-Δ[A]}{4Δt}=\frac {-Δ[B]}{3Δt}\) To find the disappearance of B, we have to rearrange the equation so \(\frac{Δ[B]}{Δt}=-3\frac{-Δ[A]}{4Δt})\) then we plug in the values to find that \[\frac{Δ[B]}{Δt}=-3(-(-5.1 x 10^{-5}M/s))\] which gives \[-1.53x10^{-4}M/s\]

3. The rate of formation of C is \(\frac{Δ[C]}{Δt}\). Since we know that \(\frac{-Δ[A]}{aΔt}=\frac{Δ[C]}{cΔt}\) we can plug in the stoichiometric values (a and c) and get \(\frac{-Δ[A]}{4Δt}=\frac{Δ[C]}{2Δt}\). To find the formation of C, we would rearrange the equation so \[\frac{Δ[C]}{Δt}=2(-(-5.1 x 10^{-5}M/s))\] which gives \[1.02x10^{-4}M/s\].

Q24.43A

Does the half-life of a reaction get longer or shorter as initial reactant concentration increases and why? Please answer for a) zero-order reactions b) second-order reactions.

Tutorial

To solve this problem, we would first need to know what reaction order the reaction is because the rate depends on this order. To find rate order you can make a proportion with concentrations and set them equal to a proportion with the initial rates from 2 experiments. Place an undefined exponent on the concentration and solve for this exponent and this will be your reaction order for this reaction. The rate law for a zero order reaction is equal to \(k[R]^0\) where k is the rate constant and R is the reactant. The rate law for a second order reaction is equal to \(k[R]^2\). K is a value that has to be determined experimentally but will remain constant for the same reaction even if concentrations change. Given these formulas, we can manipulate them to get the half life formulas for each order. The half life for a zero order reaction is \(t_{1/2}=[R]_0/2k\) where \([R]_0\) is the initial concentration. The half life function for a second order reaction is \(t_{1/2}=1/k[R]_0\)

Solution

a. According to the rate law for zero order reactions, the half life would increase as the initial concentrations increase because the half life is directly proportional to the initial concentration.

b. According to the rate law for a second order reaction, the half life would decrease as the initial concentration increases because the half life is inversely proportional to the initial concentration.

Q25.21A

The disintegration rate for a sample containing Sr9038 as the only radioactive nuclide is \(8754 dis h^{-1}\). The half life of Sr9038 is 27.7 years. Estimate the number of Sr9038 atoms in the sample at this time.

Tutorial

To solve this problem, we need to use the first order equation for Activity that \(A=λN\), where A is the activity, λ is the proportionality constant, and N is the number of atoms that have not decayed. The activity (A) can also be called the rate of disintegration and it is defined as the decrease in the number of radioactive nuclei over time. We must also know that half life and the proportionality constant are related by the equation \(t_{1/2}=ln(2)/λ\). To solve, we must plug all of the information into these formulas.

Solution

First find λ by rearranging the equation \(t_{1/2}=ln(2)/λ\) so \(λ=ln(2)/t_{1/2}\). Then plug in the half life of 27.7 years. \(λ=ln(2)/27.7y=0.02502y^{-1}\). Next to be able to enter the information into the formula \(A=λN\), we must convert A into dis/y from dis/h. \((8754 dis/h)(8760 h/y)=7.67 x10^7 dis/y\). Then we can rearrange |(A=λN\) so \(N=A/λ\). When the information is entered into this equation, we get \[N=(7.67x10^7 dis/y)/(0.02502 y^{-1})=3.065x10^9 dis\].

Q19.3

What is the half-life of an isotope that is 75% decayed after 16 days?

Tutorial

A half life (t_1/2) is the amount of time it takes for half of the original concentration to decay. If one half life has gone by then \([A]=1/2[A]_0\). If two half lives have gone by then \([A]=1/2*1/2[A]_0\) meaning \(1/4[A]_0\)_. If three half lives have gone by then \([A]=1/2*1/2*1/2[A]_0\) meaning \(1/8[A]_0\).

Solution

Given the information above, it is clear two half lives have gone by, the first decaying 50% of the of the original and the second decaying an additional 50% of the half remaining, which would mean that 75% of the original amount would be decayed overall. Given that 75% is 2 half lives and 16 days have gone by, the half live would be half of the given time. \[\frac{16}{2}= 8 days\]

The half life of this isotope is 16 days.

Q21.3.2

How does a star produce such enormous amounts of heat and light? How are elements heavier than Ni formed?

Tutorial

To answer this question we must know how nuclear chemistry works. Nuclear chemistry does not follow the law of conservation of mass because some of the mass is converted into massive amounts of energy. The formula to calculate this is \[ΔE=Δmc^2\] where \(Δm\) is the (mass of free nucleons - mass of nucleus), and c is the speed of light (2.998 x 10^8 m/s). Also, we must know the difference between nuclear fusion vs nuclear fission. Nuclear fusion is when small nuclei join together to form a larger one. Nuclear fission is where a larger nucleus divides to form two large fragments because it absorbed a neutron.

Solution

The center of the star has a lot of pressure which creates kinetic energy and temperatures that are high enough to undergo a process called nuclear fusion. Nuclear fusion is when multiple small nuclei come together to form heavier ones. In stars, fusion occurs with four protons that fuse into one alpha particle. This process releases of two positrons, two neutrinos, and a substantial amount of energy. This process of nuclear fusion of four protons is how the star can produce so much light and heat.

Elements larger than nickel are produced by nuclear fusion, when two small nuclei come together to form heavier ones. When two nuclei larger than nickel are fused, energy is absorbed by the element.