Extra Credit 5

Q19.1D

From the observation listed, estimate the value of \(E^∘\) for the half reaction.

\[M^{2+}_{(aq)}+2e^- → M_{(s)}\]

1. The metal M reacts with \(HNO_3\)and \(HCl\), and \(M\) can replace \(Ag^+\) but not \(Cu^{2+}\).
2. The metal reacts with \(HNO_3\) and it can not produce \(H_2\); however, it can interact with solid Iron.

Tutorial

This problem provides us with two metal-metal redox reactions (shown below), one of which is spontaneous and one of which is non-spontaneous. From this information we can determine a range of possible \(E^°\) values for the metal.

\[ 1. M_{(s)} + 2Ag^{+}_{(aq)} → M^{2+}_{(aq)} + 2Ag_{(s)} \qquad E^°_{(cell)} > 0 \\ 2. M_{(s)} + Cu^{2+}_{(aq)} → M^{2+}_{(aq)} + Cu_{(s)} \qquad E^°_{(cell)} < 0 \]

Because we can use a table of Standard Reduction Potentials (SRPs) for silver and copper, we can determine a range of possible \(E^°\) values that correspond to the spontaneity of the given reactions.

We are also given that the metal reacts with \(HNO_3\) and \(HCl\). So, we can examine the SRP values for these species to determine what the SRP value of the metal may be.

We are given that the metal ion can react with solid iron, which has the half reaction \(Fe^{2+}+2e^- → Fe_{(s)} \qquad E^°=-0.44\) Therefore, the SRP of the metal M must be greater than -0.44 because we know that

Solution

For the reaction of metal M and silver, we know that the \(E^°_{cell}\) is positive because the reaction proceeds spontaneously. The half reaction of silver: \(Ag^+e^-→Ag]) has \(E^°= 0.80V\). We are given that the silver reaction is the reduction half, which means that the metal M must have a lower SRP than silver does, meaning that \(E^°_{metal}<0.80V\).

For the reaction of metal M and copper, \(E^°_{cell}\) is negative because the reaction is non-spontaneous. The copper half reaction \(Cu^{2+}+2e^- → Cu\) has \(E^°= 0.34V\). We know that the copper reaction is the reduction half, which means that the metal M must have a SRP higher than the value for copper because this reaction is non-spontaneous. So, \(E^°_{metal}>0.34V}\).

From both of these metal reactions, we know that \(0.80V > E^°_{metal} > 0.34V}\).

We know that the metal reacts with aqueous nitric acid, which has the half reaction \(NO_{3}^{-} + 4H^+ + 3e^- → NO + 2H_2O \qquad E^°=0.96 \)

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According to statement a: Metal M can replace Ag+ but not Cu2+, thus we can assume its reduction potential is lower than Ag+ and higher than Cu2+, which gives us a rage of 0.34<X<0.8.

Then, according to statement b: Metal M reacts with HNO3 but cannot produce H2, which indicates that its potential is above zero, fitting the range indicated by statement A. Also, being able to interact with Fe indicates that it is able to either replace Fe or be replaced by Fe. In this case, it is safe to assume that it can be replaced by Fe because it fits the range indicated by statement A.

Answer: 0.34< X< 0.8

Q19.31B

Calculate a theoretical voltage of silver-lead button cell at 298K.

Cell reaction: Pb(s)+Ag2O(s)→PbO(s)+2Ag(s)

ΔG∘f for PbO(s)=−187.9kJ/mol

ΔG∘f for Ag2O(s)=−11.20kJ/mol

ΔG∘f for 2Ag(s)=0kJ/mol

ΔG∘f for Pb(s)=0kJ/mol

Because question did not state otherwise, we assume this reaction takes place under standard condition, thus E = Edot

First, we should split the whole equation up into two half reactions:

Reduction half reaction: Ag₂O(s) + 2e^- → 2Ag(s)

Oxidation half reaction: Pb(s)→PbO(s)+2e^-

Second, calculate the overall delta G.

Overall deltaG = deltaG(PbO)+deltaG(2Ag) - (deltaG(Ag₂O)+deltaG(Pb))

= 187.9Kj/mol+0Kj/mol-(11.2KJ/mol+0KJ/mol)

= 176.7KJ/mol

Third, using the equation”delta Gdot = -nFEdot”, where n is the mole of electrons transferred during the reaction, and F is Faraday’s constant, which is measured in coulombs per mole: F=9.649×104 C/mol, we can then solve for Edot:

Let Edot be X

176.7KJ/mol = -2*(9.649E4 C/mol)*X

X = -9E-4V

Answer: Edot = -9E-4V

Q20.15A

Describe Chelation and how it relates to polydentates. Give an example of a polydentate.

Chelation is a phenomenon in which a polydentate ligand bonds to a transition metal ion, forming a ring, some describe this ring as “two claws of a lobster clinging to one metal ion.” One of the most well-known example of polydentates is EDTA, which is a hexadentate ligand.

Q21.23B

If the complex [Fe(NH3)6]2+ is paramagnetic, can you tell whether it is an octahedral or tetrahedral complex? What if the complex were diamagnetic?

First, we need to determine the spin of this compound. Since NH3 is a strong field ligand, it is a low spin complex.

Then, we need to determine how many electrons are there in the d orbital in the metal ion. Since NH3 has an oxidation number of 0, Fe would have to have a 2+ oxidation number to satisfy the 2+ charge of the complex. Since Fe originally has 2 electrons in 4S orbital and 6 electrons in 3D orbital, Fe2+ has 0 electrons in 4S orbital and 6 Electrons in 3D orbital.

Since the complex is paramagnetic, it has unpaired electrons. Now we can look at the layout of octahedral and tetrahedral ligand field.

tetrahedral: (L.S. stands for low spin, H.S. stands for high spin)

octahedral:

From these two images, we can see that if the complex is low spin and has 6 electrons in the d orbital, then it would have unpaired electrons in the tetrahedral field, which causes this complex to be paramagnetic. However, if this complex is diamagnetic, low spin and have 6 electrons in the d orbital, then it would have to be in an octahedral field.

Q24.34A

A reaction 50% complete in 40.0 min. How long is the start will the reaction by 75% complete if it is (a) first order (b) zero order?

(a)First, we need to use the integrated first order reaction equation to solve for k, the reaction constant. The equation: ln[A]=−kt+ln[A]o, where [A]o is the concentration of reactant at the time 0 and [A]t is the concentration of reactant at time t. Rearrange the equation and we can get: ln([A]t/[A]o)=−kt. Plugging in the concentration ratio(50% complete = 1:2 ratio of [A]t to [A]o) and time, we get k = 0.01733mol/min

Now we can plug in the concentration ratio(0.25:1) and k(0.01733) to solve for t using the same equation, t = 79.99 mins.

(b) If the reaction is zero order, then the reaction rate is independent of the reactant concentrations. And since 75% is the second half life of a reaction(75% = 50%+25%), it will take 40min*2 = 80mins for this reaction to be 75% complete.

Q25.11D

Write equations for the following nuclear reactions:

1. Alpha decay of Bi210: \(\ce{^210_83Bi \rightarrow ^{4}_2He + ^{206}_81Tl}\)
2. Beta decay of Bi210: \(\ce{^210_83Bi \rightarrow ^0_{-1}\beta + ^{210}_84Po}\)
3. Fusion of Californium-252 and Boron-10: \(\ce{^252_98Cf + ^10_{5}B \rightarrow ^{262}_103Lr}\)

Q25.11E

Convert the word in to equations for the below nuclear reactions.

1. Bombardment of Be47 with protons to produce B58 and γ rays: \(\ce{^7_4Be + ^1_1H \rightarrow ^0_{0}\gamma + ^{8}_5B}\)
2. Bombardment of Mg1217 with H13 to produce Al1319 and n01: \(\ce{^17_12Mg + ^3_{1}H \rightarrow ^19_13Al + ^{1}_0n}\)
3. Bombardment of P1522 with neutrons to produce Si1422: \(\ce{^22_15P + ^1_{0}n \rightarrow ^22_14Si + ^{1}_1H}\)

Q25.42E

Of the following pairs, explain which of the two atoms is more stable and why?

1. Ti4422 or Ti4622: Isotopes are believed to be more stable if their neutron-proton ratio is roughly within the range of 1:1 - 1:1.5, therefore

\(\ce{^44_22Ti}\) is more stable because its neturon-proton ratio is 1:1 whereas \(\ce{^46_22Ti}\)'s neutron-proton ratio 1.09:1.

1. Cl3417 or Cl3517: \(\ce{^34_17Cl}\) is more stable because the neutron-proton ratio is 1:1, whereas \(\ce{^35_17Cl}\)'s neutron-proton is 1.06:1
2. Ge6632 or Ge6532: \(\ce{^65_32Ge}\) is more stable because the neutron-proton ratio is 1.03, whereas \(\ce{^66_32Ge}\) is 1.06:1