Extra Credit 5B

17.1.5) A) We have the half reactions:

Ca->Ca²⁺+2e^-

F₂+2e^-->2F^- When we combine these two reactions together the electrons cancel out and we are left with:

Ca+F₂->2F^-+Ca²⁺

B) We have the half reactions:

(Li->Li⁺+e^-)*2 which me multiply by two in order to cancel out with the two electrons on the second reaction

Cl₂+2e^-->2Cl^- We combine these two reactions together and the electrons cancel out leaving us with:

2Li+Cl₂->2Cl^-+2Li⁺

C) We have the half reactions

(Fe->Fe³⁺+3e^-)*2 We multiply by two in order to cancel out electrons

(Br₂+2e^-->2Br^-)*3 We multiply by three in order to cancel out electrons

We combine these two equations we get 2Fe+3Br₂->6Br^-+2Fe³⁺

D) We have the half reactions

MnO₄^-+4H⁺+3e^-->MnO₂+2H₂O

(Ag->Ag⁺+e^-)*3 Which we multiply by three to cancel out the electrons when we combine. We are left with the following reaction:

MnO₄^-+4H⁺+3Ag->3Ag⁺2H₂O+MnO₂

19.1.3) A) The electron configuration for La is [Xe] 6s² 5d¹ because after Xe on the periodic table the first two electrons go to the 6s shell and the final electron goes to the 5d shell.

The configuration for La³⁺ is simply [Xe] because the three electrons from the La disappear in order to give the 3+ charge. It would first disappear from the 6s shell, then the 5d shell.

B) The configuration for Sm is [Xe] 6s² 4f⁶ as the total of eight extra electrons first fill the 6s shell and then go to the 4f shell.

The configuration of Sm³⁺ is [Xe] 4f⁵ because the three electrons that are lost from (Sm) are first taken from the 6s shell because they are farthest away from the nucleus and therefore are the easiest to take away. Then the last electron is taken from the 4f shell.

C) The electron configuration of Lu is [Xe] 4f¹⁴ 5d¹ 6s² because the electrons first fill the 6s shell then go to the 4f shell but in order to not completely fill the shell the last electron moves to the 5d orbital.

The configuration for Lu³⁺ is [Xe] 4f¹⁴ because the three electrons that are again lost in order to make the ion 3+ are taken from the highest energy orbitals that are farthest from the nucleus which are the 5d and 6s orbitals.

12.3.18) Reaction: A->B+C

A) The reaction is 2^nd order because as [A] doubles the rate quadruples, we see this by looking at the inequality

(.356/.23)=2(99,900/41,700) The left side comes from [A] and the right side comes from the given rates and therefore the inequality shows that the reaction is second order. Therefore, the reaction would be written as rate=k[A]².

B) The rate constant of the reaction is .0018L-mol^-1-s^-1

This is found when we take into the account that it is a second order reaction in regards to [A] and therfore we set up the second order reaction as

Rate=k[A]² where k is the rate constant. The [A] is squared because it is a second order reaction and the final SI units of the rate constant, k, are [L-mol^-1-s^-1] in order to cancel with the units of [A] notice that the units of A would originally be [mol/L] but since it is second order the [A] is squared and therefore so are the units leaving you with [mol/L]²

In the equation we then isolate K giving us the equation

K=[A]²/Rate

We then chose a rate and a concentration THAT MUST CORRESPOND from the given data. For example use 41,700mol/(L-s) and .23mol/L

We plug in giving K=(.23mol/L)²/41,700mol/L-s yielding the answer .0018s^-1

12.6.9) The step out of the four given reactions that does not terminate the chain reaction is

CH₃+HCl->CH₄+Cl

This because this reaction yields two products that first can continue to react but more importantly all of the products aren't stable unlike the other three reactions. The other three reactions all form stable ion and therefore they do not want to continue to react meaning that the total chain reaction is terminated.

21.4.21) A) The equation used to calculate the time is

t=-(1/k)(ln(A_t/A₀) where t=time k=the decay constant A_t is a certain concentration and A₀ is the initial concentration.

the decay constant k is found via

k=(ln(2))/t_.5 where t_.5 represents the half-life given as 4.7*10¹⁰years...plugging this value in we get

k=1.47*10^-11years

We infer that the strontium if formed via the decomposition of rubidium and therefore each mole of rubidium decomposed equals the same number in moles of strontium produced.

Therefore the initial number of moles of rubidium=

moles or rubidium left+moles of strontium

which when we plug in the given values divided by the atomic mass

(.00823g)/(87g/mol)+(.00047g)/(87g/mol)=

9.96*10^-5moles

now substituting back into the original equation we get

t=-(1/1.47*10^-11)(ln((9.96*10^-5)/(10^-4))

3,800,000,000 billion years

B) Time and amount of strontium are directly related. The amount produced via decay can be found by subtracting the initial amount from the amount that is currently being observed. Having a large initial amount will make this number smaller and seeing the age of the rock is proportionally related to the age then rock will be younger.

20.3.9) The reaction: Pb_s+2VO_aq²⁺+4H⁺->Pb_aq²⁺+2V_aq³⁺+2H₂O_liq

For the two half reactions we must first identify the oxidation numbers starting with Pb_s with starts with an oxidation number of 0 but then when it reacts and becomes Pb_aq²⁺ its oxidation numbers increases to 2+ which is a loss of electrons and there is the oxidation half reaction yielding

Pb_s->Pb_aq²⁺+2e^- Because this is the oxidation half reaction this must then occur at the anode because the oxidation half reaction always occurs at the anode in a galvanic cell. Seeing that this reaction is at the anode we can also conclude that this electrode is negative because the electrons, which are negative, leave the anode. (think opposites attract)

The oxidation number of V in VO_aq²⁺ is +4 because the O inherently has a charge of -2 and in order to make the overall charge =+2 we must set up the equation (+2)=(-2)+x where x is substituted for the oxidation number of V. In the products side the oxidation number of V³⁺ is 3+ because it has a 3+ charge. Therefore the oxidation number is being REDUCED so this is part of the reduction half reaction which is:

(e^-+2H⁺+VO_aq²⁺->V_aq³⁺+H₂O_liq)*2 which we multiply by two in order to cancel out the electrons with the reduction half reaction. The reduction half reaction always occurs at the cathode so therefore this reaction occurs at the cathode. Since the electrons flow from the anode to the cathode the electrode at the cathode must be positive in order to attract the electrons.