Extra Credit 23

Q19.7

Write and balance the net reaction using the two following half reactions. Find the Voltage for this reaction

BrO−(aq)+H2O(l)+2e−→Br−(aq)+2OH−(aq)BrO−(aq)+H2O(l)+2e−→Br−(aq)+2OH−(aq)

Mg(s)+2OH−(aq)→Mg(OH)2(s)+2e−Mg(s)+2OH−(aq)→Mg(OH)2(s)+2e−

A 19.7

half reaction 1: \(\ce{[BrO^{-}_{aq}+H_{2}0_{l}+2e^{-}\rightarrow Br^{-}_{aq}+2OH^{-}_{aq}]}\)

SRP=0.776 volts (reduction/cathode reaction)

half reaction 2: \(\ce{[Mg_{(s)}+2OH^{-}_{(aq)}\rightarrow Mg(OH)_{2}_{(s)}+2e^{-}]}\)

SRP=-2.69 volts (oxidation/anode reaction)

net equation: \(\ce{[BrO^{-}_{(aq)}+H_{2}O_{(l)}+2e^{-}+Mg_{(s)}+2OH^{-}_{(aq)}\rightarrow Br^{-}_{(aq)}+2OH^{-}_{(aq)}+Mg(OH)_{2}_{(s)}+2e^{-}]}\)

simplified net equation: \(\ce{[BrO^{-}_{(aq)}+H_{2}O_{(l)}+Mg_{(s)}\rightarrow Br^{-}_{(aq)}+Mg(OH)_{2}_{(s)}]}\)

voltage: Ecell=Ecathode-Eanode=(0.766)-(-2.69)=3.456 volts

Q19.63B

How many grams of metal are deposited at cathode by passage of 2.30 A of current for 75 minutes in electrolysis of aqueous solution containing:

1. Mg2+Mg2+
2. Sn4+Sn4+
3. Fe3+Fe3+
4. Ni2+Ni2+

S19.63B

to solve for moles of electrons:

\(\ce{[mole^{-}=75mins\times \frac{60sec}{1min}\times \frac{2.3Coulombs}{1sec}\times \frac{1mole^{-}}{96485Coulombs}]}\)

Mole ratio for Mg=24.31g per mole

Mole ratio for Sn=118.71g per mole

Mole ratio for Fe=55.85g per mole

Mole ratio for Ni=58.69g per mole

1. \(\ce{[MassMg=75mins\times \frac{60sec}{1min}\times \frac{2.3Coulombs}{1sec}\times \frac{1mole^{-}}{96485Coulombs}\times \frac{1moleMg^{2+}}{2mole^{-}}\times \frac{24.31gMg}{1molMg}=1.34gMg]}\)
2. \(\ce{[MassSn=75mins\times \frac{60sec}{1min}\times \frac{2.3Coulombs}{1sec}\times \frac{1mole^{-}}{96485Coulombs}\times \frac{1molSn^{4+}}{4mole^{-}}\times \frac{118.71gSn}{1molSn}=3.26gSn]}\)
3. \(\ce{[MassFe=75mins\times \frac{60sec}{1min}\times \frac{2.3Coulombs}{1sec}\times \frac{1mole^{-}}{96485Coulombs}\times \frac{1moleFe^{3+}}{3mole^{-}}\times \frac{55.85gFe}{1molFe}=2.05gFe]}\)
4. \(\ce{[MassNi=75mins\times \frac{60sec}{1min}\times \frac{2.3Coulombs}{1sec}\times \frac{1mole^{-}}{96485Coulombs}\times \frac{1moleNi^{2+}}{2mole^{-}}\times \frac{58.69gNi}{1molNi}=2.05gNi]}\)

The length of electrolysis time is converted from minutes to seconds. This answer is then multiplied by the ratio of Coulombs to seconds (even though the question is asked in Amps, Coulombs and Amps are in a 1:1 ratio therefore a conversion calculation is unnecessary). This answer is then converted to moles of electrons and then moles of the ion and then to grams of the element using its mole ratio.

Q21.4C

1. How many unpaired e- would you find in the octahedral complex [ZnCl6]4−[ZnCl6]4−?
2. How many unpaired electrons would you expect to find in the tetrahedral complex [CuI4]2−[CuI4]2−? Would you expect more unpaired electrons in the octahedral complex [Mn(NH3)8]3+[Mn(NH3)8]3+?

S21.4C

1. Zn has oxidation number +2, so Zn2+ has electron configuration [Ar]3d10. There are zero unpaired electrons.
2. Cu has oxidation number +2, so Cu2+ has electron configuration [Ar]3d9. There is one unpaired electron. Mn has oxidation number +3, so Mn3+ has electron configuration [Ar]3d4. There are four unpaired electrons. Therefore, this octahedral complex has more unpaired electrons than the tetrahedral complex.

Once the charge of the metal, electron configurations, type of spin, and number of electrons are determined, the crystal field splitting can be drawn out. Both crystal field splitting diagrams are attached below.

Charged metal ions, when they have d electrons, loses S electrons first. This is why Zn2+ is still able to keep all of its d electrons, because it only lost 2 electrons and there are 2 s electrons. However, this is also why Mn3+ did lose a d electron, because it only had 2 s electrons to use, but it lost 3 total, therefore taking the last one from the d orbital.

In part B, the octahedral complex (despite having fewer electrons overall) contained a weak field ligand (NH3). A weak field ligand forces the metal complex to have a "high spin" meaning as many electrons are left unpaired as possible while still following Hund's rule. This is why [Mn(NH3)8]3+ had more unpaired electrons than the other complex.

Q24.12B

For the reaction A + B àC+ D the following initial rate of reaction were found. What is the rate law for this reaction?

S24.12B

Rate = (2.16 x 10-3)[A]1[B]2

1st step: choose two reactions that have a reactant concentration that cancel each other out.

\(\ce{[\frac{rxn1}{rxn3}=\frac{4.2\times 10^{-3}}{1.7\times 10^{-2}}=\frac{0.6^{A}}{0.6^{A}}\times \frac{1.8^{B}}{3.6^{B}}]}\)

2nd step: solve for an exponent

\(\ce{[0.25=0.5^{B}]}\)

\(\ce{[B=2]}\)

3rd step: do the same for the other exponent

\(\ce{[\frac{rxn1}{rxn2}=\frac{4.2\times 10^{-3}}{1.3\times 10^{-2}}=\frac{0.6^{A}}{1.8^{A}}\times \frac{1.8^{B}}{1.8^{B}}]}\)

\(\ce{[0.3=0.3^{A}]}\)

\(\ce{[A=1]}\)

4th step: create the rate law

\(\ce{[rate=k[A]^{1}[B]^{2}]}\)

5th step: choose a reaction and plug in values to solve for k

rxn1: \(\ce{[(4.2\times 10^{-3})=k(0.6)^{1}\times (1.8)^{2}]}\)

\(\ce{[(4.2\times 10^{-3})=k(1.944)]}\)

\(\ce{[k=2.16\times 10^{-3}]}\)

\(\ce{[rate=(2.16\times 10^{-3})[A]^{1}[B]^{2}]}\)

Q24.59B

1. Do catalysts take part in the reaction they catalyze? Do catalysts always speed up a reaction? Explain.
2. What is the function of a catalyst?

S24.59B

1. No, catalysts do not take part in the reactions they catalyze. No, catalysts do not always speed up a reaction; some negative catalysts, called inhibitors, can slow down the rate of a reaction.
2. Catalysts main function is to provide an alternative pathway for a reaction. By changing the mechanism for a reaction, a catalyst provides a pathway with a lower activation energy, resulting in a faster reaction.

1. Catalysts do not react with the elements of a reaction or alter any chemical properties of a reaction, only its speed. There are inhibitors (negative catalysts) that can slow the rate of a reaction rather than speeding it up.

2. Catalysts provide a new pathway with a lower activation energy for the reaction to take. The reaction can more easily occur, speeding up the reaction time.

Q25.27E

Approximately how old is a wooden object given that the object has a disintegration rate of 11.0 dis/minxg? It is given that the half-life of this particular wood is 5730 years.

S25.27E

Solve for k

\(\ce{[t1/2=\frac{ln2}{k}]}\)

\(\ce{[(5730)=\frac{0.6931}{k}]}\)

\(\ce{[k=1.21\times 10^{-4}]}\)

solve for t

\(\ce{[k=\frac{-1}{t}\times ln(\frac{Acurrent}{Ainitial})]}\)

\(\ce{[(1.21\times 10^{-4})=\frac{-1}{t}\times ln(\frac{11dis/min}{15dis/min})]}\)

\(\ce{[(1.21\times 10^{-4})=\frac{-1}{t}\times (-0.310155)]}\)

\(\ce{[t=2565.26 years old]}\)

This problem is solved using two equations for finding half life reactions. The first step of this solution is solving for the K constant using the equation: t(half life)=ln(2)/k. Once the value of k is known, it can be applied to another equation using the time as well as current and initial amounts of the substance: k=-1/t x ln(Acurrent/Ainitial). Solve for t by plugging in the rest of the known values.

Q21.1.2

Describe the differences between nonionizing and ionizing radiation in terms of the intensity of energy emitted and the effect each has on an atom or molecule after collision. Which nuclear decay reactions are more likely to produce ionizing radiation? nonionizing radiation?

S21.1.2

Ionizing: gives a substantial (and commonly dangerous) amount of energy/radiation, caused by unstable molecules, can take/"knock" electrons off other molecules it interacts with

associated with: gamma rays, alpha particles, beta particles

Non-ionizing: carries less energy, can only have effect on few certain molecules (ex: H2O) in order to remove electrons

associated with: X-rays, not associated with many nuclear decay reactions.

Q24.3.1

Write the name of the following complexes

1. [CoCl3(NH3)3]
2. [Co(ONO)3(NH3)3]
3. [Fe(ox)2(H2O)2]-
4. Ag2[HgI4]

S24.3.1

1. triamminetrichlorocobalt(III)
2. triamminetrinitrito-O-cobalt(III); or triamminetrinitritocobalt(III)
3. diaquadioxalatoferrate(III) ion
4. silver(I) tetraiodomercurate(II)

1. Put the ligands in alphabetical order

2. Know the root form of each ligand name

3. Put prefixes indicating amounts in front of each non-metal ligand (ex: tri)

4. End with the metal and its charge

specials: if the complex has an overall charge add the word ion to the end of the name, if there are two parts to the complex split the name into a cation and an anion.