Extra Credit 16

Q17.2.5

Question: Identify the species oxidized, species reduced, and the oxidizing agent and reducing agent for all the reactions in the previous problem.

S17.2.5

First separate the half-reactions and then balance them.

For acidic solutions: add \(H_2O\) to balance oxygen and add \(H^+\) ions to balance the protons.

For basic solutions: add \(H_2O\) to balance oxygen, add \(H^+\) ions to balance the protons, add \(OH^-\) to balance \(H^+\), combine \(OH^-\) and \(H^+\) to form water, and finally cancel common terms.

Remember, oxidation/reducing agent loses electrons and reduction/oxidizing agent gains electrons.

a) Step 1: \(Al(s)+Zr^{4+}(aq)\)\(\rightarrow\)\(Al^{3+}(aq)+Zr(s)\)

Step 2: \(Al(s)\)\(\rightarrow\)\(Al^{3+}(aq)+3e^{-}\) \(Zr^{4+}+4e^{-}\)\(\rightarrow\)\(Zr(s)\)

Therefore we can tell that \(Al(s)\) is oxidized and \(Zr^{4+}(aq)\) os reduced.

b) Step 1: \(Ag^+(aq)+NO(g)\)\(\rightarrow\)\(Ag(s)+NO_3^-(aq)\) (acidic solution)

Step 2: \(Ag+(aq)+e^-\)\(\rightarrow\)\(Ag(s)\)

Step 3: \(2H_2O(l)+NO(g)\)\(\rightarrow\)\(NO_3^-(aq)+4H^+(aq)+3e^-\)

Therefore we can tell that \(NO(g)\) is oxidized and \(Ag^{+}(aq)\) is reduced.

c) Step 1: \(SiO_3^{2-}(aq)+Mg(s)\)\(\rightarrow\)\(Si(s)+Mg(OH)_2(s)\) (basic solution)

Step 2: \(4e^-+3H_2O(l)+SiO_3^{2-}(aq)\)\(\rightarrow\)\(Si(s)+OH^-(aq)\) \(OH^-(aq)+Mg(s)\)\(\rightarrow\)\(Mg(OH)_2(s)+2e^-\)

Therefore we can tell that \(Mg(s)\) is being oxidized and \(SiO_3^{2-}(aq)\) is being reduced.

d) Step 1: \(ClO_3^-(aq)+MnO_2(s)\)\(\rightarrow\)\(Cl^-(aq)+MnO_4^-(aq)\) (basic solution)

Step 2: \(7e^-+3H_2O(l)+ClO_3^-(aq)\)\(\rightarrow\)\(Cl^-(aq)+6OH^-(aq)\) \(4OH^-(aq)+MnO_2(s)\)\(\rightarrow\)\(MnO_4^-(aq)+2H_2O(l)+4e^-\)

Therefore we can tell that \(MnO_2(s)\) is being oxidized and \(ClO_3^{-}(aq)\) is being reduced.

Q19.1.14

Question: A 2.5624-g sample of a pure solid alkali metal chloride is dissolved in water and treated with excess silver nitrate. The resulting precipitate, filtered and dried, weighs 3.03707 g. What was the percent by mass of chloride ion in the original compound? What is the identity of the salt?

S19.1.14

\(?Cl(aq)+AgNO_3(aq)\)\(\rightarrow\)\(?NO_3(aq)+AgCl(s)\)

\((3.03707g AgCl(s))\)\(\times\)\((\frac{1mol}{143.32g}AgCl(s))\)\(\times\)\((\frac{1mol ?Cl(aq)}{1mol AgCl(s)})\)\(=0.0212mol ?Cl(aq)\)

\((2.5624g?Cl(aq))\)\(\times\)\((\frac{1}{0.0212mol?Cl(aq)})\)\(=120.92\)\(\frac{g}{mol}?Cl\)

\(120.92\frac{g}{mol}?Cl-35.45\frac{g}{mol}Cl=85.47\frac{g}{mol}Rb\)

Therefore the salt is \(RbCl(aq)\).

\((2.5624g RbCl(aq))\)\(\times\)\((\frac{1mol}{120.92g}RbCl)\)\((\frac{1molCl}{1molRbCl})\)\((35.45\frac{g}{mol}Cl)\)\(=0.7512g Cl\)

\(0.7512g Cl\)\(\div\)\(2.5624g RbCl\)\(\times\)\(100=29.32\)\(\%\) by mass of \(Cl^{-}\)

Q19.3.6

Question: How many unpaired electrons are present in each of the following?

1. \([CoF_6]^{3-}\) (high spin)
2. \([Mn(Cn)_6]^{3-}\) (low spin)
3. \([Mn(Cn)_6]^{4-}\) (low spin)
4. \([MnCl_6]^{4-}\) (high spin)
5. \([RhCl_6]^{3}\) (low spin)

S19.3.6

It is helpful to remember that at high spin we will be filling in all the spots before pairing up unpaired \(e^-\) and at low spin we will be filling up as much of the lower level before moving on to the top level.

1. \(Co^{3+}\rightarrow3d^{6}\) therefore there are 4 unpaired \(e^{-}\)
2. \(Mn^{3+}\rightarrow3d^{4}\) therefore there are 2 unpaired \(e^{-}\)
3. \(Mn^{2+}\rightarrow3d^{5}\) therefore there are 1 unpaired \(e^{-}\)
4. \(Mn^{2+}\rightarrow3d^{5}\) therefore there are 5 unpaired \(e^{-}\)
5. \(Rh^{3+}\rightarrow4d^{6}\) therefore there are 0 unpaired \(e^{-}\)

Q12.4.6

Question: What is the half-life for the first-order decay of phosphorous-32 \((^{32}_{15}P\rightarrow^{32}_{16}S+e^{-})\). \(K=4.85\times10^{-2}day^{-1}\)

S12.4.6

\(t_\frac{1}{2}=\frac{ln2}{K}\)

\(t_\frac{1}{2}=\frac{ln2}{4.85\times10^{-2}d^{-1}}\)

\(t_\frac{1}{2}=14.3 days\)

Q21.2.1

Question: Write the following isotopes in hyphenated form (e.g., "carbon-14")

1. \(^{24}_{11}Na\)
2. \(^{29}_{13}Al\)
3. \(^{73}_{36}Kr\)
4. \(^{194}_{77}Ir\)

S21.2.1

1. sodium-24
2. aluminum-29
3. krypton-73
4. iridium-194

Q21.5.4

Question: Cite the conditions necessary for a nuclear chain reaction to take place. Explain how it can be controlled to produce energy, but not produce an explosion.

S21.5.4

The conditions necessary for a nuclear chain reaction to occur are to increase or decrease the non-fission neutron absorption in the system depending on the number of further fission produced. Therefore a neutron absorbing element must be present to control the amount of free neutron so that the reactor doesn't die out nor explode

Q20.4.2

Question: List two factors that affect the measured potential of an electrochemical cell and explain their impact on the measurements.

S20.4.2

Two factors that affect the measured potential of an electrochemical cell are concentration (Q) and temperature. With an increase in temperature or concentration, there's a decrease in the voltage of the cell.

Q20.5.30

Question: The silver-silver bromide electrode has a standard potential of 0.07133 V. What is the \(K_{sp}\) of \(AgBr\)?

S20.5.30

We are given the following:

Cathode: \(AgBr(s)+e^1\)\(\rightarrow\)\(Ag(s)+Br^-(aq)\) \(E^\circ\)\(=0.07133V\)

Anode: \(Ag^+(aq)+e^-\)\(\rightarrow\)\(Ag(s)\) \(E^\circ\)\(=0.8V\)

\(K_{sp}=[Ag^{+}][Br^{-}]\)

\([Ag^{+}]=\frac{K_{sp}}{[Br^{-}]}=\frac{K_{sp}}{[1]}=K_{sp}\)

To find the value of \(E^\circ_{cell}\), plug in the given values into the following equation.

\(E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}\)

\(E^\circ_{cell}=0.01733V-0.8V\)

\(E^\circ_{cell}=-0.7287V\)

Once we've found the \(E^\circ_{cell}\) value, we can use the Nernst Equation to find the \(K_{sp}\) value.

\(E_{cell}=E^\circ_{cell}-(\frac{0.0591V}{n})logQ\)

\(0V=-0.7287V-(\frac{0.0591V}{1})logK_{sp}\)

\(0.7287V=-0.0591VlogK_{sp}\)

\(logK_{sp}=-12.33\)

\(K_{sp}=4.7\times10^{-13}\)