Extra Credit 9

Q17.1.8

Identify the species that was oxidized, the species that was reduced, the oxidizing agent, and the reducing agent in each of the reactions of the previous problem.

First step, make sure all equations are balanced.

i.e.

1. \(SO_3^{2-} (aq) +Cu(OH)_2(s)\rightarrow SO_4^{2-}(aq)+Cu(OH)(s)\)
   1. Balanced: \(SO_3^{2-}(aq) +2Cu(OH)_2(s) \rightarrow SO_4^{2-}(aq) + 2Cu(OH)(s) + H_2O(l)\)
2. \(O_2(g)+Mn(OH)_2(s) \rightarrow MnO_2(s)\)
   1. Balanced: \(O_2(g) + 2Mn(OH)_2(s) \rightarrow 2MnO_2 (s) + 2H_2O(l)\)
3. \(NO_3^{-}(aq)+H_2(g) \rightarrow NO(g)\)
   1. Balanced: \(NO_3^-(aq) +4H^+(aq) +H_2(g) \rightarrow NO(g) + 2H_2O(l)\)
4. \(Al(s)+CrO_4^{2-}(aq) \rightarrow Al(OH)_3(s)+Cr(OH)^{4-}(aq)\)
   1. Balanced: \(8H_2O(l) + Al(s) + 2CrO_4^{2-}(aq) \rightarrow Al(OH)_3(s) + 2Cr(OH)_4^-(aq) +5OH^-(aq)\)

S17.1.8 (Solutions)

Oxidized = reducing agent:

1. \(SO_3^{2-}\)

2. \(Mn(OH)_2\)

3. \(H_2\)

4. \(Al\)

Reduced = oxidizing agent:

1. \(Cu(OH)_2\)

2. \(O_2\)

3. \(NO_3^{-}\)

4. \(CrO_4^{2-}\)

OIL RIG is a good acronym to use for remembering what oxidation and reduction is and it stands for, "Oxidation is Loss of electrons and Reduction is Gain of electrons. A reducing agent is oxidized, because it loses electrons in the redox reaction. An oxidizing agent, or oxidant, gains electrons and is reduced in a chemical reaction.

Correction: These balanced equations are correct. Some of these reactions are being balanced in acidic conditions and some in basic. In acidic you had \(H_2O\) and \(H^+\) while in basic you had \(H_2O\) and \(OH^-\).

Q19.1.7

Which of the following elements is most likely to form an oxide with the formula \(MO_3\): Zr, Nb, or Mo?

S19.1.7 (Solutions)

Mo because it is the only one to have 6^+ electrons to have Krypton configuration, which needs 6 electrons. Nb would have to take from the 3p orbital and lose its Krypton configuration so it is unfavored. Zr would need 4 electrons to reach Krypton configuration.

Correction : Mo is correct because of the electron configuration rules and the need to complete it.

Q19.2.9

Predict whether the carbonate ligand \(CO_3^{2-}\) will coordinate to a metal center as a monodentate, bidentate, or tridentate ligand.

S19.2.9 (Solutions)

\(CO_3^{2-}\) will coordinate to a metal center as a monodentate ligand.

It only bonds to the carbon not the 3 oxygens making it monodentate. Side note: The only bidentate ligands are (ox) and (en). EDTA is a hexadentate ligand.

Correction: This is correct because it is only bonded to the carbon not the 3 oxygen's making it monodentate, as stated in the answer

Q12.3.22

The following data have been determined for the reaction:

\[I^{-}+OCl^{-} \rightarrow IO^{-}+Cl^{-}\]

Determine the rate equation and the rate constant for this reaction.

Q12.3.22 (Solutions)

Steps:

1. Rate will have to equal = k\([I^{-}]^m\)\([OCl^{-}]^n\)

2. To derive the rate, it may easier to rearrange the data in this way:

3. Looking at the table of data, choose two of the three to compare. Let's choose 1 vs 2. Notice \(OCl^{-}\) has the same concentration for 1 and 2. That means you will be using the concentrations for 1 and 2 of \(I^{-}\) which are 0.10 and 0.20 respectively as well as the rates given for 1 and 2 which are \(3.05\cdot 10^{-4}\) and \(6.2\cdot 10^{-4}\) respectively.

With that information you do as follows:

\((\frac{.2}{.1})^m=(\frac{6.2\cdot 10^{-4}}{3.05\cdot 10^{-4}})\) which equals \(2^{m}\)=2.0327 (The purpose of setting up the information this way is to find the order of \(I^{-}\) )

To solve for the order (aka m) do the natural log on both sides. This gives you \(mln(2)=ln(2.0328)\) which in turn gives m a value of 1 when you solve for m.

With that, we have half of the rate... \(rate= k[I^{-1}]^1[OCl^{-}]^?\) Now, it is time to find the order of \(OCl^{-}\).

The only problem here is that there are no values for I that are equivalent which we need to be equivalent to calculate \(OCl^{-}\) and so I suspect there is a typo.

Correction: There was a typo being that there was no constant concentration for \(I^-\) to be able to determine the concentration for \(OCl^-\). So assuming 0.10 is constant from experiment 1 to experiment 3 for \(I^-\) is the only way to full solve this problem (following the same steps depicted above). Once you've found the power of each by determining the order of each reactant you find the rate equation. Once that is done use the equation which in this situation would be \(Rate=K[reactant]^{order}[reactant]^{order}\). Then to find the rate constant (k) plug in the values choosing either experiment 1,2, or 3 then solve from there.

Q12.7.2

Compare the functions of homogeneous and heterogeneous catalysts.

Q12.7.2 (Solutions)

One more thing that could be added as a side not is that homogenous catalyst is less practical then the heterogeneous if trying to get an expirement done in good time and less effort.

Correction : The main idea behind the Heterogeneous and Homogenous catalyst is that homogenous catalyst are of the same phase making the modification very easy and the reaction mechanism well understood. For heterogeneous catalyst the modification is very difficult and the reaction mechanism is poorly understood.

Q21.4.25

A \(_{5}^{8}\textrm{B}\) atom (mass = 8.0246 amu) decays into a \(_{4}^{8}\textrm{B}\) atom (mass = 8.0053 amu) by loss of a \(β^{+}\) particle (mass = 0.00055 amu) or by electron capture. How much energy (in millions of electron volts) is produced by this reaction?

S21.4.25 (Solutions)

First, set up the proper equation so we can identify the products and reactants.

\(_{5}^{8}\textrm{B}+ _{-1}^{0}\textrm{B}\rightarrow_{4}^{8}\textrm{B}\).

We are given the mass of everything within the reaction. The equation we must use is \(E=mc^2\).

We know c to be \(2.998\cdot 10^8 \frac{m}{s}\) and to find m, we use \(\Delta u= Product Mass-Reactant\)

Using the masses given, you get \(\Delta u=[8.0053 amu-(8.0246 amu+.00055 amu)]=-.0195 amu\).

1 amu= \(1.66 \cdot 10^{-27} kg\) so converting -.0195 amu to kg will give you \(-3.238\cdot 10^{-29}kg\).

Using the equation, \(E=\Delta mc^2\), input all values:

\(E=-3.238\cdot 10^{-29}kg(2.998\cdot 10^8 \frac{m}{s})^2 = -2.91\cdot 10^{-12}kg\cdot (\frac{m}{s})^2\)

To convert from Joules to eV= \(-2.91\cdot 10^{-12}J \cdot \frac{1eV}{1.6\cdot 10^{-19}}J = -18189758.46 eV\)

From eV to MeV = \(-18189758.46 eV \cdot \frac{.000001 MeV}{1eV}\) = -18.19 MeV

Correction: This answer is correct after using the equation \(E=mc^2\) and plugging in values into the equation \(\Delta U=Product Mass- Reactant\).

Q20.3.12

Write the spontaneous half-reactions and the overall reaction for each proposed cell diagram. State which half-reaction occurs at the anode and which occurs at the cathode.

1. \(Pb(s)∣PbSO_4(s)∣SO_4^{2-}(aq)∥Cu^{2+}(aq)∣Cu(s)\)
2. \(Hg(l)∣Hg_2Cl_2(s)∣Cl^{-}(aq) ∥ Cd^{2+}(aq)∣Cd(s)\)

Q20.3.12 (Solutions)

Remember that in cell notation the anode is always written on the left side and the cathode is always written on the right side.

1. \(Pb(s) + SO_4^{2-}(aq) \rightarrow PbSO_4(s) +2e^{-}\) This half-reaction occurs at the anode because \(SO_4^{2-}(aq)\) loses electrons.

\(Cu^{2+}(aq) + 2e^{-} \rightarrow Cu(s)\) This half-reaction occurs at the cathode because \(Cu^{2+}(aq)\) gains electrons.

Overall: \(Pb(s) + SO_4^{2-}(aq) + Cu^{2+}(aq) \rightarrow Cu(s) + PbSO_4(s)\)

2. \(2Hg(l) + 2Cl^{-}(aq) \rightarrow Hg_2Cl_2(s) + 2e^{-}\) This half-reaction occurs at the anode because \(Cl^{-}(aq)\) loses electrons.

\(Cd^{2+}(aq) + 2e^{-} \rightarrow Cd(s)\) This half-reaction occurs at the cathode because \(Cd^{2+}\) gains electrons.

Overall: \(2Hg(l) + 2Cl^{-}(aq) + Cd^{2+}(aq) \rightarrow Hg_2Cl_2(s) + Cd(s)\)

The anode is where oxidation occurs (where there is a loss of electrons in the reaction), and the cathode is where reduction occurs (where there is a gain of electrons in the reaction). A good trick to remember is remembering these two monikers: AnOx and RedCat.

Correction: These are all correct and answered correctly by doing the half reaction and determining anode(oxidation) and cathode(reduction) half of the reactions.

Q20.5.24

The biological molecule abbreviated as NADH (reduced nicotinamide adenine dinucleotide) can be formed by reduction of NAD+ (nicotinamide adenine dinucleotide) via the half-reaction NAD+ + H+ + 2e− → NADH; E° = −0.32 V.

1. Would NADH be able to reduce acetate to pyruvate?
2. Would NADH be able to reduce pyruvate to lactate?
3. What potential is needed to convert acetate to lactate?

\(acetate + CO_2 + 2H^{+} +2e^{-} \rightarrow pyruvate +H_2O\) E° = −0.70 V

\(pyruvate + 2H^{+} + 2e^{-} \rightarrow lactate\) E° = −0.185 V

Q20.5.24 (Solutions)

Use electrochemistry concept of galvanic cells, we may use the equation \(E_{cell}= E_{cathode}-E_{anode}\)
Reminder: The cathode is where reduction takes place, and the anode is where oxidation takes
place.

1. If you want to reduce acetate to pyruvate using NADH, you are
putting acetate as the cathode, and NADH as the anode, so the equation will look like this: \(E_{cell}= -0.7V-(-0.32V)=-0.28V\) Our end value is negative, making the reaction nonspontaneous, and thus, you cannot use NADH to reduce acetate to pyruvate.

2. Likewise, you do \(E_{cell}= -0.185V-(-0.32V)=-0.135V\) Our answer dictates that the reaction is spontaneous, and thus, NADH can reduce pyruvate to lactate.

3. For question 3, you will want to convert acetate to lactate. First, you must know the standard cell potential, which is \(-0.7 V + (-0.185V)=-0.885 V\) So, you need at least 0.885 V potential to convert it.

If you want to take into account NADH (which means using NADH to convert it), then subtract the cell potential value of NADH from what the previous answer: \(-0.885 - (-0.32)= -0.565V\)

You will then need at least .565V to convert acetate to lactate.

Correction: This question is solved using\(Ecell=Ecathode-Eanode\).