Extra Credit 5

Solution 17.1.5

Here are the steps to balance any redox reaction given their half reactions in acidic solution.

a)Balance elements other than Oxygen or Hydrogen

b)For each unbalanced Oxygen add one \(H_{2}O\) to the other side

c)For each unbalanced Hydrogen add a proton (\(H^{+}\)) to the other side

d)Add all the charges on each side of the half reaction

e)Add the necessary number of electrons to the most positive side to make sure the charges equal each other

d)Before combining both half reactions, make sure the electrons cancel each other. If not, multiply one half reaction(or both) by the necessary coefficient.

e)Add both half reactions and cancel out similar terms on both sides (electrons, water, protons).

1. OHR: \({Ca} \rightarrow {Ca^{2+}}+{2e}^{−}\) , RHR: \({F}_{2}+{2e}^{−} \rightarrow {2F}^{−}\)

a) Elements are already balanced:

One Calcium for each side the oxidation half reaction

Two Fluorines for each side of the reduction half reaction.

b)c) No other elements present so we can skip step b and c

d)e) Charges are already balanced

OHR: \(0 \rightarrow {2^{+}} + {2^{-}} \)

RHR: \(0 + {2^{-}} \rightarrow 2(1^{-})\)

f) There are two electrons per reaction, so we are able to combine each half reaction without further calculations

e) Cancel out electrons to obtain overall reaction

\[{Ca} + {F}_{2} \rightarrow {Ca^{2+}} + {2F}^{−}\]

2. OHR: \({Li} \rightarrow {Li^+}+{e^−}\), RHR: \({Cl_2}+{2e^−} \rightarrow {2Cl^−}\)

a) Elements are already balanced:

One Lithium for each side the oxidation half reaction

Two Chlorines for each side of the reduction half reaction.

b)c) No other elements present so we can skip step b and c

d)e) Charges are already balanced

OHR: \(0 \rightarrow {1^{+}} + {1^{-}} \)

RHR: \(0 + {2^{-}} \rightarrow 2(1^{-})\)

f) Because there are two electrons on the RHR and only one electron on the OHR, it's necessary to multiply the OHR by a factor of two in order to cancel out the electrons.

\(2* ({Li} \rightarrow {Li^+}+{e^−}) \)

\({2Li} \rightarrow {2Li^+}+{2e^−}\)

e) Combine half reactions. Cancel out electrons to obtain overall reaction

\[{2Li} + {Cl}_{2} \rightarrow {2Li^{+}} + {2Cl}^{−}\]

3. OHR: \({Fe} \rightarrow {Fe^{3+}}+{3e^−}\), RHR: \({Br_2}+{2e^−} \rightarrow {2Br^−}\)

a) Elements are already balanced:

One Iron for each side the oxidation half reaction

Two Bromines for each side of the reduction half reaction.

b)c) No other elements present so we can skip step b and c

d)e) Charges are already balanced

OHR: \(0 \rightarrow {3^{+}} + {3^{-}} \)

RHR: \(0 + {2^{-}} \rightarrow 2(1^{-})\)

f) Because there are three electrons on the OHR and only two electron on the RHR, it's necessary to multiply the OHR by a factor of two and multiply the RHR by a factor of 3. This way we'll have 6 electrons on opposite sides of each reaction and we'll be able cancel out the elctrons.

\(2* ({Fe} \rightarrow {Fe^{3+}}+{3e^−}) \)

\({2Fe} \rightarrow {2Fe^{3+}}+{6e^−}\)

\(3* ({Br_2}+{2e^−} \rightarrow {2Br^−})\)

\({3Br_2}+{6e^−} \rightarrow {6Br^−}\)

e) Combine half reactions. Cancel out electrons to obtain overall reaction

\[{2Fe} + {3Br}_{2} \rightarrow {2Fe^{3+}} + {6Br}^{−}\]

4. OHR: \({Ag} \rightarrow {Ag^+}+{e^−}\) , RHR: \({MnO^{−}_{4}}+{4H^+}+{3e^−} \rightarrow {MnO_2}+{2H_2O}\)

a) Elements are already balanced:

One Silver for each side the oxidation half reaction

One Manganese for each side of the reduction half reaction.

b) While there is nothing left to do for the OHR, the RHR does require more balancing.

Since there are four oxygens on the left side and two oxygens on the right side, two \(H_{2}O\) were added to the right side.

c) Because of the addition of 2 water molecules, there are 4 unbalanced Hydrogens. 4 protons were then added to the left side.

d) Charges are already balanced for the OHR

OHR: \(0 \rightarrow {1^{+}} + {1^{-}} \)

For the RHR however, there's a charge of 3+ on the left side and 0 on the right side

RHR: \(1^{-} + {4^{+}} \rightarrow 0)\)

e) So there needs to be 3 electrons added to the left side (most positive) in order to equal out the charge

RHR: \(1^{-} + {4^{+}} + 3^{-} \rightarrow 0)\)

f) Because there are three electrons on the RHR and only one electron on the OHR, it's necessary to multiply the OHR by a factor of three in order to cancel out the electrons.

\(3* ({Ag} \rightarrow {Ag^+}+{e^−}) \)

\({3Ag} \rightarrow {3Ag^+}+{3e^−}\)

e) Combine half reactions. Cancel out electrons to obtain overall reaction

\[{MnO^{−}_{4}}+{4H^+}+{3Ag} \rightarrow {3Ag^+}+{MnO_2}+{2H_2O}\]

*edit: the solution is correct, greatly explained.

Solution 19.1.3

In order to write the electron configuration, we begin by finding the element on the periodic table. Since La, Sm, and Lu are all a period below the noble gas Xenon, we can abbreviate \({1s^2}{2s^2}{2p^6}{3s^2}{3p^6}{3d^{10}}{4s^2}{4p^6}{4d^{10}}{5s^2}{5p^6}\) as [Xe] when writing the orbital configurations. We then find the remaining of the orbital configurations using the Aufbau Principle. For other elements not just those in period 6, the shorthand notation using noble gases would be the noble gas in the period above the given element.

1. La has three additional electrons. Two of them fill the 6s shell and the other single electron is placed on the 5d shell.

\(La:\) [Xe] \({6{s}^2} {5{d}^1}\)

2. Sm has eight more electrons. The 6s orbital is filled as previously and the 4f orbital receives 6 electrons because pairing electrons requires lower energy on the 4f shell than on the 5d shell.

\(Sm:\) [Xe] \({6{s}^2} {4{f}^6}\)

3. Lu has seventeen more electrons. Two electrons fill the 6s orbital, 14 electrons fill the 4f orbital, and extra single electron goes to the 5d orbital .

\(Lu:\) [Xe] \({6s^2}{4f^{14}}{5d^1}\)

To find the 3+ ion electron configuration, we remove 3 electrons from the neutral configuration, starting with the 6s orbital.

1. The ionization of La³⁺ removes the three extra electrons. So it reverts back to the stable Xenon configuration.

\({La^{3+}:}\) [Xe]

2. The ionization of Sm³⁺ removes two electrons from the 6s shell and one from the outermost (4f) shell

\({Sm^{3+}}:\) [Xe] \({4f^5}\)

3. The ionization of Lu³⁺ removes its two 6s shell and one from the outermost (5d) shell, leaving only a full 4f shell

\(Lu^{3+}:\) [Xe] \(4f^{14}\)

Solution 19.2.5

1. en is bidentate. This isomer has a coordination of 6.

2. Notice this is an octahedral complex, so it has a coordination of 6. En is bidentate. It can have both cis and trans geometric isomers.

3. Notice this is an octahedral complex, so it has a coordination of 6.

cis-NH3 mer- Cl

4. En is bidentate, so it's an octahedral complex with a coordination of 6.

5. Notice that this a square planar, so it has a coordination of 6. There could be both cis and trans (geometrical) isomers.

Left: trans Right: cis

Q12.3.18

For the reaction A⟶B+CA⟶B+C, the following data were obtained at 30 °C:

1. What is the order of the reaction with respect to [A], and what is the rate equation?
2. What is the rate constant?

Solution 12.3.18

1. The rate equation for an \(n\) order reaction is given as \(\frac{dr}{dt}={k}{[A]^n}\). Where \([A]\) is the concentration in M, and \(\frac{dr}{dt}\) is the rate in M/s.

We can then use each set of data points, plug its values into the rate equation and solve for \(n\). Note you can use any of the data points as long as the concentration corresponds to its rate.

Rate equation 1: \(4.17 \times {10}^{-4}={k}{[0.230]^n}\)

Rate equation 2: \(9.99 \times {10}^{-4}={k}{[0.356]^n}\)

We divide Rate equation 1 by Rate equation 2 in order to cancel out k, the rate constant.

\({\frac{4.17 \times {10}^{-4}}{9.99 \times {10}^{-4}}} = {\frac{k[0.230]^n}{k[0.356]^n}} \)

\({0.417}={0.646^n}\)

Now the only unknown we have is \(n\). Using logarithm rules one can solve for it.

\(ln{\: 0.417}={n \cdot ln{\: 0.646}}\)

\(\frac{ln{\: 0.417}}{ln{\:0.646}}=n=2\)

The rate equation is second order with respect to A and is written as \(\frac{dr}{dt}={k}{[A]^2}\).

2. We can solve for \(k\) by plugging in any data point into our rate equation \(\frac{dr}{dt}={k}{[A]^2}\).

Using the first data points for instance \( [A]=0.230 \frac{mol}{L}\) and \( \frac{dr}{dt} = 4.17 \times {10}^{-4} \frac{mol}{L \cdot s}\)] we get the equation \(4.17 \times {10}^{-4} \frac{mol}{L \cdot s}={k}{[0.230 \frac{mol}{L}]^2}\)

Which solves for \(k=7.88 \times {10}^{-3} \frac{L}{mol \cdot s}\)

Since we know this is a second order reaction the appropriate units for \(k\) can also be written as \( \frac{1}{M \cdot s}\)

Solution 12.6.9

Chain reactions are mechanisms were the elementary steps supply the reactants necessary to carry on the chain reaction until all reactants are exhausted. By definition, a termination step is one which consumes all radicals. Radicals are molecules that contain at least one unpaired electron.That is a step which started off with free radicals and produces a stable molecule.

As we can see step 1, 3, and 4 do not have radicals as their final product. So they are determined to be termination steps.

Step 2 on the other hand can be categorized as a chain propagation step because the number of free radicals consumed is equal to the number of free radicals generated. In other words this is the only step that doesn't terminate the chain reaction.

Solution 21.4.21

1. For this first part of the problem, we assume that the rock sample only contained Rubidium originally.

Through beta decay, the rubidium in the rock eventually decays to strontium.

\[{^{87}_{37}Rb} \rightarrow {^0_{-1}\beta}+{^{87}_{38}Sr}\]

Note that the loss of a \( \beta ^{-}\) particle doesn't affect the number of nucleos, so mass is conserved. Beta decay only affects the atomic number. We can use this to calculate the original amount of rubidium in the rock by simply summing the two current weights of rubidium-87 and strontium-87

\({8.23mg}+{0.47mg} = {8.70mg}\)

Since we know that all nuclear reactions are first order reactions we can use kinematics to calculate the decay constant \(k\).

\( {t_{1/2}} = \frac{ln{(2)}}{k} \)

We know the half-life (\({t_{1/2}}\)) of the decay of rubidium by \( \beta \) emission is 4.7 × 10¹⁰ years, so we have

\( {4.7 \times 10^{10}} = \frac{ln{(2)}}{k} \)

Solving for \(k\) gives,

\( {k} = 1.47 \times 10^{-11} \)

We can use the decay constant \( k \) to calculate how old the rock is using the radioactive decay equation:

\(N=N_o{e}^{- {k} {t}} \)

\(\ln{\frac{N}{N_o}}=-kt \)

We substitute our known values and solve for t

\(\ln{\frac{8.23}{8.70}}=-1.47 \times 10^{-11}t \)

\(t=3.77 \times 10^9 years\)

The rock sample is about 3.8 billion years old.

2. If the rock originally had some \(^{87}_{38}Sr\) present then that would mean less than 0.47 mg actually came from the decay of rubidium. This would result in a lower original mass (\(N_o\)) which in turn would increase the magnitute of \(\ln{\frac{N}{N_o}}\), but it would still be a negative number because the fraction \({\frac{N}{N_o}}\)<1, so it would actually decrease the left side of the equation. Dividing that by the same decay constant \(k\) we obtain a smaller \(t\). So the rock would be younger than the age calculated in part 1.

Solution 20.3.9

Since this reaction occurs spontaneously, this means the reaction occurs without any need of work. \(E_{cell}\) would be positive.

1. From the main reaction, we see that lead is oxidized (it loses electrons as it goes from an oxidation state of 0 to 2+), and Vanadium is reduced (it gains electrons as it goes from an oxidation state of 5+ to 3+).

OHR: \({Pb(s)} \rightarrow {Pb^{2+}(aq)}\)

RHR: \({VO^{2+}(aq)} \rightarrow {V^{3+}(aq)}\)

To balance the reactions, we first add \(H_2O\) for each unbalanced oxygen, then add \(H^+\) to balance the hydrogens. Finally we add electrons to balance the charges.

OHR: \({Pb(s)} \rightarrow {Pb^{2+}(aq)}+{2e^-}\)

RHR: \({VO^{2+}(aq)} \rightarrow {V^{3+}(aq)}+{H_2O(l)}\)

\({VO^{2+}(aq)}+{2H^+} \rightarrow {V^{3+}(aq)}+{H_2O(l)}\)

\({VO^{2+}(aq)}+{2H^+}+{e^-} \rightarrow {V^{3+}(aq)}+{H_2O(l)}\)

2. The cathode is defined as the electrode where reduction takes place. This means the reduction reaction involving Vanadium occurs at the cathode. Likewise, the anode is defined as the electrode where oxidation takes place. So, the oxidation reaction involving Lead occurs at the anode.

3. The cathode is negatively charged because electrons flow towards this electrode. The anode is positively charged because electrons flow away from this electrode. (Note that this is only true for galvanic cells and not for electrochemical cells)

Solution 20.5.20

The question asks about the pH needed this means our reaction will not be under standard conditions. We must use the Nernst equation to find the concentration of hydronium (H⁺) ions required for the reaction to be spontaneous. For a reaction to be spontaneous the \(E_{cell} > 0\).

\[E_{cell}={E^o_{cell}} − \frac{RT}{nF} lnQ \]

Since the reaction occurs at 298K, we can use the simplified version of the Nernst equation where \(R\), \(T\) and \(F\) are simplified to \(0.0592 V \) and \(\ln{Q}\) becomes \(\log{Q}\).

\[E_{cell}={E^o_{cell}} − \frac{0.0592 V}{n} log{Q} \]

A) In order to solve the Nerst equation we must first calculate the standard electric potential of the cell.

We are given the reduction half reaction:

\(Ni^{2+}(aq)+{2e^-} \rightarrow {Ni(s)} \) \(E^o=−0.257 V\)

The question states that hydrogen gas is the reductant or reducing agent. This means our second half reaction is the oxidation reaction between hydrogen gas and hydronium ions.

\({H_2(g)} \rightarrow {2H^+(aq)}+{2e^-} \) \(E^o= 0.00 V\)

We then find our overall standard cell potential by subtracting the standard cell potential of the anode (where oxidation happens) from the standard cell potential of the cathode (where reduction happens):

\(E^o_{cell}=E^o_{cathode}−E^o_{anode} \)

\(E^o_{cell}=−0.257 V − 0.00 V= −0.257 V\)

*note that \(E^o_{cell}\) is negative, so it's nonspontaneous.

B) Next step is to find \(n\) which is the total number of moles of electrons transferred. In this case \(n=2\) because two electrons are used in the cathode and likewise two electrons are produced at the anode.

C) Finally we must write the equation for \(Q\) using the law of mass action

\(Q=\frac{[H^+]^2}{(1 atm) [1 M]} \)

\(Q={[H^+]^2}\)

The Nernst equation becomes:

\(0<{−0.257} − \frac{0.0592}{n} \log{Q} \)

\(0<{−0.257} − \frac{0.0592}{2} \log{[H^+]^2} \)

\(0<{−0.257} − 0.0296 \log{[H^+]^2}\)

We can now solve for the concentration of hydronium ions:

\(\frac{−0.257}{0.0296} > \log{[H^+]^2}\)

\( {[H^+]} < 4.56 \times 10^{-5} \)

We can relate the concentration of hydronium ions to pH in order to find its value.

\(pH=−\log{[H^+]} \)

\(pH=−\log{[H^+]} > −\log{4.56 \times 10^{-5}} \)

\(pH > 4.34\)

A pH greater than 4.34 is needed for the reaction to take place at 25°C.