Extra Credit 1

Q17.1.1

If a 2.5 A current is run through a circuit for 35 minutes, how many coulombs of charge moved through the circuit?

S17.1.1

In this example, we are given that a current of 2.5 Amps (A) is run through a circuit for 35 minutes, and we are asked to find the charge in Coulombs (C). We know that Amps is the charge (in Couloumbs) per second, which gives us this equation: \[A =\frac{C }{s}\].

To solve for Coulombs, we can rewrite the equation:

\(C ={A}*{s}\)

When we plug in the values, we need to pay attention to units. Since we were given time in terms of minutes, we need to convert it to seconds. Subsequently, we get...

\(C={\frac {2.5 \ Coloumbs}{second}}*35 \ minutes*{\frac{60 \ seconds}{minute}}\)

Therefore...

\(C={5.25}*{10^3 \ Coulombs}\)

With correct significant figures, the answer should be 5.3 * 10^3 Coulombs of charge is moved through the circuit.

Q17.7.4

A current of 2.345 A passes through the cell shown in Figure for 45 minutes. What is the volume of the hydrogen collected at room temperature if the pressure is exactly 1 atm? Assume the voltage is sufficient to perform the reduction. (Hint: Is hydrogen the only gas present above the water?)

S17.7.4

The overall equation for this reaction is represented below:

\[2H_2O(l) \rightarrow 2H_2 (g)+ O_2 (g)\] where \({E^{\circ}}_{cell}=−1.229 V\)

The cathode or reduction half reaction:

\[4H^+ (aq) + 4e^- \rightarrow 2H_2( g) \] \({E^{\circ}}_{cathode}=0 V\)

The anode or oxidation half reaction:

\[2H_2O(l) \rightarrow O_2(g)+ 4H^+ (aq) + 4e^-\] \(E^{\circ}_{anode}=+1.229 V\)

As four electrons are present in both half reactions, we know there are four moles of electrons transferred.

Using the current in Amps, the time interval the current is applied and the moles of electrons we found, we can find the amount of \(H_2\) gas that formed.

Assuming standard conditions (1 atm as given and 298 K), we can use three equations to find the volume of hydrogen gas formed:

\[Coulombs ={Amps}\cdot{seconds}\]

\[96,485 \ Coulombs = 1 \ mole \ of \ electrons\]

\[PV=nRT\]

Where: P is the Pressure (in atm), V is the Volume (in Liters), n is the number of moles of \(H_2\) gas, R is the universal gas constant [in (L * atm) / (mol * K)] and T is the temperature in Kelvin (K).

The first equation will convert current (Amps) and time (minutes or seconds) into charge (Coulombs).

\[C ={A}*{s}\]

\[C= {2.345A}*{45 \ minutes}*{\frac{60 \ seconds}{1 \ minute}}\]

C= 6331.5 Coulombs

*Keep in mind to carry significant digits until the very end of the problem to get the most accurate results*

The second equation converts Coulombs to moles of electrons which can then be converted to moles of H₂ gas. Using the oxidation half reaction above, we find for every four moles of electrons, two moles of hydrogen gas are formed.

\[{6331.5 \ C}\cdot{\frac{1 \ mole \ e-}{96485 \ C}}\cdot{\frac{2 \ mol \ H_2}{4 \ mol \ e-}} = mol \ H_2 \]

= 0.03281 moles H₂

The final equation uses the given pressure, temperature and the moles of hydrogen gas we calculated to find the volume of hydrogen gas that was formed.

if we rearrange \(PV=nRT\) by solving for V, we get: \[V= \frac{nRT}{P}\]. Using this form we find:

\[V= \frac { {0.03281 \ mol \ H_2} \ * \ {0.08206 \ (L \ * \ atm ) / (mol \ * \ K)} \ * \ {298K} }{1 \ atm} \]

V= 0.80 L of H₂ gas is collected

Q19.2.1

Indicate the coordination number for the central metal atom in each of the following coordination compounds:

1. [Pt(H₂O)₂Br₂]
2. [Pt(NH₃)(py)(Cl)(Br)] (py = pyridine, C₅H₅N)
3. [Zn(NH₃)₂Cl₂]
4. [Zn(NH₃)(py)(Cl)(Br)]
5. [Ni(H₂O)₄Cl₂]
6. [Fe(en)₂(CN)₂]⁺ (en = ethylenediamine, C₂H₈N₂)

S19.2.1

The coordination number indicates the the number of covalent bonds one or more ligands have to a metal to form a coordination complex. Ligands can be monodentate (binds to the metal at one location) which form one coordinate covalent bond or polydentate (binds to the metal at more than one location) which form multiple coordinate covalent bonds. Bidentane ligands form two covalent bonds with the central metal atom, tridentane form three, tetradentane form four, pentadentance form five and hexadentane form six.

1.) The metal identified in the first complex is Platinum (Pt). It forms a covalent bond between to two aqua \((H_2O)\) ligands and two bromo \((Br^-)\) ligands. As both the aqua and bromo ligands are monodentane, this complex has a coordination number of four.

2.) The metal identified in the second complex is also Platinum. It forms a covalent bond between one amine ligand \((NH_3)\), one pyridine ligand (abbreviated as (py) ), one chloro ligand \((Cl^-)\), and one bromo ligand. All of these ligands are also monodentane, so the complex has a coordination number of four.

3.) The metal identified in the third complex is Zinc (Zn). It forms a covalent bond between two amine ligands, and two chloro ligands. All of the ligands are monodentane, thus the coordination number is four.

4.) The metal identified in the fourth complex is also Zinc. It forms a covalent bond between an amine ligand, a pyridine ligand, a Chloro ligand, and a Bromo ligand. All four of these ligands are monodentante, so the coordination number is four.

5.) The metal identified in the fifth complex is Ni. It is bonded to 4 aqua ligands and 2 chloro ligands. Therefore the metal forms 4+2= 6 coordinate covalent bonds with the ligand, so the coordination number is 6.

6.) The metal identified in the sixth complex is Fe. Iron forms a covalent bond with two ethylenediamine ligands (written as (en) ), and two cyano ligands \((CN^-)\). Ethylenediamine is bidentane, meaning each ligand will make two covalent bonds with the central metal atom. Since cyano ligands are monodentane, we find the coordination is six ((\(2(2)_{en} + 2(1)_{cyano} \)).

Q12.3.13

Nitrosyl chloride, NOCl, decomposes to NO and Cl₂.

\(2NOCl(g) \rightarrow 2NO (g) + Cl_2(g)\)

Determine the rate equation, the rate constant, and the overall order for this reaction from the following data:

S12.3.13

In this question, we are asked to find the values the rate constant, rate order, and rate equation. Let us start by defining each of these terms. The rate coefficient relates a chemical reaction to the concentration of the reactants. This is determined via experiments or experimental data. The rate equation an expression showing the relationship of the reaction rate to the concentrations of each reactant. The order of a reaction is found by summing the exponents of which the concentration of the reactants are raised to. The order of a reaction gives us information about the mixture. (Libretexts. "18.8: Rate Law and Specific Rate Constant." Chemistry LibreTexts. Libretexts, 26 July 2016. Web. 11 June 2017.)

We will first begin by determining the rate order. In order to find the rate constant, we need to compare rates at two different concentrations of the reactants. because we are assuming the reaction is a forward reaction. For convenience, we will use the rates of NOCl at .1 M and .2 M.

To find the rate order we can use the general form:

\[{\frac {[M_1]}{[M_2]}}^n=\frac{rate_1}{rate_2}\]

Where \(M_1\) and \(rate_1\) are the concentration and rate of one experiment and \(M_2\) and \(rate_2\) are the concentration and rate of another.

If we designate the experiment at .2 M as experiment 1 and the experiment at .1 M as experiment two we find \(\frac{0.2}{0.1}\) = 2 : this shows that the NOCl concentration is being doubled.

Now lets compare the ratio of the respective rates for the NOCl concentrations.

\(\frac{3.2 \cdot 10^{-9}}{8 \cdot 10^{-10}} = 4 \) : this shows that the rate is being quadrupled

From the calculations above, we conclude that when the NOCl concentration is doubled, the rate is quadrupled.

Therefore, in order to find the order of the reaction, we have to determine what it takes in order for the doubling of the concentration to quadruple the rate:

\(2^x=4\)

When solving for x, we find that x=2, meaning the NOCl is second order.

We can check this process using the other data. Let us assume Experiment 1 uses .3 M and Experiment 2 uses .2 M:

\(\frac{7.2 \cdot 10^{-9}}{3.2 \cdot 10^ {-9}}= 2.25 \)

\(\frac{0.3}{0.2}\) =1.5

We find that the rate order of NOCl remains two: \(1.5^2=2.25\)

Therefore the rate equation is rate= \(k \ [NOCl]^2\)

As there NOCl is the only reactant, the sum of the exponents of the reactants is two and, therefore, the order for this equation is a second order.

Using the rate law found above, we can use the rate at a given concentration of NOCl to find the rate constant, k.

\(8.0 \cdot 10^{-10} (\frac {M}{h}) = k \ [0.1 M]^2\)

solving for k, we get that k= 8.0 × 10⁻⁸ \(/M \cdot hour \)

In order to determine the units for k consider the units for the rate and the concentration:

\[\frac {M}{h} = M^2 \cdot x\] and thus: \[x = \frac{M}{M^2 \cdot h} = \frac {1}{M \cdot h}\]

Molarity is squared due to the rate order of NOCl being two.

Q12.6.5

What is the rate equation for the elementary termolecular reaction \(A+2B \rightarrow products \)? For \(3A \rightarrow products\)?

S12.6.5

We are given that both of these reactions are elementary termolecular. The molecularity of a reaction refers to the number of reactant particles that react together with the proper and energy and orientation.Termolecular reactions are often split into two reactions, called elementary steps, that often involve a bimolecular reaction and a unimolecular reaction, as seen in this case. The rate law for elementary reations is determined by the stochiometry of the reaction without needed experimental data.

The basic rate form for the elementary step is what follows:

\(rate= {k} \cdot {reactant \ 1}^{i} \cdot {reactant \ 2}^{ii} \cdot ... \) Where i and ii are the stochiometric coefficient from reactant 1 and 2 respectively.

For: \(3A \rightarrow products \)

\({k} \cdot {A}^3 = rate\)

For: \(A + 2B \rightarrow products \)

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

Note that the order of these reactions are both three.

Q21.4.17

If 1.000 g of \({^{226}_{88}}Ra\) produces 0.0001 mL of the gas \({^{222}_{86}}Rn\) at STP (standard temperature and pressure) in 24 h, what is the half-life of ²²⁶Ra in years?

S21.4.17

In this problem, we are asked to find the half life of ²²⁶Ra.

The equation used is \(ln(N_T) \ = \ ln(N_0) \ - \ kt\)

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

We can use the two equations above by substituting the second equation for k. After doing this, we should get:

\(ln(N_T) \ = \ ln(N_0) - (\frac{ln(2)}{t_{1/2}})t \) where N_T is the amount of isotope remaining at time t, N₀ is the initial amount of the isotope (thus amount at t=0), and k is the activity constant. In order to work with this equation, we will need to find the amount of moles of Ra and Rn.

\( N_0 = {\frac{1 mol \ Ra }{226 g \ Ra}} \ \cdot \ {1 g \ Ra}\) = 0.00442478 moles Ra

* remember at STP, we are working at 1 atm, 298 K, and 22.4 L/mol

\( N_1 = {0.0001 \ mL \ Rn} \cdot \frac{1 \ L}{1000 \ mL} \cdot {\frac{1 \ mole}{22.4 \ L}} \cdot \frac {1 \ mol \ Ra}{1 \ mol \ Rn} = \ {4.464} \cdot \ {10^{-9}} \ mol \ Ra\)

We can simply subtract the moles lost from decay into Rn (N1) from the initial moles of Ra (N₀), to find N_T.

\( N_T = 0.00442478 \ - \ (4.464 \ \cdot \ 10^{-9}) \)= 0.004424774

Next, we will be utilizing the following equation: Now we can use \(ln(N_T) \ = \ ln(N_0) \ - \ kt\) to find k, where N_t is our Ra remaining (.004424774 mol) and N₀ is the initial Ra before decay (.00442478 mol ) and our t is 24 hours.

ln(.004424774)-ln(.00442478) = -k*\({24 \ hours}*{\frac{1 \ day}{24 \ hours}}*{\frac{1 \ year}{365 \ days}}\)

after isolating k, we get that k = 3.7 x 10^-4 /year

Next, we will be utilizing the following equation: k= -ln(2)/t_1/2

We get that t_1/2 is equal to 2.00 x 10³ years

Q20.3.5

One criterion for a good salt bridge is that it contains ions that have similar rates of diffusion in aqueous solution, as K⁺ and Cl⁻ ions do. What would happen if the diffusion rates of the anions and cations differed significantly?

S20.3.5

The purpose of a salt bridge in an electrochemical cell is to allow electrons to move between cell compartments to complete the electrical circuit. This way the oxidation and reduction reactions at the anode and cathode are connected and maintain the charge neutrality that prevents the cell from reaching equilibrium. Without the maintenance of this flow, the respective sides of the cell would get polarized with a positive or negative charge, which defeats the purpose of an electrochemical cell because electricity will not be able to be produced/flow through the cell. Likewise, if the diffusion rates of the ions participating in the salt bride/salt neutrality maintenance occurred at different rates, one side of the cell which is receiving the charged ions at a faster rate than the other half of the cell would become polarized with charge, thus preventing an overall neutral charge. This will slowly prevent electricity from being able to flow through the cell and will in turn, "kill" the cell.

Q20.5.16

For each reaction, calculate E°_cell and then determine ΔG°. Indicate whether each reaction is spontaneous.

1. 2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g)
2. K₂S₂O₆(aq) + I₂(s) → 2KI(aq) + 2K₂SO₄(aq)
3. Sn(s) + CuSO₄(aq) → Cu(s) + SnSO₄(aq)

S20.5.16

The \(E^ \circ_{cell}\) can be determine the spontaneity of a oxidation reduction reaction at standard conditions. If \(E^ \circ_{cell}\) is greater than zero, the reaction is spontaneous, while if it is negative, the reaction is non-spontaneous. These redox reactions occur at electrochemical cells: with the reduction reaction occurring at the cathode and the oxidation reaction occurring at the anode. The table of standard reduction potentials is essential for solving such problems because it gives us the information about the cathodic and anodic reduction potentials that are needed to find the overall potential for the electrochemical cell.

*In this question we are asked to find the overall reaction potential for the cell using the reaction potential values for the cathode and anode using the standard reduction potential table. First let us determine the reduction and oxidation half reactions and their respective potential values. Then we will use the equation below to find the overall equation potential.

Consider this equation: \({E}^°_\text{cell} = \text{E}^°_\text{cathode} - \text{E}^°_\text{anode}\) Note \(E^ \circ \) is the standard reduction potential of a reaction, and is found using the reduction reaction, regardless of whether it is a cathode or anode. The anode reaction is negated because the \(E^\circ\) is reversed to find the \(E^\circ_{cell}\) of the oxidation reaction.

First Set of Reactions:

anode for reaction 1: \(H_2(g) \rightarrow 2H^+(aq)+ 2e^-\) \(\text{E}^°_\text{cathode}\) = 0 V

cathode for reaction 1: \(Na^+(aq) + e^- \rightarrow Na(s)\) \(\text{E}^°_\text{anode}\) = -2.71 V

\({E}^°{cell} = {-2.71 \ V}-{0 \ V}\)

= 2.71 V

We can use the following equation to find the Gibbs Free Energy : \[ΔG^{\circ} = -nFE^°_{cell}\]

Where n is the number of moles of electrons transferred and F is Faraday's Constant. As two moles of electrons are used in the balanced equation we use 2 for "n".

\({ΔG} = -(2 \ mole \ e^-) \cdot (96.485 \ \frac{kJ}{mol \ \cdot \ V}) \cdot (2.71 \ V)\)

\({ΔG}\) = -522.9 kJ

* Since the delta G is negative, the reaction is spontaneous

Second Set of Reactions: Note: \(K_2S_2O_6\) can be decomposed into \(2K^+\), and \(2SO_3^{2-}\) .

anode for reaction 2: \(2SO_4^{2-}(aq) + 2H_2O + 4e^- \rightarrow 2SO_3^{2-}(aq)+ 4OH^- \) \(\text{E}^°_\text{anode}\) = -.93 V

cathode for reaction 2: 2I₂(s)+ 4e^- → 4I^-(aq) \(\text{E}^°_\text{cathode}\) = 0.535 V (note: \(E^{/circ}_{cell}\) is not dependent on stochiometry

\(E^°_{cell} = \ 0.535 \ V \ - (-.93) \ V) \)

= 1.465 V

\({ΔG} = {nF}\text{E}^°_\text{cell}\)

\({ΔG} = -(4 \ mole \ e^-) \cdot (96.485 \ \frac{kJ}{(mol \cdot \ V)}) \cdot (1.465 \ V) \)

\({ΔG}\) = -565.4 kJ

* Since the delta G is negative, the reaction is spontaneous

Third Set of Reactions:

anode for reaction 3: Sn²⁺(aq) + 2e^- → Sn(s) \(\text{E}^°_\text{anode}\) = -0.14 V

cathode for reaction 3: Cu²⁺(aq) + 2e^- → Cu(s) \(\text{E}^°_\text{cathode}\) = 0.337 V

\({E}^°_{cell} = {0.337 \ V}-{-0.14 \ V}\)

\({E}^°_{cell} \ = \ 0.477 \ V \)

= 0.477 V

\({ΔG}\) = \(nF{E}^°_{cell}\)

\({ΔG} = -(2 \ mole \ e^-) \cdot (96.485 \ \frac{kJ}{mol \ \cdot \ V}) \cdot (0.477 \ V) \)

\({ΔG}\) = -93kJ

* Since the delta G is negative, the reaction is spontaneous