Extra Credit 17

Question 19.5 B Solution:

A voltaic cell represented by the following cell diagram has Ecell=−0.119V

Ag(s)| Ag⁺(aq)(0.075 M) || Hg²⁺(aq)(x M)|Hg(l)

The first step in solving for the molarity of mercury in solution is to find the two half reactions represented in the voltaic cell, and then to balance them and add them to represent the overall reaction. The oxidized species should lose electrons going from reactants to products, and in this case, is represented on the left of the voltaic cell using solid and aqueous silver. Oxidation occurs at the anode, which is indicated on the left side of the double lines, indicating a salt bridge. The reduced species should gain electrons when going from reactants to products, and is represented to the right of the two bars using aqueous mercury to liquid mercury. Liquid mercury represents the inert cathode of the voltaic cell and corresponds to the reduction reaction.

Oxidation Reaction(Anode): Ag(s) → Ag⁺(aq)+ e^-

Reduction Reaction(Cathode): Hg²⁺(aq) +2e^- → Hg(s)

After the two reactions have been isolated, balance the two reactions so that the number of electrons are equal. Once the reactions are balanced, add them together, making sure to cancel all species present on both the reactant and product sides, in this case, the electrons.

2*[(Ag(s) → Ag⁺(aq) + e^- )] = 2Ag(s) → 2Ag⁺(aq) + 2e^-

[Hg⁺(aq) +2e^- → Hg(s)]

Overall Reaction: 2Ag(s)+Hg⁺(aq) → Hg(s)+2Ag⁺(aq)

Once you have found the overall reaction, you must find the standard cell potential E° for the battery using the given cell potential standards on the table of cell potentials. Use the values in the formula E°_cell=E°_cathode-E°_anode in order to find the standard reduction potential. (When plugging in cell potentials into the equation, use only reduction potentials, so you would need to convert the oxidation reaction at the anode to a reduction reaction in order to find the corresponding cell potential. The resulting reduction reaction at the anode is Ag⁺(aq) + e^- → Ag(s) because it gains electrons.

E°_cell=E°_cathode-E°_anode

=0.845V-0.800V

=0.045V

Then use the Nerst equation (E_cell = E°_cell -(0.0592V/n)*logQ in order to solve for the concentration of the mercury in solution by substituting all of the givens into the equation. Do not include Ag(s) and Hg(l) in solving for Q, because solids and liquids have no effect on equilibrium constants.

E_cell = E°_cell - (0.0592V/n)*log[Ag]²/[Hg], where [Hg] is unknown

Where n = moles of electrons transferred in the redox reaction, in this case n = 2 mol e^-;

Given E_cell = -0.119 V; [Ag⁺] = 0.075 M;

Calculated E°_cell = 0.045 V

-0.119V = 0.045V - (0.0592V/2mol electrons)*log[0.075M]²/[x]

Solve for x.

-0.164 V=-0.0296V/mol e^-*log0.005626/[x]

5.5405=log0.005626/[x]

x=1.6202E-8

Therefore there is a concentration of 1.6202E-8 M of aqueous mercury initially in the cell.

Question 19.45A Solution:

Solve for Ecell of the following voltaic cell

Cu(s)| Cu²⁺(aq)(0.01 M) || Cu²⁺ (aq)(0.1 M)|Cu(s)

The first step in solving for the molarity of mercury in solution is to find the two half reactions represented in the voltaic cell, and then to balance them and add them to represent the overall reaction. The reduced (oxidation reaction) species should lose electrons going from reactants to products, and in this case, is represented on the left of the voltaic cell. The oxidized (reduction reaction) species should gain electrons when going from reactants to products, and is represented to the right of the two bars. Then find the standard cell potentials for each half reaction. Because these are the same species with the reactions flipped to create the oxidation and reduction sites, the two values should be the same, one positive and the other negative.

Oxidation Reaction(Anode): Cu(s) → Cu²⁺(aq)+ 2e^- E°_cell =-0.340 V

Reduction Reaction(Cathode): Cu²⁺(aq) +2e^- → Cu(s) E°_cell =0.340 V

Once you have found the standard potentials of the two half reactions, you must find the standard cell potential E° for the overall reaction using the given cell potential standards on the table of cell potentials. Use the values in the formula E°_cell=E°_cathode-E°_anode in order to find the standard cell potential.

E°_cell=E°_cathode-E°_anode

=0.340 V+0.340 V

=0.680 V

Then use the nerst equation in order to solve for the overall cell potential for the battery by subsituting all of the givens into the equation.

E_cell = E°_cell - (0.0592V/n)*log[oxidation]/[reduction]

E_cell= 0.680 V - (0.0592V/2 mol electrons)*log[0.01 M]/[0.1 M]

Solve for the cell potential.

E_cell=0.7096 V

Question 19.45B Solution
(Solution for assigned question done by Christine Bonilla)

To calculate the E_cell of any voltaic cell, use the equation: E_cell = E^o_cell - (0.0592/n)*logQ

Where E^o_cell of a voltaic cell is always zero, n = moles of electrons transferred in the redox reaction, and Q = equilibrium constant under the given conditions

Q = [Na⁺(saturated solution)/[Na⁺]

We are given [Na⁺] = 0.130, and [Na⁺] is unknown, so calculate [Na⁺] using the value of K_sp = 1.3E-12

And K_sp = [Na⁺]²[CrO₄^2-] = (2s)²(s)

4s³ = 1.3E-12, so s = 6.9E-5

[Na⁺(saturated)] at the anode = (2s)² = 1.90E-8

Use n=1 in the equation, because 1 mole of electrons are transferred in the redox reaction and [Na⁺] = 0.130 is given

Now you have everything you need to solve for E_cell = E^o_cell - (0.0592/n)*logQ

E_cell = 0 V - (0.0592/1)*log(1.90E-8/.130)

E_cell = 0.4046 V

Question 21.3B Solution:

Sketch the geometric isomers of [CuF₂(NH3)₂(NO)₂]^-.

A geometric isomer is where there are trans and cis isomers of a single complex. A cis isomer is where the like ligands are bonded in positions that are adjacent to one and other. A trans isomer is where the like ligands are bonded in locations that are opposite to each other, resulting in that no two like ligands are positioned next to each other.

When drawing the isomers, it is first important to determine the likely structure of the complex. In this case, it will be an octahedral complex because there are six ligands to occupy the six octahedral bonding sites--the coordination number is 6.

Although octahedral complexes can have mer and fac isomers, this particular octahedral complex does not. It cannot have mer nor fac isomers because they must include three of the same ligands bonded to the metal, and there are only two of the same ligands in this example.

I don't know how to create a picture on here, so here is the best that I can do.

Question 24.5A Solution:

In the reaction A→products, 4.50 minutes after the reaction is started, [A]=0.587 M. The rate of reaction at this point is

rate=−Δ[A]/Δt = 2.1×10^-2M*min^-1

Assume that this rate remains constant for a short period of time.

• What is [A][A] 6.00 minutes after the reaction is started?

In order to find the concentration of A after six minutes of the reaction continuing, just substitute the given values into the given rate law as defined by reactant A.

rate = -Δ[A]/(6 minutes-4.5 minutes) =2.1 x 10^-2 M*min^-1

Δ[A]= -0.0315 M

Once you have found the change in molarity of A, add it to the original amount to find the concentration of the reactant after 6 minutes.

0.587 M - 0.0315 M =0.5555 M

0.5555 M is the new concentration of A after 6 minutes of reaction.

• At what time after the reaction is started with [A]=0.56M

In order to find the time at which the reaction started, you must find the change in concentration of A.

0.587 M - 0.560 M = 0.027M = change in concentration.

Then substitute this value into the rate law to solve for the change in time. Then solve for the x value of time.

rate = -0.027 M/(4.5 minutes-x minutes) =2.1 x 10^-2 M*min^-1

4.5min - x min= -1.286

x = 5.786 minutes.

Therefore, when the concentration is 0.56 M, the reaction has occurred for 5.786 minutes.

Question 24.7A Solution

(Solution for assigned question done by Christine Bonilla)

For the reaction A+2B→2CA+2B→2C, the rate of reaction is 1.75 x 10-5 M s-1 at the time when [A]=0.3575M[A]=0.3575M.

a.What is the rate of formation of C?

b.What will [A] be 1 minute later?

c. Assume the rate remains at 1.75 x 10-5M s-1. How long would it take for [A] to change from 0.3580 to 0.3500M?

Rates of formations and disappearance are related, where: rate = -(1/a)*(Δ[A]/Δt) = -(1/b)*(Δ[B]/Δt) = (1/c)*(Δ[C]/Δt] where a, b, and c, are the coefficients for the balanced chemical equation. The rates of the individual products and reacts are equal to each other in an elementary reaction because the rate of formation of the product must take place at the same rate reactants are reacting. Products cannot form without the reactants.

Given this information:

a. Calculate the rate of formation of C. Use the equation: rate = (1/c)*(Δ[C]/Δt), where Δ[C]/Δt represents the rate of formation of t in M/s

1.75E-5 M/s = (1/2)*(Δ[C]/Δt)

Δ[C]/Δt = 3.5E-5 M/s

b. [A] after 1 minute can be represented by this equation: rate = -(Δ[A]/Δt)

First, 1 minute must be converted to seconds, because the rate is in M/second:

1 minute*(60 seconds/1 minute) = 60 seconds

And Δ[A] = [A]_final - [A]_initial, and the initial concentration of A is 0.3575 M

Now plug these values in the above equation:

1.75E-5 M/s = -([A]_final - 0.3537M)/(60 seconds - 0 seconds)

[A]_final = 0.353 M

c. Use the same equation as above to calculate the time it takes for [A] to change from 0.3580 M to 0.3500 M

1.75E-5 M/s = -(0.3500 M - 0.3580 M)/Δt

Assuming that t_initial = 0 seconds, Δt = 457 seconds, and therefore t_final = 457 seconds

Convert to minutes: (457 seconds)*(1 minute/60 seconds) = 7.62 minutes

Question 24.48A Solution:

By an appropriate draft, indicate why there is some relationship between the enthalpy change of the activation energy for an endothermic reaction but not for the exothermic reaction.

There is a relationship between enthalpy change and activation energy because the enthalpy change in an endothermic reaction is part of the activation energy. The activation energy is the total amount of energy required to make the reaction happen, and the enthalpy change is the amount of energy from the activation of the reaction that is stored after the products are formed. As shown on the graph, the enthalpy change is a part of the activation energy for an endothermic reaction.

However, an exothermic reaction shows that activation energy and enthalpy change are completely independent of one and other. The enthalpy change is the total amount of energy released form the reaction (energy of products minus the energy of reactants), so it is not factored into the activation energy of the reaction. No matter how the activation energy is altered due to catalysts, the enthalpy change will remain independent, since the transition from reactants to products loses energy.

Question 25.25D Solution:

Suppose the activity of a sample containing ³⁵S has an activity 500 times the detectable limit. How long before the radioactivity can no longer be detected if the experiment is run continuously?

To determine time with exponential decay, use the equation:

ln(N_t/N_o) = -λt This equation is similar to first order chemical reactions, because nuclear decay is also first order.

Where N_t = number of atoms for activity at time t, N_o = initial number of atoms for activity, λ = decay constant, and t_1/2 = half life of sulfur-35

To first solve this question, you must find out the half life of sulfur-35. If you check on the table of half lives, it is noted that it has a half life of 88 days, so

³⁵S t_1/2= 88 days

Then to find the decay constant λ, use the equation t_1/2=ln(2)/λ, because nuclear decay is first order:

λ=ln(2)/88 days^-1

λ=0.007877 days^-1

Then use the formula where activity is equal to e^λ*t and then solve for time.

A=e^λ*t

And to compare two activities, use ln(N_t/N_o) = -λt

Because the question says the sample contains an activity of 500 times the detectable limit, N_o = 500 and N_t = 1, and λ = 0.007877 days^-1

ln(1/500)= -0.007877 * t

t=789.2 days before the substance activity cannot be detected.

Question 18.4 Solution:

For each of the following reactions:

1) identify the oxidation and reduction half-equations,

2) balance the equation (adding H⁺ and H₂O as needed),

3) find E^o (in volts)

4) determine whether the reaction is spontaneous under standard conditions.

a. Ag(s)+Cu²⁺(aq)→Ag⁺(aq)+Cu(s)

b. Ni(s)+MnO₄^-(aq)→Ni²⁺(aq)+Mn²⁺(aq)

c. Mn²⁺(aq)+NO₃^-(aq)→MnO₂(s)+NO(g)

1) identify the oxidation and reduction half-equations

In order to identify the oxidized and reduced species, you must assign oxidation numbers to the elements in the equation. Whichever element loses electrons (has an oxidation number that is bigger in the reactants than in the products) is oxidized, and whichever gains electrons is the reduction reaction.

Once the correct elements have been identified, separate them into their corresponding oxidation and reduction reactions.

a. Oxidation: Ag(s)→Ag⁺(aq)

Reduction:Cu²⁺(aq)→Cu(s)

b. Oxidation: Ni(s)→Ni²⁺(aq)

Reduction: MnO₄^-(aq) →Mn(s)

c. Oxidation: Mn²⁺(aq) →MnO₂ (s)

Reduction: NO₃^-(aq) →NO(g)

2) balance the equation

In order to balance the equation, you must first start by balancing the elements themselves on either side of the equation. Balance oxygen with water molecules, and balance the hydrogens with H⁺ ions. This will change the charge for each reduction and oxidation reaction, but this is easily balanced by adding electrons to the reactant side in the reduction reaction and to the product side of the oxidation reaction. Make sure the charges are balanced! Then multiply each half reaction by a number that will make it so that the amount of electrons transferred are equal. In order to get the overall reaction, add the reduction and oxidation reactions together. If there are unequal amount of of electrons for each reaction, multiply each one by a common factor to have equal electrons. Then, add the two equations back together, cancelling any species that exist on both the reactant and product side. This should cancel out the electrons transferred:

a. Oxidation: (Ag(s)→Ag⁺(aq) +e^- )*2 E°=-0.80 V

Reduction:Cu²⁺(aq)+ 2e^-→Cu(s) E°= 0.16 V

Overall Reaction: 2Ag(s)+Cu²⁺(aq)→2Ag⁺(aq)+Cu(s)

b. Oxidation: (Ni(s)→Ni²⁺(aq) +2e^- )*5 E°= 0.23 V

Reduction: (MnO₄^-(aq)+8H⁺+5e^-→Mn(s)+4H₂O(l))*2 E°= 1.49

Overall Reaction: 5Ni(s)+2MnO₄^-(aq)+8H⁺→5Ni²⁺+(aq)+2Mn²⁺(aq)+4H₂O(l)

c. Oxidation: (Mn²⁺(aq)+2H₂O(l)→MnO₂(s)+4H⁺+2e^-)*3 E°=-1.21

Reduction: (NO₃^-(aq)+4H⁺+3e^-→NO(g)+2H₂O(l))*2 E°=0.96

Overall reaction: 3Mn²⁺(aq)+2NO₃^-(aq)+2H₂O(l)→3MnO₂(s)+2NO(g)+4H⁺

3) find E∘ for the equation

In order to find the standard cell potential, look for the standard reduction potentials of each of the half reactions. Then, substitute the values into the equation E°_cell=E°_cathode-E°_anode. Reduction occurs at the cathode, and oxidation occurs at the anode, so use the standard reduction potentials of each reduction and oxidation reaction as shown above. These values can be found on a chart for standard reduction potential.

1. 0.16 V - (-0.80 V) =E°, so E^o=0.96 V
2. 1.49 V - (0.23 V) = E°, so E^o= ​​​​1.26 V
3. 0.96 V - (-1.21 V) =E°, so E^o=3.17 V

4) determine whether the reaction is spontaneous under standard conditions.

Reactions are spontaneous if ΔG^o(Gibb's free energy) < 0

Given the reaction for standard cell potential: ΔG^o = -nFE^o_cell, then E^o_cell is positive when ΔG^o is negative, which indicates a spontaneous reaction.

Since a cell reaction is spontaneous when the standard cell potential is positive.

1. E°=0.96 V, Therefore the reaction is spontaneous.
2. E°=1.26 V Therefore the reaction is spontaneous.
3. E°=3.17 V Therefore the reaction is spontaneous

Question 21.4.1 Solution:

What do chemists mean by the half-life of a reaction?

When a chemist refers to the half life of a reaction, they are asking for the amount of time it takes for half of the original reactant to react and form products. This half life should remain constant throughout the entire reaction, where after each half life, another half of the substance has turned into a product--the ratio of decay of reactants to time should remain constant. For example, if one half life goes by, then half of the product remains. If two have gone by, then a fourth of the product remains, and if three have gone by, then an eighth of the original reactant remains. This demonstrates exponential decay, which can be modeled by the equation:

[A]=0.5ⁿ[A₀] where A is the amount of original reactant remaining, [A₀] is the original amount of reactant, and n is the number of half lives that have happened.