Extra Credit 33

Q19.11D

You may assume that the reactants and products in the equations are in their standard states. Use the information from Table P2 to predict if a spontaneous reaction will occur in the forward direction as drawn for the following cases.

1. \(\mathrm{ Zn (s) + Cu^{2+} \rightarrow Zn^{2+} + Cu (s)}\)
2. \(\mathrm{2Hg^{2+} (aq) + 2Br^- (aq) \rightarrow Hg_2^{2+}(aq) + Br_2 (l)}\)
3. \(\mathrm{2Fe^{2+} (aq) + Cl_2 (g) \rightarrow 2Fe^{3+}(aq) + 2Cl^-(aq)}\)

S19.11D

a. E^°_cell=E^°_{(reduction half cell) -} E^°_{(oxidation half cell)}

E^°_{(reduction half cell)} = 0.340V

E^°_{(oxidation half cell)}=-0.763V

E^°_cell=E^°_{(reduction half cell) -} E^°_{(oxidation half cell)}= 0.340V-(-0.763V)=1.103V

Because the reduction potential of Cu²⁺ is greater than that of Zn²⁺ and E^°_cell is a positive number, a spontaneous reaction will occur in the forward direction as drawn.

b.

E^°_cell=E^°_{(reduction half cell) -} E^°_{(oxidation half cell)}

E^°_{(reduction half cell)} = 0.911V

E^°_{(oxidation half cell)}=1.087V

E^°_cell=E^°_{(reduction half cell) -} E^°_{(oxidation half cell)}= 0.911V-1.087V=-0.126V.

Because the reduction potential of the oxidation half cell is greater than that of the reduction half cell and E^°_cell is negative, a spontaneous reaction will not occur.

c.

E^°_cell=E^°_{(reduction half cell) -} E^°_{(oxidation half cell)}

E^°_{(reduction half cell)} = 1.358V

E^°_{(oxidation half cell)}=0.771V

E^°_cell=E^°_{(reduction half cell) -} E^°_{(oxidation half cell)}= 1.358V-0.771V=0.587V.

Because the reduction potential of the reduction half cell is greater than that of the oxidation half cell and E^°_cell is positive, a spontaneous reaction will occur.

Q19.90A

\(\ce{Pb^2+(aq)}\) and \(\ce{Ag+(aq)}\) were produced when 0.978 g sample of nitric acid was dissolved. After diluting the solution to 250.0 mL of deionized water, a silver electrode was placed in solution. The potential difference read 0.273 V. Calculate the mass of \(\ce{Ag}\) in the sample as a percent.

S19.90A

The equation for the oxidation is \(\mathrm{H_2(g) \rightarrow 2H^+(aq) + 2e^- \hspace{38 pt}\quad {-E}^\circ=0.00\:V}\).

The equation for reduction is \(\mathrm{(Ag^+(aq) + e^- \rightarrow Ag(s) ) \times2 \hspace{23 pt}\quad E^\circ=0.800\:V}\).

The overall reaction is \(\mathrm{H_2(g) + 2Ag^+(aq) \rightarrow 2H^+(aq) + 2Ag(s) \quad E^\circ_{cell}= 0.800\:V}\).

The equation to calculate Ecell is

E_cell= E^º_cell - (0.0592/2) x log([H⁺]²/[Ag⁺]²)

[H⁺]= 1.00

0.273V= 0.800V - 0.0592/2 log((1.00)²/[Ag+]²)

-0.527V=-0.0592/2 log ((1.00)²/[Ag+]²)

17.804= log 1/[Ag+]²

10^17.804=1/[Ag+]²

[Ag⁺]²=1/10^17.804

[Ag⁺]=1.25x10^-9M

Mass Ag= (1.25x10^-9Ag⁺ mol/L) x (1L/1000mL) x (107.87gAg/1molAg) x (1mol Ag/1mol Ag⁺) x 250mL= 3.38x10^-8% Ag.

Q21.7A

Draw:

1. \(\ce{[PtI4]^2-}\)
2. \(\textrm{fac-}\ce{[Cu(H2O)3(NH3)3]^2+}\)
3. \(\ce{[CoCl(H2O)5]^2+}\)

S21.7A

For b and c, the H₂O attaches to the central atom by the oxygen atom. For b, the structure was determined because it is a fac isomer, therefore the three H2O groups must all form 90 degree angles from the central atom and the three NH₃ groups must also from 90 degree angles from the central atom.

Q24.16A

One of the following statements is true and the other is false regarding the first-order reaction \(\ce{4A \rightarrow B + C}\). Identify the true statement and the false one, and explain your reasoning.

1. A graph of \(\ce{[A]}\) versus time is a straight line.
2. The rate of the reaction is one half the rate of disappearance of \(\ce{A}\).

S24.16A

1. False because the graph of [A] versus time is a curved line for first order reactions.

2. False: Since the rate is dependent on balanced coefficients [A] is four times the amount in rate = k[A] based on the coefficients, so the rate of reaction would be one fourth the rate of the disappearance of A.

Q24.87B

Find the general rate law and the magnitude of \(\ce{k}\) for the overall reaction.

1. \(\ce{2(N2O5 \rightarrow NO2 + NO3)}\)
2. \(\ce{NO2 \rightarrow NO3 \rightarrow NO2 + O2 + NO}\)
3. \(\ce{NO + NO3 \rightarrow 2NO2}\)

S24.87B

First add up all of the equations to get the overall equation.

\(\ce{2N2O5 + NO2 + NO3 + NO + NO3 \rightarrow 2NO2 + 2NO3 + NO2 + O2 + NO + 2NO2}\)
\(\ce{\Rightarrow 2N2O5 \rightarrow 4NO2 + O2 \: (overall\: reaction)}\)

The rate of the reaction is based on the rate of the formation of the products. Rate = k[NO₂]⁴[O₂].

Q25.33C

What is the energy (in joules) associated with the destruction of 9.56x10^-24 g of matter?

S25.33C

To calculate the energy associated with the destruction of 9.56x10^-24 grams of matter, you would use the equation

E= mc²

m= 9.56x10^-24 grams and c= 2.998x10⁸m/s

E=(9.56x10^-24 grams)x(1kg/1000grams)x(2.998x10⁸m/s)²

= 2.87x10^-18kg m²s²

= 2.87x10^-18J

Q21.2.1

S21.2.1

Nuclear reactions violate the law of conservation of mass whereas the law is not violated in chemical reactions. In chemical reactions, the mass changes are not noticeable. In nuclear reactions, the mass changes, even if they are very slight, are noticeable and there are large energies involved with mass changes.

Q24.6.8

For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present.

1. [Cu(NH₃)₄]²⁺
2. [Ni(CN)₄]²⁻

S24.6.8

a. Based on the Cu²⁺atom, we know that there are 9 valence electrons available. Because the NH₃ is a strong field ligand, this produces a low spin, and leaves one unpaired electron.

b. Based on the Ni²⁺atom, we know that there are 8 valence electrons available. Because CN^-is a strong field ligand, a low spin is produced and no electrons are left unpaired.