Extra Credit 22

Question 19.7 B: Determine the value of ΔG^o for the following reactions carried out in voltaic cells.

a. \[Al(s)+Cu^{2+}\rightarrow Cu(s)+Al^{3+}\]

b. \[Fe^{3+}+Ag(s)\rightarrow Fe^{2+}+Ag^{+}\]

c. \[Cu(s)+2Fe^{3+}\rightarrow Cu^{2+}+2Fe^{2+}\]

Answer 19.7B

Because these are redox reactions, the first step is to balance the equations:

a. The oxidation half reaction is \[Al(s)\rightarrow Al^{3+}\]

To balance the charge, add electrons to the products: \[Al(s)\rightarrow Al^{3+} +3e^{-}\]

The reduction half reaction is: \[Cu^{2+} \rightarrow Cu(s) \]

To balance this, add electrons to the reactants: \[Cu^{2+} +2e^{-} \rightarrow Cu(s) \]

Now in order to combine the two half reactions, the electrons have to cancel out. Therefore, multiply the oxidation reaction by 2 and the reduction reaction by 3:

\[2Al(s) \rightarrow 2Al^{3+} +6e^{-} \]

\[3Cu^{2+} +6e^{-} \rightarrow 3 Cu(s) \]

Now that the electrons will cancel, combine the two reactions:

\[2Al(s) +3Cu^{2+} \rightarrow 3Cu(s) + 2Al^{3+} \]

Now that the reaction is balanced, utilize the Gibbs free energy equation:

\[\Delta G^{o}=-nFE^{o}_{cell}\]

To determine \(E^{o}_{cell}\) , use the standard reduction potential chart and add the two potentials:

\[E^{o}_{cell}=E_{cathode}-E_{anode}\]

(.34)-(-1.66) = 2.0 = \(E^{o}_{cell}\)

Furthermore, n is the number of electrons it takes to balance the redox:

n=6

and F is the Faraday's constants which is 96,485 C/mol

plugging these values into the Gibbs Free energy equation will result in:

\[\Delta G^{o}=-6 mol\times (\tfrac{96485C}{mol})\times 2.0V\]

= -1.16E6 J

b. The same process occurs for this reaction: Fe³⁺ + Ag(s) -----> Ag⁺ + Fe²⁺

Oxidation half reaction: Ag(s) --> Ag⁺

balanced: Ag(s) --> Ag+ e^-

reduction half reaction: Fe³⁺ --> Fe²⁺

balanced: Fe³⁺ + e^- --> Fe²⁺

Because the electrons have a 1:1 ratio, the equations do not have to be multiplied by a coefficient:

Ag(s) + Fe³⁺ --> Ag⁺ + Fe²⁺

Therefore, n=1

To find E^o_cell find the standard reduction potentials for both the reactions:

( .8)-(.771)= .029 V = E^o_cell

Therefore, plugging into the Gibbs free energy equation:

ΔG^o= -1 mol* 96,485 C/mol * .029V

= -2798.1 V

c. Cu(s) + 2Fe³⁺ ---> Cu²⁺ +2Fe²⁺

oxidation reaction: Cu(s) ---> Cu²⁺

balanced: Cu(s) --> Cu²⁺ + 2e^-

reduction reaction: 2Fe³⁺ --> 2Fe²⁺

balanced: 2Fe³⁺ +e^- --> 2Fe²⁺

To cancel out the electrons, multiply the reduction reaction by 2:

2*(2Fe³⁺ +e^- --> 2Fe²⁺ )

=4 Fe³⁺ + 2e^- --> 4Fe²⁺

now that they both have 2 electrons, combine to get complete redox:

Cu(s) + 4Fe³⁺ --> Cu²⁺ + 4 Fe²⁺

Therefore, n=2

E^o_cell= (.771) -(.153) = .618V

ΔG^o= -2 mol * 96,485 C/mol * .618V

= -119255.5 J

Q 19.63A: Calculate the amount in grams of a metal that is deposited at the cathode by running a current of 3.15 A for 78 min in an electrolysis reaction fro an aqueous solution containing a) Zn²⁺ b) Sn²⁺ c) Fe³⁺ and d) Ni²⁺

In order to solve this problem, you have to be aware of some important conversion factors:

1 coulomb = 1 ampere * 1 second

78 min = 4680 seconds

1 mol e- = 96485 Coulombs

Therefore, use dimensional analysis to solve for moles:

(3.15A) * (1 C/A*s) * (4680 s) * (1mol e- / 96485 C ) = .153 mol e-

a) To calculate the grams of Zn²⁺ that is deposited, write the half reaction that would occur in the hydrolysis reaction to get stoichiometric coefficients:

Zn²⁺ + 2e^- ---> Zn(s)

Now use the molar ratio of Zn to electrons and its molar mass, 65.41 g/mol, to determine the amount deposited:

( .153 mol e-) * ( 1 mol Zn2+/ 2 mol e-) * ( 65.41 g/ 1 mol) = 5.0 g Zn

b) To find the grams of Sn²⁺ deposited, find the balanced half reaction:

Sn²⁺ +2e^- --> Sn (s)

Using the same method as above, and utilizing the molar mass of Sn, 118.7g/mol, we can used dimensional analysis to solve:

.153 mol e- * ( 1 mol Sn/ 2 mol e-) * (118.7g/mol) = 9.08 g Sn

c) To find the grams of Fe deposited:

The half reaction is Fe³⁺ + 3e^- --> Fe(s)

The molar mass of Fe is: 55.847 g/mol

Therefore,

0.152 mol e^- * ( 1 mol Fe/ 3 mol e-) * (55.847 g/ mol) = 2.85 g Fe

d) To determine the grams of Ni deposited:

The half reaction is Ni²⁺ + 2e^- --> Ni(s)

The molar mass of Ni is 58.693 g/mol

So,

.153 mol e- * ( 1 mol Ni / 2 mol e-) * ( 58.693 g Ni/ 1 mol)= 4.49 g Ni

Q 21.4 B : a. How many unpaired electrons would you expect to find in the octahedral complex [CoF₆]^3-

b. How many unpaired electrons would you expect to find in a tetrahedral complex [FeBr₄]^2-? Would you expect more, fewer, or the same number of unpaired electrons as in the octahedral complex [Co(NH₃)₆]³⁺

Answer 21.4B

a. To determine how many unpaired electrons you would expect to find in the octahedral complex, you have to determine the complex has low spin or high spin. The spin is dictated by whether or not the ligand is strong or weak field. To determine this, you have to look at the spectrochemical series:

I^- < Br^- < Cl^- < SCN^- < NO^3- < F^- < OH^- < H₂O < NCS^- < gly < py < NH₃ <en<NO₂^-<PPh₃ <CO<CN^-

Where I^- is a weak field ligand and CN^- is the strongest field ligand. Therefore, the ligand F^- is a weak field ligand because it is on the left of the spectrochemical series. Weak field ligands have high spin because the energy barrier to get to the e_g energy level is easy to overcome. High spin means that the electron will fill both levels in the energy diagram before they pair with each other.

The oxidation state for the metal, Co, in this octahedral complex, is Co³⁺ . It has an electron configuration of [Ar]3d⁶ because complex ions give up the s-orbital first. The energy diagram would then look like this:

⥠ ⥠

⥮ ⥠ ⥠

There are 4 unpaired electrons