Extra Credit 47

Q17.7.1 Identify the reaction at the anode, reaction at the cathode, the overall reaction, and the approximate potential required for the electrolysis of the following molten salts. Assume standard states and that the standard reduction potentials in Table P1 are the same as those at each of the melting points. Assume the efficiency is 100%.

1. CaCl₂
2. LiH
3. AlCl₃
4. CrBr₃

1) CaCl₂

To figure out which half reaction occurs at the anode or cathode, you must look at the E^o values of each half reaction on the SRP table. The one with the bigger E^o on that table is the one being oxidized (therefore anode) and the one with the smaller E^o on that table is the one being reduced (therefore cathode.)

(Another way to look at it is by using the SRP table. If one reaction is higher on the table compared to the other reaction, it will be reduced.)-Phase 2 addition

\(Ca^{2+}{(aq)} +2e^-\rightleftharpoons Ca(s)\) E^o= -2.84 V (therefore this is the cathode)

\(Cl_2(g) +2e^-\rightleftharpoons 2Cl^-(aq)\) E^o=1.396 V (therefore this is the anode)

Overall reaction:

a) Balance electrons

\(Ca^{2+}{(aq)} +2e^-\rightleftharpoons Ca(s)\)

\(Cl_2(g) +2e^-\rightleftharpoons 2Cl^-(aq)\)

b) Add the two half reactions together to get the overall reaction:

\(CaCl_2\rightleftharpoons Ca(s) + Cl_2(g)\)

E^o_Cell= E^o_cathode- E^o_anode = -2.84 - 1.396 = -4.236 V

2) LiH

\(Li^{+}(aq) + e^-\rightleftharpoons Li(s)\) E^o=-3.04 V (therefore this is the cathode)

\(2H^{+} + 2e^-\rightleftharpoons H_2(g)\) E^o=0.00 V (therefore this is the anode)

Overall reaction:

a) Balance electrons:

\(2Li^{+}(aq) + 2e^-\rightleftharpoons 2Li(s)\)

b) Add the two half reactions together to get the overall reaction:

\(2LiH\rightleftharpoons 2Li(s) +H_2(g)\)

E^o_Cell= E^o_cathode- E^o_anode =-3.04-0.00= -3.04 V

3) AlCl₃

\(Al^{3+} +3e^-\rightleftharpoons Al(s)\) E^o=-1.676 V(therefore this is the cathode)

\(Cl_2(g) +2e^-\rightleftharpoons 2Cl^-(aq)\) E^o=1.396 V (therefore this is the anode)

Overall reaction:

a) Balance electrons:

\(2Al^{3+} +6e^-\rightleftharpoons 2Al(s)\)

\(3Cl_2(g) +6e^-\rightleftharpoons 6Cl^-(aq)\)

b) Add the two half reactions together to get the overall reaction:

\(2AlCl_3\rightleftharpoons 2Al(s)+3Cl_2(g)\)

E^o_Cell= E^o_cathode- E^o_anode= -1.676-1.396=-3.072V

4) CrBr₃

\(Cr^{3+}+e^-\rightleftharpoons Cr^{2+}\) E^o=-0.424 V (therefore this is the cathode)

\(Br_2+2e^-\rightleftharpoons 2Br^{-}\) E⁰=1.087 V (therefore this is the anode)

Overall reaction:

a) Balance electrons:

\(2Cr^{3+}+2e^-\rightleftharpoons 2Cr^{2+}\)

b) Add the two half reactions together to get the overall reaction:

\(2CrBr_3\rightleftharpoons 2Cr(s)+3Br_2(g)\)

E^o_Cell= E^o_cathode- E^o_anode =-0.424-1.087=-1.511V

Q12.3.10 The decomposition of acetaldehyde is a second order reaction with a rate constant of 4.71 × 10−8 L/mol/s. What is the instantaneous rate of decomposition of acetaldehyde in a solution with a concentration of 5.55 × 10−4 M?

k=\(4.71*10^{-8}\) L/mol/s

[C₂H₄O] =\(5.55*10^{-4}\) M

Rate= k[C₂H₄O]²

Rate= (\(4.71*10^{-8}\))(\(5.55*10^{-4}\))²

Rate= \(1.45*10^{-14}\) M/s

Q12.6.2 In general, can we predict the effect of doubling the concentration of A on the rate of the overall reaction A+B⟶C? Can we predict the effect if the reaction is known to be an elementary reaction?

No, you can not predict the effect of doubling the concentration of A on the rate of the overall reaction A+B⟶C because you do not know the rate equation (aka you do not know what order A and B are.) Also, to add on, the rate order can only be determined experimentally in a lab.-Phase 2 addition

However, if you know that the reaction is an elementary reaction, you know that A and B are both first order, so the rate law would be rate=k[A][B], so therefore doubling [A] would double the rate of the reaction.

Q21.4.14 A 1.00 × 10–6-g sample of nobelium, No-254, has a half-life of 55 seconds after it is formed. What is the percentage of No-254 remaining at the following times?

1. 5.0 min after it forms

2. 1.0 h after it forms

The reactions are first order, so therefore you use the equation: t_1/2=\(\frac{ln2}{k}\) and e^-kt= \(\frac{[A]}{[A_o]}\)

5 min after it forms (5 min=300sec):

55=\(\frac{[ln2]}{[k]}\)

k=0.0126 s^-1

e^{-(0.0126)(300)}=0.0228

(0.0228)(100%)=2.28%

1 hour after it forms (1 hour=3600sec)

55=\(\frac{[ln2]}{[k]}\)

k=0.0126 s^-1

e^{-(0.0126)(3600)}=1.997 x 10^-20

(1.997 x 10^-20)(100%)= 1.997 x 10^-18%

Q20.3.2 If two half-reactions are physically separated, how is it possible for a redox reaction to occur? What is the name of the apparatus in which two half-reactions are carried out simultaneously?

A galvanic cell is the name of the apparatus in which two half reactions are carried out simultaneously. In a galvanic cell, the anode and the cathode solutions are separate and electrons flow, when a current is applied, from anode to cathode. However there is a salt bridge that closes the circuit, therefore there is not an electron buildup in one of the half reactions.

Q20.5.13 For the cell represented as Al(s)∣Al3+(aq)∥Sn2+(aq), Sn4+(aq)∣Pt(s), how many electrons are transferred in the redox reaction? What is the standard cell potential? Is this a spontaneous process? What is ΔG°?

First, you must write the half reactions:

1) \(Al^{3+}(aq)+ 3e^- \rightarrow Al(s)\)

2) \(Sn^{4+}(aq)+ 2e^- \rightarrow Sn^{2+} (aq)\)

Then, you have to balance the half reactions by charge. Do so by multiplying equation 1 by 2, and equation 2 by 3.

This leaves you with: \(2Al(s)+3Sn^{4+}(aq)+ 6e^- \rightarrow 3Sn^{2+} (aq)+2Al^{3+}(aq)+6e^-\) for your overall reaction

6 electrons were transferred in this reaction

E^o_cell=E^o_cathode- E^o_anode (look up values in SRP table)

For equation 1, E^o=-1.676 V

For equation 2, E^o=0.154 V

Equation 1 is anode, because it is being oxidized. Equation 2 is cathode because it is being reduced

E^o_cell=0.154-(-1.676)=1.83 V

ΔG^o=-nFE^o_cell, where n= number of electrons transferred, F=96,485 C/mol, E^o_cell=standard cell potential

ΔG^o=-(6)(96,485)(1.83)=-1,059,405 J/mol = -1,060 kJ/mol: ΔG^o is a negative number, therefore it is spontaneous

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⁻

[Cu(NH₃)₄]²⁺: Cu has a coordination number of 4, since it has 4 (NH₃)'s attached to it. This means that it is either tetrahedral(if it has 0 or 10 d electrons), or square planar (if it has 1-9 d electrons.) The NH₃ ligand has 0 charge, and the over all charge of the complex is 2+, therefore the transition metal, Cu, has a charge of +2, so it has 9 d electrons. Therefore the shape is square planar. Since Cu has 9 d electrons, it is neither high spin nor low spin (if you were to draw the energy diagram they are the same for both high and low spin.) For the enegy diagram, you completely fill the d_xz, d_yz, d_z², and d_xy orbitals, and you have 1 electron in the dx²-y² orbital, which means that you have 1 unpaired electron.

[Ni(CN)₄]²⁻: Ni has a coordination number of 4, since it has 4 (CN)'s attached to it. This means that it is either tetrahedral(if it has 0 or 10 d electrons), or square planar (if it has 1-9 d electrons.) The CN ligand has a -1 charge and there are 4 of them so there is a -4 charge from that. The overall charge of the complex is -2. Therefore the transition metal, Ni, has a charge of +2, so it has 8 d electrons. Therefore the shape is square planar. CN is a strong field ligand, therefore it is low spin. If you were to draw the energy diagram for [Ni(CN)₄]^2-, you would completely fill the d_xz, d_yz, d_z², and d_xy orbitals, leaving the dx²-y², empty which means that you would have no unpaired electrons.

Q14.4.6 Iodide reduces Fe(III) according to the following reaction:

2Fe³⁺(soln) + 2I⁻(soln) → 2Fe²⁺(soln) + I₂(soln)

Experimentally, it was found that doubling the concentration of Fe(III) doubled the reaction rate, and doubling the iodide concentration increased the reaction rate by a factor of 4. What is the reaction order with respect to each species? What is the overall rate law? What is the overall reaction order?

Doubling [Fe³⁺] doubles the the reaction rate, so it is first order. Doubling [I^-] increases the rate by a factor of 4, which means that it is second order. The overall reaction order is 1+2=3. The overall rate is law is: rate=k[Fe³⁺][I^-]²