Name: _________________________________________
Date:____________________________
Homework 4
Put your answers on a separate sheet of paper OR insert spaces in this document – do NOT try and cram
your answers in the small space provided here!!
1.) Combine the following three half-reactions into THREE spontaneous, balanced equations and
calculate Eocell for each. Use pairs of equations only (e.g. equation 1 with equation 2, then equation 1
with equation 3 and so on). Rank the relative strengths of oxidizing agents and reducing agents
(hint: electrons, H+1, and water are not involved in determining the relative strengths of the
oxidizing and reducing agents they were used only to balance the redox reactions)
(1) NO3-1 (aq) + 4H+1 (aq) + 3e-1 → NO (g) + 2H2O
Eo = 0.96 V
(2) N2 (g) + 5H+1 (aq) + 4e-1 → N2H5+1 (aq)
Eo = -0.23 V
+1
-1
+2
(3) MnO2 (s) + 4H (aq) + 2e → Mn (aq) + 2H2O (l) Eo = 1.23 V
First Set: The more positive/larger number stays written as the reduction, therefore, reaction 1 stays as is,
reaction 2 will be flipped
(1) NO3-1 (aq) + 4H+1 (aq) + 3e-1 → NO (g) +
(2) N2 (g) + 5H+1 (aq) + 4e-1 → N2H5+1 (aq)
2H2O
Eo = 0.96 V
Eo = -0.23 V
(1) NO3-1 (aq) + 4H+1 (aq) + 3e-1 → NO (g) + 2H2O
(2) N2H5+1 (aq) →
N2 (g) + 5H+1 (aq) + 4e-1
Eo = 0.96 V
Eo = +0.23 V
Each reaction needs to be multiplied by a factor to make the number of electrons lost = number electrons
gained but we do NOT multiply the Eo values!
(1) 4(NO3-1 (aq) + 4H+1 (aq) + 3e-1 → NO (g) + 2H2O)
(2) 3(N2H5+1 (aq) →
N2 (g) + 5H+1 (aq) + 4e-1 )
Eo = 0.96 V
Eo = +0.23 V
(1) 4NO3-1 (aq) + 16H+1 (aq) + 12e-1 → 4NO (g) + 8H2O
(2) 3N2H5+1 (aq) →
3N2 (g) + 15H+1 (aq) + 12e-1
Eo = 0.96 V
Eo = +0.23 V
4NO3-1 (aq) + 16H+1 (aq) + 3N2H5+1 (aq) → 3N2 (g) + 15H+1 (aq) + 4NO (g) +
8H2O
REDUCE!!
4NO3-1 (aq) + H+1 (aq) + 3N2H5+1 (aq) → 3N2 (g) + 4NO (g) + 8H2O Eo cell = 1.19 V
Since reaction 1 stayed written as the reduction, then NO3-1 is the stronger oxidizing agent than N2
Since reaction 2 was reversed, N2H5+1 is a stronger reducing agent than NO
Second Set: the larger/more positive Eo value stays as the reduction. Thus, reaction 3 stays as the reduction
while reaction 2 is flipped
(2) N2 (g) + 5H+1 (aq) + 4e-1 → N2H5+1 (aq)
(3) MnO2 (s) + 4H+1 (aq) + 2e-1 → Mn+2 (aq) + 2H2O (l)
Eo = -0.23 V
Eo = 1.23 V
(2) N2H5+1 (aq) → N2 (g) + 5H+1 (aq) + 4e-1
(3) MnO2 (s) + 4H+1 (aq) + 2e-1 → Mn+2 (aq) + 2H2O (l)
Eo = 0.23 V
Eo = 1.23 V
The number of electrons lost must equal the number of electrons gained so we need to multiply equation 3
by a factor but DO NOT multiply the E value!
(2) N2H5+1 (aq) → N2 (g) + 5H+1 (aq) + 4e-1
(3) 2(MnO2 (s) + 4H+1 (aq) + 2e-1 → Mn+2 (aq) + 2H2O (l) )
Eo = 0.23 V
Eo = 1.23 V
(2) N2H5+1 (aq) → N2 (g) + 5H+1 (aq) + 4e-1
(3) 2MnO2 (s) + 8H+1 (aq) + 4e-1 → 2Mn+2 (aq) + 4H2O (l)
Eo = 0.23 V
Eo = 1.23 V
N2H5+1 (aq) + 2MnO2 (s) + 8H+1 (aq) → N2 (g) + 5H+1 (aq) + 2Mn+2 (aq) + 4H2O (l)
REDUCE!!
N2H5+1 (aq) + 2MnO2 (s) + 3H+1 (aq) → N2 (g) + 2Mn+2 (aq) + 4H2O (l) Eocell = 1.46 V
Since reaction 3 stayed written as the reduction, then MnO2 is the stronger oxidizing agent than N2
Since reaction 2 was reversed, N2H5+1 is a stronger reducing agent than Mn+2
Third Set: the larger/more positive E values stays written as the reduction, so reaction 3 stays and reaction
1 is flipped
(1) NO3-1 (aq) + 4H+1 (aq) + 3e-1 → NO (g) + 2H2O
(3) MnO2 (s) + 4H+1 (aq) + 2e-1 → Mn+2 (aq) + 2H2O (l)
Eo = 0.96 V
Eo = 1.23 V
(1) NO (g) + 2H2O → NO3-1 (aq) + 4H+1 (aq) + 3e-1
(3) MnO2 (s) + 4H+1 (aq) + 2e-1 → Mn+2 (aq) + 2H2O (l)
Eo = -0.96 V
Eo = 1.23 V
The number of electrons lost must = the number of electrons gained so multiply each reaction by a factor
but DO NOT multiply the E values!
(1) 2(NO (g) + 2H2O → NO3-1 (aq) + 4H+1 (aq) + 3e-1 )
(3) 3(MnO2 (s) + 4H+1 (aq) + 2e-1 → Mn+2 (aq) + 2H2O (l))
(1) 2NO (g) + 4H2O → 2NO3-1 (aq) + 8H+1 (aq) + 6e-1
(3) 3MnO2 (s) + 12H+1 (aq) + 6e-1 → 3Mn+2 (aq) + 6H2O (l)
Eo = -0.96 V
Eo = 1.23 V
Eo = -0.96 V
Eo = 1.23 V
2NO (g) + 4H2O + 3MnO2 (s) + 12H+1 (aq) → 2NO3-1 (aq) + 8H+1 (aq) + 3Mn+2 (aq) + 6H2O (l)
REDUCE!!!
2NO (g) + 3MnO2 (s) + 4H+1 (aq) → 2NO3-1 (aq) + 3Mn+2 (aq) + 2H2O (l) Eocell = 0.27 V
Since reaction 3 stayed written as the reduction, then MnO2 is the stronger oxidizing agent than NO3-1
Since reaction 1 was reversed, NO is a stronger reducing agent than Mn+2
Oxidizing Agents: the larger more positive E value for the reduction reaction contains (as the reactant) the
oxidizing agent)
NO3-1 > N2
MnO2 > N2
MnO2 > NO3-1
So: MnO2 > NO3-1 > N2
Reducing Agents: the larger more positive E value for the oxidation (!!) reaction (the reverse of the
reduction reaction) contains (as the reactant!!) the reducing agent. This works only if the reactions are
REVERSED!!! OR, if we look at the reduction reactions, the oxidizing agents are the reactants, the reverse
reaction would be oxidation, so the reducing agents are found on the PRODUCT side of a REDUCTION
reaction and the strength is OPPOSITE to the trend for oxidizing agents. Whatever the product is of the
reaction that has the N2 in it is the strongest because N2 was the weakest and so on.
N2H5+1 > NO
N2H5+1 > Mn+2
NO > Mn+2
So: N2H5+1 > NO > Mn+2
2.) A concentration cell consists of two Ag/Ag+1 half cells. In half-cell A, electrode A is placed in 0.010
M AgNO3. In half-cell B, electrode B is placed in 4.0 x 10-4 M AgNO3. What is the cell potential at
298 K? Which half cell is acting like the anode? Which half cell is acting like the cathode?
Half cell A: 0.010 M which is a stronger (higher) concentration than Half cell B: 4.0 x 10-4 M
In the half cell we have two possible reactions: Ag+1 + 1e-1 → Ag (s) or Ag (s) → Ag+1 + 1e-1
The stronger [Ag+1] needs to be lowered. The Ag+1 need to “go away”. The reaction that shows the Ag+1
ions “going away” is the reduction reaction.
The less [Ag+1] needs to be increased. The Ag+1 needs to be “created or formed”. The reaction that shows
the Ag+1 ions forming is the oxidation reaction.
Ag+1 (conc) + 1e-1 → Ag (s) Eo = +0.80 V
Ag (s) → Ag+1(dilute) + 1e-1 Eo = -0.80 V
Ag+1 (concentrated) → Ag+1 (dilute) Eocell = 0.00V
Ecell =
Ecello
[Ag +1 ]dilute
0.0592V
log
1
[Ag + 1 ]concentrated
Ecell = 0.00 V -
[Ag +1 ]dilute
0.0592V
log
1
[Ag + 1 ]concentrated
Ecell = -
[Ag +1 ]dilute
0.0592V
log
1
[Ag + 1 ]concentrated
Ecell = -
[ 4.0 x 10 -4 ]dilute
0.0592V
log
1
[0.010 ]concentrated
Ecell = 0.0828 V =
+ 0.083 V
3.) What is the coordination number of the central metal ion in the following complexes?
a. [Pt(NH ) Cl ]
3 3
3
+1
: 6 ligands therefore
6
b. [Cu(en)2]+2 2 ligands – 2 binding sites each therefore
c.
4
[Fe(EDTA)]-2 1 ligand with 6 binding/grabbing sites therefore
6
4.) Write the formula for each – if it is an ion – be sure to include its charge:
a. Hexaaquairon(II) ion:
[Fe(H2O)6]+2
b. Tetraamminecopper (II) tetrachloroplatinate (II): [Cu(NH3)4]+2 [PtCl4]-2
[Cu(NH3)4][PtCl4]
c. Tris(ethylenediamine)chromium (III) hexacyanocobaltate (III): [Cr(en)3]+3 [Co(CN)6]-3
[Cr(en)3][Co(CN)6]
5.) Name the following species:
a. K3[Fe(CN)6] - hint – think about which species is the cation and which is the anion!
Potassium hexacyanoferrate(III)
b. [Co(NH3)5(CN)2]NO3
Pentaamminedicyanocobalt (III) nitrate
c. [Pt(NH3)3Cl]2[PtCl4]
Triamminechloroplatinum (II) tetrachloroplatinate (II)
6.) What is the maximum number of unpaired d electrons that an atom or ion can possess? BRIEFLY
explain how you made your choice.
The maximum number of unpaired d electrons is 5. A filled d orbital holds 10 electrons. Each
electron fills an orbital singly at first before they pair up. Using the orbital box diagram, the
most unpaired electrons occurs when each orbital is holding 1 electron.
Any more than 5 single electrons (unpaired electrons) and they must start pairing up!
7.) A complex in solution absorbs green light. What is the color of solution? RED if the species
absorbs green light, it reflect red light and that’s what we see!
8.) Sketch the orientation of all 5 of the d orbitals please put each on their OWN set of axes.
Ha ha . . .
9.) Using 3 equations, how does the speed of light relate to frequency and wavelength, how does
energy relate to frequency, how does energy relate to wavelength?
c=λυ
E=hυ
E = hc/λ
10.) Put the visible region of the electromagnetic spectrum in order from longest wavelength (on the left)
to shortest wavelength (on the right).