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CHEMISTRY THE CENTRAL SCIENCE

CHAPTER 20: ELECTROCHEMISTRY


#3a
I2O5(s) + 5 CO2(g) --> I2(s) + 5 CO2(g)
This means:
2 I5+ + 5 C2+ --> 2 I0 + C4+
I is reduced: 5+ --> 0
C is oxidized: 2+ --> 4+

#3b
2 Hg2+(aq) + N2H4(aq) --> 2 Hg(l) + N2(g) + 4 H+(aq)
2 Hg2+ + N2- --> 2 Hg0 + N0
Hg is reduced: 2+ --> 0
N is oxidized: 2- --> 0

#3c
3 H2S(aq) + 2 H+(aq) + 2 NO3-(aq) --> 3 S(s) + 2 NO(g) + 4 H2O(l)
3 S2- + 2 N5+ --> 3 S0 + 2 N2+
N is reduced: 5+ --> 2+
S is oxidized: 2- --> 0

#5a
TiCl4(g) + 2 Mg(l) --> Ti(s) + 2 MgCl2(l)

#5b
Ti is reduced: 4+ --> 0
Mg is oxidized: 0 --> 2+

#7a
Sn2+(aq) --> Sn4+(aq)
Add electrons to the most positive side,

Sn2+(aq) --> Sn4+(aq) + 2 e-
This is an oxidation: Sn2+ --> Sn4+

#7b
TiO2(s) --> Ti2+(aq) in acidic solution
TiO2(s) + 2 e- --> Ti2+(aq) + 2 O2-
Redox in acidic medium generates H2O, let each O2- make one H2O,
TiO2(s) + 2 e- --> Ti2+(aq) + 2 H2O(l)
and then balance the negative charges with protons,
TiO2(s) + 4 H+ + 2 e- --> Ti2+(aq) + 2 H2O(l)
This is a reduction: Ti4+ --> Ti2+

#7c
ClO3-(aq) --> Cl-
add electrons
ClO3-(aq) + 6 e- --> Cl- + 3 O2-
balance the charges and generate H2O

ClO3-(aq) + 6 e- + 6 H+ --> Cl- + 3 H2O(l)
This is a reduction: Cl5+ --> Cl-

#13a
Silver reduction: Ag+(aq) + 1e- --> Ag0(s)
Iron oxidation: Fe0 --> Fe2+ + 2 e-

#13b
CATHODE: where the reduction occurs --> Ag
ANODE: where the oxidation occurs --> Fe

#13c
CATHODE: positive --> Ag
ANODE: negative --> Fe

#13d
Electrons are negative. Since opposite charges attract, electrons will be attracted by the positive electrode (CATHODE).
electrons will flow from Fe towards Ag

#13e
Using the same reasoning as above:
Cations migrate towards Ag
Anions migrate towards Fe

#17a
The cathode is where reduction occurs,
2 H+(aq) + 2 e- --> H2(g)

#17b
The component conditions are standard in a standard electrode:
[H+] = 1.00 M
PH2 = 1.00 atm

#17c
The platinum foil is an electron carrier that does not get oxidized nor reduced in the process (inert). The reaction from part a occurs at the surface of the electrode.

#18a
The anode is where oxidation occurs,
H2(g) --> 2 H+(aq) + 2 e-

#18b
Since the platinum electrode serves as the reaction surface, it is best to make it have a large surface area so that more reactants and products can be adsorbed onto it.

#18c
The Standard Hydrogen Electrode


#21a
Reduction of Ti3+, reduction potential unknown
Ti3+(aq) + 2 e- --> Ti+(aq)
Oxidation of Cr2+, oxidation potential of +0.41 V
2 [Cr2+(aq) --> Cr3+(aq) + e-]
Tables list reduction potentials, you would find that 0.41 V is the reduction potential of the reaction:
Cr3+(aq) + e- --> Cr2+

#21b
Ecell = oxidation potential + reduction potential
+1.19 V = [+0.41 V] + reduction potential
+1.19 V - [+0.41 V] = reduction potential = +0.78 V

#21c
Voltaic cell for the reaction:
Ti3+(aq) + 2 Cr2+(aq) --> Ti+ + 2 Cr3+(aq)


#23a
Cl2(g) + 2 I-(aq) --> 2 Cl-(aq) + I2(s)
I- is oxidized, Cl2 is reduced,
Ecell = oxidation potential + reduction potential
Ecell = [-0.536] + [+1.359] = +0.823 V

#23b
Ni(s) + 2 Ce4+(aq) --> Ni2+(aq) + 2 Ce3+(aq)
Ni0 is oxidized, Ce4+ is reduced
Ecell = oxidation potential + reduction potential
Ecell = [+0.28] + [+1.61] = +1.89 V

#23c
Fe(s) + 2 Fe3+(aq) --> 3 Fe2+(aq)
Fe0 is oxidized, Fe3+ is reduced
Ecell = oxidation potential + reduction potential
Ecell = [+0.440] + [+0.771] = +1.211 V

#23d
2 Al3+(aq) + 3 Ca(s) --> 2 Al(s) + 3 Ca2+(aq)
Ca0 is oxidized, Al3+ is reduced
Ecell = oxidation potential + reduction potential
Ecell = [+2.87] + [-1.66] = +1.21 V

#37a
To sort in order of increasing oxidizing strength, sort in order of increasing reduction potential:
Cu2+(aq) < O2(g) < Cr2O7- < Cl2(g) < H2O2(aq)

#37b
To sort in order of increasing reducing strength, sort in order of decreasing reduction potential:
H2O2(aq) < I-(aq) < Sn2+ < Zn(s) < Al(s)





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