Reviewing Electrochemistry

Chem I | Chem II AP
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1)
Given: Fe/Fe+2 (1 M) // MnO4- (1 M), Mn+2/Pt        (acidic environment)
a)
Diagram/label this voltaic cell.
A:
B:
C:
D:
E:
F:
G:
H:
I:
b)
Write a balanced equation. Calculate E°.
Fe
+
H+
+
MnO4-
Fe+2
+
Mn+2
+
H2O
E° =
5 ( Fe → Fe+2 + 2e- ) E° = 0.44 V
2 ( 5e- + 8 H+ + MnO4- → Mn+2 + 4 H2O ) E° = 1.51 V
5 Fe + 16 H+ + 2 MnO4- → 5 Fe+2 + 2 Mn+2 + 8 H2O E° = 0.44 + 1.51 = 1.95 V
c)
What is the oxidation number of Mn in MnO4-?
d)
If the salt bridge contains KNO3, to which side will the NO3- ion flow?
e)
In which direction do the electrons flow?
f)
Find E if: [Fe+2] = 0.010; [MnO4-] = 6.0; [Mn+2] = 0.10; [H+] = 1.0
V
E = E° -
0.0592
n
log (Q) = 1.95 -
0.0592
10
log(
(0.01)5 (0.1)2
(1)6 (6)2
) = 1.95 -
0.0592
10
log (2.78 x 10-14) = 1.95 - -0.08 = 2.03 v
g)
Find ΔG° at 25° C.
kJ
ΔG° = - n F E° = - (10 mol e-)(96,500 C/mol e-)(1.95 v) = - 1881750 J = - 1882 kJ
h)
Find K at 25° C.
K = 10x
   
x =
E° =
0.0592
n
log K
1.95 v =
0.0592
10 mol e-
log K
log K = 329.4 K = 10329.4
2)
10.0 g of copper are to be plated from a solution of CuSO4. How many minutes are required for this process if a current of 5 A is used?
min
Cu+2 + 2 e- → Cu
10.0 g Cu x
1 mol Cu
63.5 g Cu
x
2 mol e-
1 mol Cu
x
96500 C
1 mol e-
x
1 sec
5 C
x
1 min
60 sec
= 101.3 min
3)
A current of 5.0 A is passed through a cell containing Ni+2 for 1.5 hours. How many grams of Ni metal will be plated out on the cathode?
g Ni
Ni+2 + 2 e- → Ni
1.5 hr
3600 sec
1 hr
x
5.0 C
1 sec
x
1 mol e-
96500 C
x
1 mol Ni
2 mol e-
x
58.7 g Ni
1 mol Ni
= 8.2 g Ni
4)
Electrolysis of a molten metal chloride (MCl3) using a current of 4.19 A for 1397 seconds deposits 1.41 g of the metal at the cathode. Identify the metal.
M+3 + 3 e- → M
1397 sec x
4.19 C
1 sec
x
1 mol e-
96500 C
x
1 mol M
3 mol e-
= 0.202 mol M
g-MM =
mass
mol
=
1.41 g
0.202 mol
= 69.7 g/mol = molar mass of Ga
5)
a)
MnO4- + F- → F2 + Mn+2
E° =
   
Spont?
Oxidizing agent:
   
Reducing agent:
E° = 1.51 + -2.87 = -1.36 v
Since E° is negative, the reaction is not spontaneous.
Since Mn in MnO4- is +7 and is going to +2, MnO4- is being reduced and is the oxidizing agent.
Since F- is going to F2, F- is being oxidized and is the reducing agent.
b)
F2
+
KCl
Cl2
+
KF
E° =
   
Spont?
Oxidizing agent:
   
Reducing agent:
   
Spectator ion:
E° = 2.87 + -1.36 = 1.51 v
Since E° is positive, the reaction is spontaneous.
Since F2 is going to F-, F2 is being reduced and is the oxidizing agent.
Since Cl- is going to Cl2, Cl- is being oxidized and is the reducing agent.
Since K+ appears on each side it is the spectator ion.
6)
Given:
Br2 + 2 e- → 2 Br-
 
Al+3 + 3e- → Al
a)
Write a balanced equation. Calculate E°.
Br2
+
Al
Br-
+
Al+3
E° =
V
Multiply the bromine equation by 3 and the aluminum equation by 2 to make the number of electrons cancel.
E° = 1.09 + 1.66 = 2.75 v
b)
Find E when [Br2] = 2.0; [Br-] = 0.10; [Al+3] = 0.20.
   
E° =
V
E = E° -
0.0592
n
log (Q) = 2.75 -
0.0592
6
log(
(0.2)2 (0.1)6
(2)3
) = 1.95 -
0.0592
10
log (1.48 x 10-9) = 2.75 - -0.087 = 2.84 v
7)
Given: Al+3 + Mg → Mg+2 + Al
   
E° =
V
E° = -1.66 + 2.37 = 0.71 v
a)
Find ΔG°.
kJ
ΔG° = - n F E° = - (6 mol e-)(96,500 C/mol e-)(0.71 v) = - 411090 J = - 411 kJ
b)
Find K.
E° =
0.0592
n
log K
0.71 v =
0.0592
6 mol e-
log K
log K = 71.9  K = 1071.9 = 9.1 x 1071
8)
List the following in order of increasing strength as reducing agents:
   
Zn, Li, F-, H2
(weak)
1
2
3
(strong)
4

Increasing strengths as reducing agents means 'in the order they will be oxidized'. Li is at the bottom of the chart and is the easiest oxidized. F- is at the top of the chart and is the easiest reduced.

9)
Given: Cs, F2, Cs+, Br2
a)
Substance which will reduce K+
b)
Substance which will cause Au to be oxidized
c)
Substance which is the strongest reducing agent
d)
Substance most easily oxidized
10)
Chlorine dioxide (ClO2) is produced by the following reaction:
2 NaClO2 + Cl2 → 2 ClO2 + 2 NaCl
It has been tested as a disinfectant for municipal waste treatment. Calculate E°, ∆G° and K at 25°C.
E° =
E° = -0.954 + 1.36 = 0.406 v
ΔG° =
ΔG° = - n F E° = - (2 mol e-)(96,500 C/mol e-)(0.406 v) = - 78358 J = - 78.36 kJ
K =
E° =
0.0592
n
log K
0.406 v =
0.0592
2 mol e-
log K
log K = 13.7  K = 1013.7 = 5.2 x 1013
11)
Given: Al + Fe+2 → Al+3 + Fe
a)
Write the line notation for the voltaic cell represented above. Calculate E°.
/
//
/
E° =
V
E° = 1.66 + -0.44 = 1.22 v
Line notation goes from what is oxidized to what is reduced.
b)
If the Al+3 concentration is increased from 1.0 M to 3.0 M, will the voltage increase or decrease?
Equilibrium will shift to the left in the nonspontaneous direction.
c)
If the Fe+2 concentration is decreased from 1.0 M to 0.1 M, will the equilibrium shift right or left? Will the voltage increase or decrease?
d)
Why does the voltage of a voltaic cell (battery) decrease as time passes?
12)
Balance: (basic environment)
MnO4- → MnO2
Fe+2 → Fe+3
H2O
+
MnO4-
+
Fe+2
MnO2
+
OH-
+
Fe+2
3 ( Fe+2 → Fe+3 + e- )
3e- + 4 H+ + MnO4- → MnO2 + 2 H2O
4 OH- + 3 Fe+2 + 4 H+ + MnO4- → 3 Fe+3 + MnO2 + 2 H2O + 4 OH-
2 H2O + 3 Fe+2 + MnO4- → 3 Fe+3 + MnO2 + 4 OH-
13)
How many mL of 0.20 M H2O2 solution are required to react with 35.0 mL of 0.25 M K2Cr2O7? (acidic environment)
H2O2 → O2
Cr2O7-2 → Cr+3
mL
3 ( H2O2 → O2 + 2 H+ + 2 e- )
6e- + 14 H+ + Cr2O7-2 → 2 Cr+3 + 7 H2O
3 H2O2 + 8 H+ + Cr2O7-2 → 2 Cr+3 + 7 H2O + 3 O2
0.25 M Cr2O7-2 =
x mol
0.35 L
 x = 0.00875 mol Cr2O7-2
0.00875 mol Cr2O7-2 x
3 mol H2O2
1 mol 0.00875 mol Cr2O7-2
= 0.02625 mol H2O2
0.2 M H2O2 =
0.02625 mol H2O2
x L
 x = 0.131 L = 131 mL