# Reviewing Electrochemistry

Chem I | Chem II AP
1) Given: Fe/Fe+2 (1 M) // MnO4- (1 M), Mn+2/Pt        (acidic environment)
a)
Diagram/label this voltaic cell.
 A: anode cathode electrons positive ions negative ions Pt metal Fe metal Fe+2 in solution MnO4- and Mn+2 in solution B: anode cathode electrons positive ions negative ions Pt metal Fe metal Fe+2 in solution MnO4- and Mn+2 in solution C: anode cathode electrons positive ions negative ions Pt metal Fe metal Fe+2 in solution MnO4- and Mn+2 in solution D: anode cathode electrons positive ions negative ions Pt metal Fe metal Fe+2 in solution MnO4- and Mn+2 in solution E: anode cathode electrons positive ions negative ions Pt metal Fe metal Fe+2 in solution MnO4- and Mn+2 in solution F: anode cathode electrons positive ions negative ions Pt metal Fe metal Fe+2 in solution MnO4- and Mn+2 in solution G: anode cathode electrons positive ions negative ions Pt metal Fe metal Fe+2 in solution MnO4- and Mn+2 in solution H: anode cathode electrons positive ions negative ions Pt metal Fe metal Fe+2 in solution MnO4- and Mn+2 in solution I: anode cathode electrons positive ions negative ions Pt metal Fe metal Fe+2 in solution MnO4- and Mn+2 in solution
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? left (to Fe) right (to Pt)
e) In which direction do the electrons flow? left (to Fe) right (to Pt)
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
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 Zn Li F- H2 2 Zn Li F- H2 3 Zn Li F- H2 (strong) 4 Zn Li F- H2

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+ Cs F2 Cs+ Br2
b) Substance which will cause Au to be oxidized Cs F2 Cs+ Br2
c) Substance which is the strongest reducing agent Cs F2 Cs+ Br2
d) Substance most easily oxidized Cs F2 Cs+ Br2
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? left right increase 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