Electrochemistry I
Balance the following reactions. Then rank based on the total number of electrons that flow
from the oxidation half-reaction to the reduction half-reaction, from the highest to the lowest.
A
Li + Cu2+ → Cu + Li+
B
ClO2− + Fe → Fe3+ + OH- + ClO−
C
H2O2 + H+ + Br- → Br2 + H2O
D
O2 + H+ + Al → H2O + Al3+
E
Ag+ + ClO3− + OH- → ClO4− + Ag
F
Hg22+ + I− → Hg + I2
Highest 1__D__ 2_B___ 3__A_=_ 4_C__= 5__E__=_ 6__F___ Lowest
Explain your reasoning below. Use back if needed.
A
[Li → Li+ + e] x2
Cu2+ + 2 e− → Cu
2Li + Cu2+ → Cu + 2Li+
E [Ag+ + e− → Ag ] x2
ClO3− + H2O → ClO4− + 2 H+ + 2e−
2Ag + ClO3− + H2O → 2Ag + ClO4− + 2 H+
2Ag+ + ClO3− + H2O + 2OH−→ 2Ag + ClO4− + 2H++ 2OH−
2Ag+ + ClO3− + H2O + 2OH−→ 2Ag + ClO4− + 2H2O
2 Ag+ + ClO3− + 2OH− → ClO4− + H2O +2Ag
+
n=2
n=2
B
+
[ClO2 + 2H + 2e → ClO + H2O] x3
[Fe → Fe3+ + 3 e−] x2
−
−
−
3ClO2− + 6H+ + 2Fe → 3ClO− + 3H2O + 2Fe3+
3ClO2− + 6H+ + 6OH− + 2Fe → 3ClO− + 3H2O + 2Fe3+ + 6OH−
3ClO2− + 6H2O + 2Fe → 3ClO− + 3H2O + 2Fe3+ + 6OH−
3ClO2− + 3H2O + 2Fe → 3ClO− + 6OH- + 2Fe3+
n=6
C
2Br- → Br2 + 2e−
H2O2 + 2 H+ + 2e− → 2 H2O
2Br- + H2O2 + 2 H+ →Br2 + 2 H2O
n=2
D
[O2 + 4 H+ + 4e− → 2 H2O] x3
Al → Al3+ + 3e−] x4
4Al + 3O2 + 12H+ → 6H2O + 4Al3+
n = 12
F
Hg22+ + 2e− →2 Hg
2I- → I2 + 2e−
Hg22+ + 2I− →2 Hg + I2
n=2