Worksheet 4 -Electrochemical Balancing
Half-Cell Method
1. For each of the following, complete the half-reaction equation and classify it as
an oxidation or a reduction.
a)
b)
c)
d)
e)
dinitrogen oxide to nitrogen gas in an acidic solution
nitrite ions to nitrate ions in a basic solution
silver (I) oxide to silver metal in a basic solution
nitrate ions to nitrous acid in an acidic solution
hydrogen gas to water in a basic solution
2. Balance these equations for reactions occurring in acidic solution.
a) Sn + NO3-  SnO2 + NO
c) Fe3+ + NH4OH  Fe2+ + N2O
e) C2O42- + HNO2  CO2 + NO
g) H3PO2 + Cr2O72-  H3PO4 + Cr3+
i) XeF2 + Cl-  Xe + F- + Cl2
b) Ag + NO3-  NO2 + Ag+
d) HNO2 + I-  I2 + NO
f) HNO2 + MnO4-  Mn2+ + NO3h) VO2+ + Sn2+  VO2+ + Sn4+
3. Balance the following equations by the ion-electron method.
a) NBr3  N2 + Br- + HOBr (basic)
c) Cl2  Cl- + ClO3- (basic)
e) MnO2 + SO32-  Mn2+ + S2O62- (acidic)
g) HXeO4-  XeO64- + Xe + O2 (basic)
b) ClNO2  NO3- + Cl- (acidic)
d) H2SeO3 + H2S  S + Se (acidic)
f) XeO3 + I-  Xe + I2 (acidic)
h) CN  CN- + OCN- (basic)
4. Complete and balance the following reaction which occur in basic solution.
a) Al + NO3- + OH-  Al(OH)4- + NH3
c) N2H4 + Cu(OH)2  N2 + Cu
e) ClO- + Fe(OH)3  Cl- + FeO42-
b) PbO2 + Cl-  ClO- + Pb(OH)3d) Ag2S + CN- + O2  S + Ag(CN)2f) HO2- + Cr(OH)3-  CrO42- + OH-
5. Lead(IV) oxide reacts with hydrochloric acid to give chlorine and lead (II) chloride.
Balance the equation using the half reaction method. How many gram of chlorine are
formed from 150 g of PbO2?
6. Manganese(II) ion is oxidized to permanganate ion by bismuthate ion, BiO3-, in an
acidic solution. In the reaction, BiO3- is reduced to Bi+.
a) Write a balanced net ionic equation for the reaction.
b) How many grams of NaBiO3 are needed to oxidize the manganese in 15.0 g of
Mn(NO3)2?
7. A sample of a chromium-containing alloy weighing 3.000 g was dissolved in acid, and
all the chromium in the sample was oxidized to CrO42-. It was then found that 3.09 g of
Na2SO3 was required to reduce the CrO42- to Cr(OH)3 in a basic solution, with the SO32being oxidized to SO42-.
a) Write a balanced equation for the reaction of CrO42- with SO32-.
b) How many moles of CrO42- reacted with Na2SO3?
c) How many grams of Cr were in the alloy sample?
d) What is the weight percentage of Cr in the alloy?
Answers:
1a) 2e- + 2H+ + N2O  N2 + H2O (red) b) 2OH- + NO2- NO3- +2H2O + 2e- (oxid)
c) 2e- + H2O + Ag2O  2Ag + 2OH- (red) d) 2e- + 3H+ + NO3-  HNO2 + H2O (red) e) 2OH- + H2  2H2O + 2e- (oxid)
2a) 4H+ + 3Sn + 4NO3-  3SnO2 + 4NO + 2H2O
b) 2H+ + Ag + NO3-  NO2 + Ag+ + H2O
c) 8Fe3+ + 2NH4OH  8Fe2+ + N2O + 8H+ + H2O
d) 2H+ + 2HNO2 + 2I-  I2 + 2NO + 2H2O
e) 2H+ + C2O42- + 2HNO2  2CO2 + 2NO + 2H2O
f) H+ + 5HNO2 + 2MnO4-  2Mn2+ + 5NO3- + 3H2O
g) 3H3PO2 + 2Cr2O72- + 16H+  3H3PO4 + 4Cr3+ + 8H2O
h) 4H+ + 2VO2+ + Sn2+  2VO2+ + Sn4+ + 2H2O
i) XeF2 + 2Cl-  Xe + 2F- + Cl2
3a) 2NBr3 + 3OH-  N2 + 3Br- + 3HOBr
d) H2SeO3 + 2H2S  2S + Se + 3H2O
f) 6H+ + XeO3 + 6I-  Xe + 3I2 + 3H2O
h) 2CN + 2OH- CN- + OCN- + H2O
b) 2OH- + ClNO2  NO3- + Cl- + H2O
c) 6OH- + 3Cl2  5Cl- + ClO3- + H2O
e) 10H+ + 4MnO2 + 2SO32-  4Mn2+ + S2O62- + 5H2O
g) 2OH- + 2HXeO4-  XeO64- + Xe + O2 + 2H2O
4a) 18H2O + 5OH- + 8Al + 3NO3-  8Al(OH)4- + 3NH3
c) N2H4 + 2Cu(OH)2  N2 + 2Cu + 4H2O
e) 4OH- + 3ClO- + 2Fe(OH)3  3Cl- + 2FeO42- + 5H2O
b) OH- + H2O + PbO2 + Cl-  ClO- + Pb(OH)3d) 2H2O + 2Ag2S + 8CN- + O2  2S + 4Ag(CN)2- + 4OHf) 2HO2- + Cr(OH)3-  CrO42- + OH- + 2H2O
5. 44.5 g
c) 0.850 g
6b) 29.4g
7b) 0.0163 mol
d) 28.3%