# Can you write the equation for the oxidation and reduction half-reactions for the redox reactions below, and then balance the reaction equations? a. MnO2 + HCl –> MnCl2 + Cl2 + H2O b. S + HNO3 —> SO3 + H2O + NO2

See below.

Oxidation half-reaction: ##”2Cl”^(-) → “Cl”_2 + “2e”^(-)##
Reduction half-reaction: ##”MnO”_2 + “4H”^(+) + “2e”^(-) → “Mn”^(2+) + “2H”_2″O”##
Balanced equation: ##”MnO”_2 + “4HCl” → “MnCl”_2 + “Cl”_2 + “2H”_2″”##

Oxidation half-reaction: ##”S” + “3H”_2″O” → “SO”_3 + “6H”^(+) + “6e”^(-)##
Reduction half-reaction: ##”NO”_3^(-)+ “2H”^(+) + “e”^(-) → “NO”_2 + “H”_2″O”##
Balanced equation: ##”S” + “6HNO”_3 → “SO”_3 + “6NO”_2 + “3H”_2″O”##

Here’s how you do get the answers by the for Part a.
Then see if you can do Part b.

Step 1. Write the net ionic equation
Omit all spectator ions (##”Cl”^-## is both a reactant and a spectator ion). Also omit ##”H”^+##, ##”OH”^-##, and ##”H”_2″O”## (they come in automatically during the balancing procedure).
##”MnO”_2 + “Cl”^(-) → “Mn”^(2+) + “Cl”_2##

Step 2. Split into half-reactions
##”MnO”_2 → “Mn”^(2+)##
##”Cl”^(-) → “Cl”_2##

Step 3. Balance atoms other than ##”H”## and ##”O”##
##”MnO”_2 → “Mn”^(2+)##
##color(red)(2)”Cl”^(-) → “Cl”_2##

Step 4. Balance ##”O”##
##”MnO”_2 → “Mn”^(2+) + color(blue)(2″H”_2″O”)##
##color(red)(2)”Cl”^(-) → “Cl”_2##

Step 5. Balance ##”H”##
##”MnO”_2 + color(green)(4″H”^(+)) → “Mn”^(2+) + color(blue)(2″H”_2″O”)##
##color(red)(2)”Cl”^(-) → “Cl”_2##

Step 6. Balance charge
##”MnO”_2 + color(green)(4″H”^(+)) + color(cyan)(2″e”^(-)) → “Mn”^(2+) + color(blue)(2″H”_2″O”)##
##color(red)(2)”Cl”^(-) → “Cl”_2+ color(cyan)(2″e”^(-))##

Step 7. Equalize electrons transferred
Done.