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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.
Step 8. Add the half-reactions
##”MnO”_2 + color(green)(4″H”^(+)) + color(red)(2)”Cl”^(-)→ “Mn”^(2+) +”Cl”_2 + color(blue)(2″H”_2″O”)##
Step 9. Re-insert the spectator ions
##”MnO”_2 + color(green)(4″H”^(+)) + color(red)(“2Cl”^(-)) + color(red)(2)”Cl”^(-)→ “Mn”^(2+) + color(red)(“2Cl”^(-)) +”Cl”_2 + color(blue)(2″H”_2″O”)##
##”MnO”_2 + 4″HCl” → “MnCl”_2 +”Cl”_2 + 2″H”_2″O”##
Now try to balance the second equation.
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