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QUESTION

# 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 --&gt; MnCl2 + Cl2 + H2O b. S + HNO3 ---&gt; 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.