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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.

"MnO"_2 + color(green)(4"H"^(+)) + color(red)(2)"Cl"^(-)→ "Mn"^(2+) +"Cl"_2 + color(blue)(2"H"_2"O")
"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"