Question

complete and balance the following redox equation. when properly balanced using the smallest whole - number coefficients, the coefficient of s is h₂s + hno₃ → s + no (acidic solution) add your answer integer, decimal, or e notation allowed question 2 20 points complete and balance the following redox equation. what is the coefficient of h₂o when the equation is balanced using the set of smallest whole - number coefficients? mno₄⁻ + so₃²⁻ → mn²⁺ + so₄²⁻ (acidic solution) add your answer integer, decimal, or e notation allowed

Explanation:

Question 1

Step1: Write half - reactions

Oxidation: \(H_2S
ightarrow S\)
Reduction: \(HNO_3
ightarrow NO\)

Step2: Balance atoms other than \(O\) and \(H\)

Oxidation: \(H_2S
ightarrow S\) (already balanced for \(S\))
Reduction: \(HNO_3
ightarrow NO\) (already balanced for \(N\))

Step3: Balance \(O\) atoms by adding \(H_2O\)

Oxidation: \(H_2S
ightarrow S\) (no \(O\) atoms)
Reduction: \(HNO_3
ightarrow NO + 2H_2O\)

Step4: Balance \(H\) atoms by adding \(H^+\)

Oxidation: \(H_2S
ightarrow S+2H^+\)
Reduction: \(HNO_3 + 3H^+
ightarrow NO + 2H_2O\)

Step5: Balance charge by adding electrons

Oxidation: \(H_2S
ightarrow S+2H^++2e^-\)
Reduction: \(HNO_3 + 3H^++3e^-
ightarrow NO + 2H_2O\)

Step6: Make electron gain equal to electron loss

Multiply oxidation half - reaction by \(3\): \(3H_2S
ightarrow 3S+6H^++6e^-\)
Multiply reduction half - reaction by \(2\): \(2HNO_3 + 6H^++6e^-
ightarrow 2NO + 4H_2O\)

Step7: Add half - reactions

\(3H_2S+2HNO_3
ightarrow 3S + 2NO+4H_2O\)

Step1: Write half - reactions

Oxidation: \(SO_3^{2 -}
ightarrow SO_4^{2 -}\)
Reduction: \(MnO_4^-
ightarrow Mn^{2+}\)

Step2: Balance atoms other than \(O\) and \(H\)

Oxidation: \(SO_3^{2 -}
ightarrow SO_4^{2 -}\) (already balanced for \(S\))
Reduction: \(MnO_4^-
ightarrow Mn^{2+}\) (already balanced for \(Mn\))

Step3: Balance \(O\) atoms by adding \(H_2O\)

Oxidation: \(SO_3^{2 -}+H_2O
ightarrow SO_4^{2 -}\)
Reduction: \(MnO_4^-
ightarrow Mn^{2+}+4H_2O\)

Step4: Balance \(H\) atoms by adding \(H^+\)

Oxidation: \(SO_3^{2 -}+H_2O
ightarrow SO_4^{2 -}+2H^+\)
Reduction: \(MnO_4^-+8H^+
ightarrow Mn^{2+}+4H_2O\)

Step5: Balance charge by adding electrons

Oxidation: \(SO_3^{2 -}+H_2O
ightarrow SO_4^{2 -}+2H^++2e^-\)
Reduction: \(MnO_4^-+8H^++5e^-
ightarrow Mn^{2+}+4H_2O\)

Step6: Make electron gain equal to electron loss

Multiply oxidation half - reaction by \(5\): \(5SO_3^{2 -}+5H_2O
ightarrow 5SO_4^{2 -}+10H^++10e^-\)
Multiply reduction half - reaction by \(2\): \(2MnO_4^-+16H^++10e^-
ightarrow 2Mn^{2+}+8H_2O\)

Step7: Add half - reactions

\(5SO_3^{2 -}+2MnO_4^-+6H^+
ightarrow 5SO_4^{2 -}+2Mn^{2+}+3H_2O\)

Answer:

\(3\)

Question 2