Question

complete and balance the following redox equation. what is the coefficient of $oh^{-}$ when the equation is balanced using the set of smallest whole - number coefficients? $mno_{4}^{-}+i^{-}\to mno_{2}+io_{3}^{-}$ (basic solution)

Explanation:

Step1: Write half - reactions

Oxidation: \(I^{-}\to IO_{3}^{-}\)
Reduction: \(MnO_{4}^{-}\to MnO_{2}\)

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

Oxidation: \(I^{-}\to IO_{3}^{-}\) (already balanced for \(I\))
Reduction: \(MnO_{4}^{-}\to MnO_{2}\) (already balanced for \(Mn\))

Step3: Balance \(O\) atoms by adding \(H_{2}O\)

Oxidation: \(I^{-}+3H_{2}O\to IO_{3}^{-}\)
Reduction: \(MnO_{4}^{-}\to MnO_{2}+2H_{2}O\)

Step4: Balance \(H\) atoms by adding \(H^{+}\) (but since it's basic, we'll adjust later)

Oxidation: \(I^{-}+3H_{2}O\to IO_{3}^{-}+6H^{+}\)
Reduction: \(MnO_{4}^{-}+4H^{+}\to MnO_{2}+2H_{2}O\)

Step5: Balance charge by adding \(e^{-}\)

Oxidation: \(I^{-}+3H_{2}O\to IO_{3}^{-}+6H^{+}+6e^{-}\)
Reduction: \(MnO_{4}^{-}+4H^{+}+3e^{-}\to MnO_{2}+2H_{2}O\)

Step6: Multiply to equalize \(e^{-}\)

Multiply reduction half - reaction by \(2\): \(2MnO_{4}^{-}+8H^{+}+6e^{-}\to 2MnO_{2}+4H_{2}O\)

Step7: Add half - reactions

\(2MnO_{4}^{-}+I^{-}+3H_{2}O + 8H^{+}\to 2MnO_{2}+IO_{3}^{-}+4H_{2}O+6H^{+}\)

Step8: Simplify \(H^{+}\) and \(H_{2}O\)

\(2MnO_{4}^{-}+I^{-}+2H^{+}\to 2MnO_{2}+IO_{3}^{-}+H_{2}O\)

Step9: Convert to basic (add \(OH^{-}\) to both sides to neutralize \(H^{+}\))

Since \(2H^{+}+2OH^{-}=2H_{2}O\)
\(2MnO_{4}^{-}+I^{-}+2H_{2}O\to 2MnO_{2}+IO_{3}^{-}+2OH^{-}\)

Answer:

\(2\)