question 13 of 17 > attempt for a particular redox reaction, mno₂ is oxidized to mno₄⁻ and ag⁺ is reduced to…

Explanation:

Step1: Write half - reactions

Oxidation: $\ce{MnO2 -> MnO4^{-}}$ Reduction: $\ce{Ag^{+} -> Ag}$

Step2: Balance atoms other than O and H in half - reactions

Oxidation: already balanced for Mn. Reduction: already balanced for Ag.

Step3: Balance O atoms by adding $\ce{H2O}$

Oxidation: $\ce{MnO2 + 2H2O -> MnO4^{-}}$ Reduction: no O atoms to balance.

Step4: Balance H atoms by adding $\ce{H^{+}}$

Oxidation: $\ce{MnO2 + 2H2O -> MnO4^{-}+ 4H^{+}}$ Reduction: no H atoms to balance.

Step5: Adjust for basic solution by adding $\ce{OH^{-}}$

Oxidation: $\ce{MnO2 + 2H2O+4OH^{-} -> MnO4^{-}+ 4H^{+}+4OH^{-}}$ which simplifies to $\ce{MnO2 + 4OH^{-} -> MnO4^{-}+ 2H2O}$ Reduction: no change as there are no $\ce{H^{+}}$ ions.

Step6: Balance charge in half - reactions

Oxidation: $\ce{MnO2 + 4OH^{-} -> MnO4^{-}+ 2H2O + 3e^{-}}$ Reduction: $\ce{Ag^{+}+e^{-} -> Ag}$

Step7: Multiply half - reactions to equalize electrons

Multiply reduction reaction by 3: $\ce{3Ag^{+}+3e^{-} -> 3Ag}$ Oxidation: $\ce{MnO2 + 4OH^{-} -> MnO4^{-}+ 2H2O + 3e^{-}}$

Step8: Add half - reactions

$\ce{MnO2 + 3Ag^{+}+ 4OH^{-} -> MnO4^{-}+ 3Ag+ 2H2O}$

Answer:

$\ce{MnO2 + 3Ag^{+}+ 4OH^{-} -> MnO4^{-}+ 3Ag+ 2H2O}$