question 5 of 17 > in a particular redox reaction, mno₂ is oxidized to mno₄⁻ and ag⁺ is reduced to ag…

Explanation:

Step1: Write half - reactions

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

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

Already balanced for Mn: $\ce{MnO2 -> MnO4^-}$

Step3: Balance O atoms in oxidation half - reaction by adding $\ce{H2O}$

$\ce{MnO2 + 2H2O -> MnO4^-}$

Step4: Balance H atoms in oxidation half - reaction by adding $\ce{H+}$

$\ce{MnO2 + 2H2O -> MnO4^-+ 4H+}$

Step5: Balance charge in oxidation half - reaction by adding electrons

$\ce{MnO2 + 2H2O -> MnO4^-+ 4H+ + 3e^-}$

Step6: Balance charge in reduction half - reaction

$\ce{Ag+ + e^- -> Ag}$

Step7: Multiply half - reactions to equalize electrons

Multiply oxidation half - reaction by 1 and reduction half - reaction by 3. Oxidation: $\ce{MnO2 + 2H2O -> MnO4^-+ 4H+ + 3e^-}$ Reduction: $\ce{3Ag+ + 3e^- -> 3Ag}$

Step8: Add half - reactions

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

Answer:

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