homework: balance the redox reactions in both acids basic media 1) no + h₅io₆ → no₃⁻ + io₃⁻

Explanation:

Step1: Write half - reactions in acidic medium

Oxidation: $NO\rightarrow NO_{3}^{-}$ Reduction: $H_{5}IO_{6}\rightarrow IO_{3}^{-}$

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

$NO\rightarrow NO_{3}^{-}$ (N is already balanced)

Step3: Balance O by adding $H_{2}O$ in oxidation half - reaction

$NO + 2H_{2}O\rightarrow NO_{3}^{-}$

Step4: Balance H by adding $H^{+}$ in oxidation half - reaction

$NO + 2H_{2}O\rightarrow NO_{3}^{-}+4H^{+}$

Step5: Balance charge by adding $e^{-}$ in oxidation half - reaction

$NO + 2H_{2}O\rightarrow NO_{3}^{-}+4H^{+}+3e^{-}$

Step6: Balance atoms other than O and H in reduction half - reaction

$H_{5}IO_{6}\rightarrow IO_{3}^{-}$ (I is already balanced)

Step7: Balance O by adding $H_{2}O$ in reduction half - reaction

$H_{5}IO_{6}\rightarrow IO_{3}^{-}+2H_{2}O$

Step8: Balance H by adding $H^{+}$ in reduction half - reaction

$H_{5}IO_{6}+H^{+}\rightarrow IO_{3}^{-}+2H_{2}O$

Step9: Balance charge by adding $e^{-}$ in reduction half - reaction

$H_{5}IO_{6}+H^{+}+2e^{-}\rightarrow IO_{3}^{-}+2H_{2}O$

Step10: Make the number of electrons equal

$2(NO + 2H_{2}O\rightarrow NO_{3}^{-}+4H^{+}+3e^{-})$ $3(H_{5}IO_{6}+H^{+}+2e^{-}\rightarrow IO_{3}^{-}+2H_{2}O)$

Step11: Add the two half - reactions in acidic medium

$2NO+3H_{5}IO_{6}\rightarrow 2NO_{3}^{-}+3IO_{3}^{-}+3H_{2}O + 5H^{+}$

Step12: Convert to basic medium

Add $OH^{-}$ to both sides to neutralize $H^{+}$ $2NO+3H_{5}IO_{6}+5OH^{-}\rightarrow 2NO_{3}^{-}+3IO_{3}^{-}+8H_{2}O$

Answer:

Acidic medium: $2NO+3H_{5}IO_{6}\rightarrow 2NO_{3}^{-}+3IO_{3}^{-}+3H_{2}O + 5H^{+}$ Basic medium: $2NO+3H_{5}IO_{6}+5OH^{-}\rightarrow 2NO_{3}^{-}+3IO_{3}^{-}+8H_{2}O$