Balancing the following equationCr(OH)3+IO3−→OHI−+CrO42−(or)Balance the above equation basic medium.

EquationOxidation half reaction Cr(OH)3→CrO42−1) Balance the atoms other than H and OCr(OH)3→CrO42−2) Balance of oxygen atomsCr(OH)3+H2O→CrO42−3) Balance of Hydrogen atomsCr(OH)3+H2O+5OH−→CrO42−+5H2O4) Balance of chargesCr(OH)3+5OH−+H2O→CrO42−+5H2O+3e−…(1)Reduction half reaction IO3−→I−1) Balance the atoms other than H and OIO3−→I−2) Balance of oxygen atomsIO3−→I−+3H2O3) Balance of Hydrogen atomsIO3−+6H2O→I−+3H2O+6OH−4) Balance of chargesIO3−+6H2O+6e−→I−+6OH−+3H2O…(2)(1×2)2Cr(OH)3+10OH−+2H2O→2CrO4−2+10H2O+6e−.....(3)IO3−+6H2O+6e−→I−+6OH−+3H2O.....(4)eq(3) + eq(4)2Cr(OH)3+IO3−+8H2+4OH−→2CrO4−2+I−+13H2OSimplify the above equation2Cr(OH)3+IO3−+4OH−→2CrO42−+I−+5H2O

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Balancing the following equation
$Cr (OH)_{3} + {IO}_{3}^{-} \overset{OH}{\to} I^{-} + {CrO}_{4}^{2 -}$
(or)
Balance the above equation basic medium.

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Detailed Solution

Equation
Question Image
Oxidation half reaction $Cr (OH)_{3} \to {CrO}_{4}^{2 -}$
1) Balance the atoms other than H and O
$Cr (OH)_{3} \to {CrO}_{4}^{2 -}$
2) Balance of oxygen atoms
$Cr (OH)_{3} + H_{2} O \to {CrO}_{4}^{2 -}$
3) Balance of Hydrogen atoms
$Cr (OH)_{3} + H_{2} O + 5 {OH}^{-} \to {CrO}_{4}^{2 -} + 5 H_{2} O$
4) Balance of charges
$Cr (OH)_{3} + 5 {OH}^{-} + H_{2} O \to {CrO}_{4}^{2 -} + 5 H_{2} O + 3 e^{-} \dots (1)$
Reduction half reaction ${IO}_{3}^{-} \to I^{-}$
1) Balance the atoms other than H and O
${IO}_{3}^{-} \to I^{-}$
2) Balance of oxygen atoms
${IO}_{3}^{-} \to I^{-} + 3 H_{2} O$
3) Balance of Hydrogen atoms
${IO}_{3}^{-} + 6 H_{2} O \to I^{-} + 3 H_{2} O + 6 {OH}^{-}$
4) Balance of charges
${IO}_{3}^{-} + 6 H_{2} O + 6 e^{-} \to I^{-} + 6 {OH}^{-} + 3 H_{2} O \dots (2)$
$(1 \times 2)$
$2 Cr (OH)_{3} + 10 {OH}^{-} + 2 H_{2} O \to 2 {CrO}_{4}^{- 2} + 10 H_{2} O + 6 e^{-} . . . . . (3)$
${IO}_{3}^{-} + 6 H_{2} O + 6 e^{-} \to I^{-} + 6 {OH}^{-} + 3 H_{2} O . . . . . (4)$
eq(3) + eq(4)
$2 Cr (OH)_{3} + {IO}_{3}^{-} + 8 H_{2} + 4 {OH}^{-} \to 2 {CrO}_{4}^{- 2} + I^{-} + 13 H_{2} O$
Simplify the above equation
$2 Cr (OH)_{3} + {IO}_{3}^{-} + 4 {OH}^{-} \to 2 {CrO}_{4}^{2 -} + I^{-} + 5 H_{2} O$