Question

The values of coefficients to balance the following reaction are:
$Cr(OH{)}_3+Cl{O}^{-}+O{H}^{-}\to Cr{O}_4^{2-}+C{l}^{-}+H_{2}O$?

• A. $Cr(OH{)}_3-(2),Cl{O}^{-}-(3),Cr{O}_4^{2-}-(3),C{l}^{-}-(3)$
• B. $Cr(OH{)}_3-(2),Cl{O}^{-}-(4),Cr{O}_4^{2-}-(3),C{l}^{-}-(2)$
• C. $Cr(OH{)}_3-(2),Cl{O}^{-}-(4),Cr{O}_4^{2-}-(4),C{l}^{-}-(2)$
• D. $Cr(OH{)}_3-(2),Cl{O}^{-}-(3),Cr{O}_4^{2-}-(2),C{l}^{-}-(3)$

Answer 1

The correct option is D $Cr(OH{)}_3-(2),Cl{O}^{-}-(3),Cr{O}_4^{2-}-(2),C{l}^{-}-(3)$

$Cr(OH{)}_3+Cl{O}^{-}+O{H}^{-}\to Cr{O}_4^{2-}+C{l}^{-}+H_{2}O$

in a balanced chemical reaction number of all atoms at right side should be equal to the left side of the reaction.

Balancing a chemical reaction as:

Step 1: Assign oxidation numbers to each of the atoms in the equation and write the numbers above the atom as:

$\stackrel{+3}{Cr}(OH{)}_3+\stackrel{+1}{Cl}{O}^{-}+O{H}^{-}\to \stackrel{+6}{Cr}{O}_4^{2-}+\stackrel{-1}{C{l}^{-}}+H_{2}O$

Step 2: Identify the atoms that are oxidized and those that are reduced as:

Reduction: $\stackrel{+1}{Cl}{O}^{-}\to \stackrel{-1}{C{l}^{-}}$

Oxidation: $\stackrel{+3}{Cr}(OH{)}_3\to \stackrel{+6}{Cr}{O}_4^{2-}$

Step 3: oxidation-number change is:

Reduction: $\stackrel{+1}{Cl}{O}^{-}\to \stackrel{-1}{C{l}^{-}}$: Gain of 2 electron

Oxidation: $\stackrel{+3}{Cr}(OH{)}_3\to \stackrel{+6}{Cr}{O}_4^{2-}$: Loss of total 3 electrons

Step 4: Balance the total change in oxidation number as:

Reduction: $\stackrel{+1}{Cl}{O}^{-}\to \stackrel{-1}{C{l}^{-}}\times 3$: Gain of 6 electron

Oxidation: $\stackrel{+3}{Cr}(OH{)}_3\to \stackrel{+6}{Cr}{O}_4^{2-}\times 2$: Loss of total 6 electrons

OR

Reduction: $3\stackrel{+1}{Cl}{O}^{-}\to 3\stackrel{-1}{C{l}^{-}}$

Oxidation: $2\stackrel{+3}{Cr}(OH{)}_3\to 2\stackrel{+6}{Cr}{O}_4^{2-}$

Step 5: Balance O atoms in reduction reaction by adding $H_{2}O$ and then balance H by $H^{+}$ as:

Reduction: $3\stackrel{+1}{Cl}{O}^{-}+6H^{+}\to 3\stackrel{-1}{C{l}^{-}}+3H_{2}O$

Oxidation: $2\stackrel{+3}{Cr}(OH{)}_3+2H_{2}O\to 2\stackrel{+6}{Cr}{O}_4^{2-}+4H^{+}$

Step 6: For base catalysed reaction add $O{H}^{-}$ to both side to neutralize $H^{+}$ as:

Reduction: $3\stackrel{+1}{Cl}{O}^{-}+6H^{+}+6O{H}^{-}\to 3\stackrel{-1}{C{l}^{-}}+3H_{2}O+6O{H}^{-}$

Oxidation: $2\stackrel{+3}{Cr}(OH{)}_3+2H_{2}O+4O{H}^{-}\to 2\stackrel{+6}{Cr}{O}_4^{2-}+4H^{+}+4O{H}^{-}$

OR

Reduction: $3\stackrel{+1}{Cl}{O}^{-}+3H_{2}O\to 3\stackrel{-1}{C{l}^{-}}+6O{H}^{-}$

Oxidation: $2\stackrel{+3}{Cr}(OH{)}_3+4O{H}^{-}\to 2\stackrel{+6}{Cr}{O}_4^{2-}+2H_{2}O$

Thus overall reaction is:

$3\stackrel{+1}{Cl}{O}^{-}+3H_{2}O+2\stackrel{+3}{Cr}(OH{)}_3+4O{H}^{-}\to 3\stackrel{-1}{C{l}^{-}}+6O{H}^{-}+2\stackrel{+6}{Cr}{O}_4^{2-}+2H_{2}O$

OR

$3Cl{O}^{-}+H_{2}O+2Cr(OH{)}_3\to 3C{l}^{-}+2O{H}^{-}+2Cr{O}_4^{2-}$

thus the values of coefficients of balanced reaction are:

$Cr(OH{)}_3-(2),Cl{O}^{-}-(3),Cr{O}_4^{2-}-(2),C{l}^{-}-(3)$