Question

The balanced equation in basic medium is :

$H_2+{ReO}_4^{-}\stackrel{O{H}^{-}}{\to }Re{O}_2+H_{2}O$

• A. $3H_2+2{ReO}_4^{-}\to 2Re{O}_2+2H_{2}O+2O{H}^{-}$
• B. $H_2+3{ReO}_4^{-}\to 2Re{O}_2+3H_{2}O+2O{H}^{-}$
• C. $3H_2+3{ReO}_4^{-}\to 2Re{O}_2+3H_{2}O+2O{H}^{-}$
• D. none of these

Answer 1

The correct option is A $3H_2+2{ReO}_4^{-}\to 2Re{O}_2+2H_{2}O+2O{H}^{-}$

The unbalanced redox equation is as follows:
$H_2+{ReO}_4^{-}\stackrel{O{H}^{-}}{\to }Re{O}_2+H_{2}O$
All atoms other than H and O are balanced..
The oxidation number of Re changes from 7 to 4. The change in the oxidation number of Re is 3.
The oxidation number of H changes from 0 to 1. The change in the oxidation number per H atom is 1. For 2 H atoms, the change in the oxidation number is 2
The increase in the oxidation number is balanced with decrease in the oxidation number by multiplying $Re{O}_4^{-}$ and $Re{O}_2$ with 2 and by multiplying $H_2$ and $CI{O}_2^{-}$ with 3.
$3H_2+2{ReO}_4^{-}\stackrel{O{H}^{-}}{\to }2Re{O}_2+3H_{2}O$
O atoms are balanced by adding 1 water molecule on RHS.
$3H_2+2{ReO}_4^{-}\stackrel{O{H}^{-}}{\to }2Re{O}_2+4H_{2}O$
To balance H atoms, add 2 $H^{+}$ on LHS.
$3H_2+2{ReO}_4^{-}+2H^{+}\stackrel{O{H}^{-}}{\to }2Re{O}_2+4H_{2}O$
Since the reaction occurs in basic medium, add 2 hydroxide ions on both sides of the equation.
$3H_2+2{ReO}_4^{-}+2H^{+}+2O{H}^{-}\stackrel{O{H}^{-}}{\to }2Re{O}_2+4H_{2}O+2O{H}^{-}$
On LHS, 2 $H^{+}$ ions combine with 2 $O{H}^{-}$ ions to form 2 water molecules which cancel out with 2 water molecules on RHS.
$3H_2+2Re{O}_4^{-}\stackrel{O{H}^{-}}{\to }2Re{O}_2+2H_{2}O+2O{H}^{-}$
This is the balanced chemical equation.