Question

The balanced equation in acidic medium is :

$CuS+{NO}_3^{-}\stackrel{H+}{\to }C{u}^{2+}+S_8+NO+H_{2}O$

• A. $24CuS+16{NO}_3^{-}+64H^{+}\to 24C{u}^{2+}+3S_8+16NO+32H_{2}O$
• B. $212CuS+16{NO}_3^{-}+32H^{+}\to 24C{u}^{2+}+3S_8+16NO+32H_{2}O$
• C. $14CuS+8{NO}_3^{-}+64H^{+}\to 24C{u}^{2+}+3S_8+16NO+32H_{2}O$
• D. none of these

Answer 1

The correct option is A $24CuS+16{NO}_3^{-}+64H^{+}\to 24C{u}^{2+}+3S_8+16NO+32H_{2}O$

The unbalanced redox equation is as follows:
$CuS+{NO}_3^{-}\stackrel{H+}{\to }C{u}^{2+}+S_8+NO+H_{2}O$
Balance all atoms other than H and O.
$8CuS+{NO}_3^{-}\stackrel{H+}{\to }8C{u}^{2+}+S_8+NO+H_{2}O$
The oxidation number of S changes from -2 to 0. The change in the oxidation number of S is 2. Total change for 8 S aoms is 16.
The oxidation number of N changes from +5 to +2. The change in the oxidation number is 3.
The increase in the oxidation number is balanced with decrease in the oxidation number by multiplying $CuS$ and $S_8$ with 3 and multiplying ${NO}_3^{-}$ and $NO$ with 16.$C{u}^{2+}$ ions are also multiplied with 3.
$24CuS+16{NO}_3^{-}\stackrel{H+}{\to }24C{u}^{2+}+3S_8+16NO+H_{2}O$
O atoms are balanced by adding 31 water molecules on RHS.
$24CuS+16{NO}_3^{-}\stackrel{H+}{\to }24C{u}^{2+}+3S_8+16NO+32H_{2}O$
Hydrogen atoms are balanced by adding 64 $H^{+}$ atoms on the LHS.
$24CuS+16{NO}_3^{-}+64H^{+}\to 24C{u}^{2+}+3S_8+16NO+32H_{2}O$
This is the balanced chemical equation.