Discussion Forum

1)

 Consider the following reaction, $xMnO_{4}^{-}+yC_{2}O_{4}^{2-}+zH^{+}\rightarrow x Mn^{2+}+2yCO_{2}+\frac{z}{2} H_{2}O$

 The values of x, y, and z in the reaction are respectively

Answer:

Option C

Explanation:

The half equations of the reaction are

   $MnO_{4}^{-}\rightarrow Mn^{2+}$

  $C_{2}O_{4}^{2-}\rightarrow CO_{2}$

  The balanced half equation are

             $MnO_{4}^{-}+8H^{+}+5e^{-}\rightarrow Mn^{2+}+4H_{2}O$

                   $C_{2}O_{4}^{2-}\rightarrow2CO_{2}+2e^{-}$

   On equating number of electrons, we get

 $2MnO_{4}^{-}+16H^{+}+10e^{-}\rightarrow2Mn^{2+}+8H_{2}O$

  $5C_{2}O_{4}^{2-}\rightarrow10CO_{2}+10e^{-}$

 On adding both equations , we get

  $2MnO_{4}^{-}+5C_{2}O_{4}^{-}+16H^{+}\rightarrow 2Mn^{2+}+2\times 5CO_{2}+\frac{16}{2}H_{2}O$

  Thus, x,y, and z are 2,5 and 16 respectively.