Consider the following reaction :
`xMnO_(4)^(-)+yC_(2)O_(4)^(2-)+zH^(+) to xMn^(2+)+2yCO_(2)+(z)/(2)H_(2)O`
The value of x, y and z in the reaction are, respectively.

Options

A. 5,2 and 8
B. 5,2 and 16
C. 2,5 and 8
D. 2,5 and 16

Answer

d

Explanation

Reduction half reaction
`MnO_(4)^(-)+8H^()+5e^(-)rarrMn^(2+)+4H_(2)O`
Oxidation half reaction `C_(2)^(+3)O_(4)^(2-)rarrCO_(2)`
balance c atoms `C_(2)O_(4)^(2-)rarr2CO_(2)`
balance O.N by adding `2e^(-)` to R.H.S of Eq (ii) we have
`C_(2)O_(4)^(21-)rarr2CO_(2)+2e^(-)`
charge is automically balanced to cancel eletrons multily eq (i) by 2 and eq (iii) by 5 and adding we have
`2MnO_(4)^(-)+5C_(2)O_(4)^(2-)+16H^(+)rarr2Mn^(2+)+10CO_(2)+8H_(2)O`