Consider the following reaction :
xMNO₄⁻ + yC₂O₄²⁻ + zH⁺ → xMn²⁺ + 2yCO₂ + z/2 H₂O
The value’s of x, y and z in the reaction are, respectively :
(a) 5, 2 and 16 (b) 2, 5 and 8
(c) 2, 5 and 16 (d) 5, 2 and 8
Answers
Answered by
1
Answer:
As per the question values are 1,5&8 respectively.
Attachments:
Answered by
0
x, y and z are 2, 5 and 16 respectively.
Option (C) is correct.
Explanation:
- The half-equations of the reaction are
MnO4- → Mn2+
C2O42- → CO2
- Half balanced equations are:
( MnO4- 8H+ +5e- ----->Mn2+ + 4H2O ) x 2
( C2O4^2- -----> 2CO2 + 2e- ) x 5
Equating number of electrons, we get
2MnO4^2- + 16H+ + 10e- ---> 2Mn2+ + 8H2O
5C2O4^2- ----> 10 CO2 + 10e-
On adding equations we get.
2MnO4^2- + 5C2O4^2- + 16H+ --- > 2Mn2+ + 2 x 5 CO2 + 16 / 2 H2O
Thus x, y and z are 2, 5 and 16 respectively.
Also learn more
What is half cell reaction ? definition?
https://brainly.in/question/569365
Similar questions