A) 2, 5, 16
B) 8, 2, 5
C) 5, 2, 16
D) 5, 8, 4
Correct Answer: A
Solution :
[a] \[XMnO_{4}^{-}+Y{{C}_{2}}O_{4}^{2-}+Z{{H}^{+}}\rightleftharpoons \] \[XM{{n}^{+2}}+2YC{{O}_{2}}+\frac{Z}{2}{{H}_{2}}O\] First half reaction \[MnO_{4}^{-}\xrightarrow{{}}M{{n}^{+2}}\] .... (i) On balancing \[MnO_{4}^{-}+8{{H}^{+}}+5{{e}^{-}}\to M{{n}^{+2}}+4{{H}_{2}}O\] .... (ii) Second half reaction \[{{C}_{2}}O_{4}^{2-}\xrightarrow{{}}2C{{O}_{2}}\] ....(iii) On balancing \[{{C}_{2}}O_{4}^{2-}\xrightarrow{{}}2C{{O}_{2}}+2{{e}^{-}}~\] .... (iv) On multiplying eqn. (ii) by 5 and (iv) by 2 and then adding we get \[2MnO_{4}^{-}+5{{C}_{2}}O_{4}^{2-}+16{{H}^{+}}\xrightarrow{{}}2M{{n}^{+2}}+10C{{O}_{2}}+8{{H}_{2}}O\]
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