A) \[M{{n}^{2+}}\] - 5 \[C{{r}_{2}}O_{4}^{2-}\]-2 \[C{{O}_{2}}\]- 4 \[{{H}^{+}}\]-10
B) \[M{{n}^{2+}}\] -2 \[C{{r}_{2}}O_{4}^{2-}\]- 5 \[C{{O}_{2}}\]-10 \[{{H}^{+}}\]-16
C) \[M{{n}^{2+}}\] -6 \[C{{r}_{2}}O_{4}^{2-}\]-8 \[C{{O}_{2}}\]- 16 \[{{H}^{+}}\]-18
D) \[M{{n}^{2+}}\] - 10 \[C{{r}_{2}}O_{4}^{2-}\]- 12 \[C{{O}_{2}}\]-24 \[{{H}^{+}}\]-12
Correct Answer: B
Solution :
\[MnO_{4}^{-}+{{C}_{2}}O_{4}^{2-}+{{H}^{+}}\xrightarrow[{}]{{}}M{{n}^{2+}}+C{{O}_{2}}+{{H}_{2}}O\] \[{{C}_{2}}O_{4}^{2-}\xrightarrow[{}]{{}}2C{{O}_{2}}+2{{e}^{-}}\] (oxidation) ...(i) \[MnO_{4}^{-}+8{{H}^{+}}+5{{e}^{-}}\xrightarrow[{}]{{}}M{{n}^{2+}}+4{{H}_{2}}O\] (reduction) ...(ii) Equation (i) is multiplied by 5 and equation (ii) is multiplied by 2 and then on adding \[5{{C}_{2}}O_{4}^{2-}\xrightarrow[{}]{{}}10C{{O}_{2}}+10{{e}^{-}}\] \[\begin{align} & \underline{2MnO_{4}^{-}+16{{H}^{+}}+10{{e}^{-}}\xrightarrow[{}]{{}}2M{{n}^{2+}}+8{{H}_{2}}O} \\ & 2MnO_{4}^{-}+5{{C}_{2}}O_{4}^{2-}+16{{H}^{+}}\xrightarrow[{}]{{}} \\ \end{align}\] \[2M{{n}^{2+}}+10C{{O}_{2}}+8{{H}_{2}}O\]
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