A) 0.5
B) 1.0
C) 1.5
D) 2.0
Correct Answer: A
Solution :
\[C{{r}_{2}}O_{7}^{2-}+14{{H}^{+}}+6{{e}^{-}}\xrightarrow[{}]{{}}2C{{r}^{3+}}+7{{H}_{2}}O\] \[\begin{align} & \underline{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Fe{{C}_{2}}{{O}_{4}}\xrightarrow[{}]{{}}F{{e}^{3+}}+2C{{O}_{2}}+3{{e}^{-}}]\times 2} \\ & \underset{1mole}{\mathop{C{{r}_{2}}O_{7}^{2-}}}\,+\underset{2moles}{\mathop{2Fe{{C}_{2}}{{O}_{4}}}}\,+14{{H}^{+}}\xrightarrow{{}} \\ \end{align}\] \[2C{{r}^{3+}}+2F{{e}^{3+}}+4C{{O}_{2}}+7{{H}_{2}}O\] Thus, one mole of\[Fe{{C}_{2}}{{O}_{4}}\]is oxidised by 0.5 mole of\[C{{r}_{2}}O_{7}^{2-}\]in acidic medium.
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