| \[C{{r}_{2}}O_{7}^{2-}+F{{e}^{2+}}+{{C}_{2}}O_{4}^{2-}\to C{{r}^{3+}}+F{{e}^{3+}}+C{{O}_{2}}\](Unbalanced) |
A) 3
B) 4
C) 6
D) 5
Correct Answer: A
Solution :
[a] The reaction given is \[C{{r}_{2}}{{O}_{7}}^{2-}+F{{e}^{2+}}+{{C}_{2}}{{O}_{2}}^{2-}\xrightarrow{{}}C{{r}^{3+}}+F{{e}^{3+}}+C{{O}_{2}}\]\[C{{r}_{2}}{{O}_{7}}^{2-}\xrightarrow{{}}2C{{r}^{3+}}\] On balancing \[14{{H}^{+}}+C{{r}_{2}}{{O}_{7}}^{2-}\xrightarrow{{}}2C{{r}^{3+}}+7{{H}_{2}}O\] .....(i) \[F{{e}^{2-}}\xrightarrow{{}}F{{e}^{3+}}+{{e}^{-}}\] ....(ii) \[{{C}_{2}}{{O}_{4}}^{2-}\xrightarrow{{}}2C{{O}_{2}}+2{{e}^{-}}\] ...(iii) On adding all the three equations. \[C{{r}_{2}}{{O}_{7}}^{2-}+F{{e}^{2+}}+{{C}_{2}}O_{4}^{2-}+14{{H}^{+}}+3{{e}^{-}}\] \[\xrightarrow{{}}2C{{r}^{3+}}+F{{e}^{3+}}+2C{{O}_{2}}+7{{H}_{2}}O\] Hence the total no. of electrons involved in the reaction = 3
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