A) 3 : 1
B) 2 : 1
C) 1 : 1
D) 3 : 2
Correct Answer: D
Solution :
At cathode; \[F{{e}^{2+}}+2{{e}^{-}}\to Fe\] ; \[F{{e}^{3+}}+3{{e}^{-}}\to Fe\] \[{{({{E}_{Fe}})}_{1}}=\frac{\text{Atomic}\text{.weight}}{2};\,\,{{({{E}_{Fe}})}_{2}}=\frac{\text{Atomic}\text{.weight}}{3}\] Ratio of weight of Fe liberated \[=\frac{\text{Atomic weight}}{3}:\,\frac{\text{Atomic weight}}{2}=\,3\,\,:\,\,2\].You need to login to perform this action.
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