| \[F{{e}^{2+}}/Fe\] \[E{}^\circ =-0.44V\] |
| \[F{{e}^{3+}}/F{{e}^{2+}}\], \[E{}^\circ =+0.77V\] |
| \[F{{e}^{2+}},F{{e}^{3+}}\]and Fe bloks are kept together, then: |
A) \[F{{e}^{3+}}increases\]
B) \[F{{e}^{3+}}decreases\]
C) \[F{{e}^{2+}}/F{{e}^{3+}}remains\text{ }unchanged\]
D) \[F{{e}^{2+}}\ decreases\]
Correct Answer: B
Solution :
For the emf to be positive/ the following half-cell reaction will occur. \[\operatorname{Fe}\to F{{e}^{2+}}+2{{e}^{-}}, E{}^\circ =+0.44V\] \[2F{{e}^{3+}}+2{{e}^{-}}\to 2F{{e}^{2+}},\,\,{{E}^{o}}=+0.77V\] Overall reaction: \[\operatorname{Fe}+2F{{e}^{3+}},\to 3F{{e}^{2+}}, E{}^\circ \,cell=0.121V\] \[\operatorname{Hence} F{{e}^{3+}} will decrease\]
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