MCQ
The relation between Faraday constant $(F)$, chemical equivalent $(E)$ and electrochemical equivalent $(Z)$ is
- A$F = EZ$
- B$F = \frac{Z}{E}$
- ✓$F = \frac{E}{Z}$
- D$F = \frac{E}{Z^2}$
where $E$ is chemical equivalent, $Z$ is electrochemical equivalent and $F$ is faraday constant.
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Reason : Iodine is a polar compound.

$ {I_2} + 2{e^ - } \to \,2{I^ - }\,;\,\,{E^o}\, = \,\,0.54\,\,V $
$ MnO_4^ - \, + \,8{H^ + }\, + \,5{e^ - } \to \,M{n^{2 + }}\, + \,4{H_2}O\,;\,{E^{o\,}} = 1.52\,\,V $
$ F{e^{3 + }} + {e^ - } \to \,\,F{e^{2 + }}\,;\,\,{E^o}\, = \,\,0.77\,\,V $
$ S{n^{4 + }} + 2{e^ - } \to \,\,S{n^{2 + }}\,;\,\,{E^o}\, = \,\,0.1\,\,V $
The strongest reducant and oxidant respectively are