A) \[2{{H}^{1+}}=2{{e}^{-}}\xrightarrow{{}}{{H}_{2}}(g)\,(+0.000)\]
B) \[Z{{n}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Zn(s)\,(-0.762)\]
C) \[F{{e}^{3+}}+{{e}^{-}}\xrightarrow{{}}F{{e}^{2+}}(s)\,\,(-0.44)\]
D) \[C{{r}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Cr(s)\,\,(-0.540)\]
Correct Answer: B
Solution :
Key Idea Metal with least reduction potential is best to be used as anode. Because the substance which have lower reduction potentials are stronger reducing agents. The reduction potential of \[Z{{n}^{2+}}/Zn\] is minimum in the given reduction potential of half-reduction reactions. So, it is best suited to be used as anode. Note - Reduction occurs at cathode (or the electrode having higher reduction potentials). Oxidation occurs at anode (or the electrode having lower reduction potentials).
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