Answer:
The E° values of three
metals are listed as follows:
\[Z{{n}^{2+}}\,(aq)+2{{e}^{-}}\,\xrightarrow{\,}\,Zn(s);\] \[{{E}^{o}}=-0.76\,\text{V}\]
\[F{{e}^{2+}}(aq)+2{{e}^{-}}\xrightarrow{\,}\,2e(s);\] \[{{E}^{o}}=-0.44\,\text{V}\]
\[S{{n}^{2+}}(aq)+2{{e}^{-}}\xrightarrow{\,}\,Sn(s);\] \[{{E}^{o}}=-0.14\,\text{V}\]
The data shows that zinc is more
reactive than iron. This means that if a crack appears on the surface of iron
coated with zinc even then zinc will take part in the redox reaction and not
iron. In other words, zinc will be corroded in preference to iron. But same is
not the case with tin. It is less reactive than iron. If a crack appears on the
surface of iron coated with tin, then iron will take part in the redox reaction
and not tin. Therefore, iron will be corroded under these circumstances.
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