Answer:
(i) Oxidising agent. Water oxidises a number of highly electropositive metals whose electrode potential is less than \[-\] 0.83 V liberating \[{{H}_{2}}\] gas. For example, sodium with\[E{}^\circ =-2.71\,\text{V}\] reduces \[{{H}_{2}}\] to \[{{H}_{2}}\] gas
\[\underset{\text{(Reductant)}}{\mathop{2\,Na\,(s)}}\,+\underset{\text{(Oxidant)}}{\mathop{2\,{{H}_{2}}O(l)}}\,\,\xrightarrow{\,}2N{{a}^{+}}\,(aq)+\] \[2\,O{{H}^{-}}(aq)+{{H}_{2}}(g)\]
(ii) Reducing agent. Water reduces a number of electronegative elements whose electrode potential is higher than + 1.23 V liberating either \[{{O}_{2}}\] or \[{{O}_{3}}\] gas. For example, fluorine with \[E{}^\circ =+2.87\,\text{V}\] oxidises \[{{H}_{2}}O\] to \[{{O}_{2}}\] gas
\[\underset{\text{(Oxidant)}}{\mathop{2{{F}_{2}}(g)}}\,+\underset{\text{(Reductant)}}{\mathop{2{{H}_{2}}O(l)}}\,\xrightarrow{\,}\,{{O}_{2}}(g)+4\,{{H}^{+}}(aq)\]\[+4{{F}^{-}}(aq)\]
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