question_answer

Given that \[C{{o}^{3+}}+{{e}^{-}}\xrightarrow{\,}C{{o}^{2+}};\,\,{{E}^{o}}=+1.82\,\text{V}\] \[2{{H}_{2}}O\xrightarrow{\,}\,{{O}_{2}}+4{{H}^{+}}\,+4{{e}^{-}};\,\,{{E}^{o}}=-1.23\,\text{V}\] Explain why \[C{{o}^{3+}}\] is not stable in aqueous solution.

Answer:

                The \[{{E}^{o}}_{cell}\] can be calculated as follows: \[4[C{{o}^{3+}}+{{e}^{-}}\xrightarrow{\,}\,C{{o}^{2+}}];\]                                            \[{{E}^{o}}=+1.82\text{V}\] \[2{{H}_{2}}O\xrightarrow{\,}{{O}_{2}}+4{{H}^{+}}+4{{e}^{-}};\]                                \[{{E}^{o}}=-1.23\,\text{V}\] Add        \[4C{{o}^{3+}}+2{{H}_{2}}O\xrightarrow{\,}\,4C{{o}^{2+}}+4{{H}^{+}}+{{O}_{2}};\]          \[{{E}^{o}}=+1.82\,-1.23\,=0.59\,\text{V}\]   Since \[{{E}^{o}}_{\text{cell}}\] is positive, the cell reaction is spontaneous. This means that \[C{{o}^{3+}}\] ion will take part in the reaction. Therefore, \[C{{o}^{3+}}\] is not stable in aqueous solution.