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.
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