Answer:
[tex]Hg^0+2OH^-\rightarrow Hg^{2+}O+H_2O+2e^-[/tex]
Explanation:
Hello,
In this case, mercury (II) oxide (HgO) is obtained via the reaction:
[tex]Hg(l)+O_2\rightarrow HgO[/tex]
Nonetheless, since it is a reaction carried out in basic solution, mercury's half-reaction only, must be:
[tex]Hg^0+2OH^-\rightarrow Hg^{2+}O+H_2O+2e^-[/tex]
Thus, it is seen that OH ionis should be added due to the basic aqueous solution considering that 2 electrons are transferred from 0 to 2 in mercury.
Best regards.
The balanced oxidation half equation required is; [tex]Hg(s) + 2OH^-(aq) ----> HgO(s) + H2O(l) + 2e[/tex]
The oxidation half equation shows the loss of electrons. It shows the number of electrons lost during the process of oxidation of a given chemical specie in an aqueous solution.
For the oxidation of mercury to mercuric oxide, in basic solution, the balanced half-equation describing the process is written as;
[tex]Hg(s) + 2OH^-(aq) ----> HgO(s) + H2O(l) + 2e[/tex]
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