At the left side; the balanced half equation is:
[tex]\mathbf{2H_2O _{(l)} + 2e^- \to H_{2(g)} + 2OH^{-}_{(aq)}}[/tex]
At the right side, the balanced half equation is:
[tex]\mathbf{4OH^-_{(aq)} \to O_{2(g)} + 2H_2O_{(l)} + 4e^-}[/tex]
The objective of this question is to determine the balanced half-reaction for the electrodes on the left and the right side.
A half-reaction is an oxidation-reduction reaction part of a redox reaction where we make sure both half-reactions are balanced by multiplication of the reactions with appropriate numbers in order to have an equal number of electrons.
According to the given information:
The overall cell reaction can be expressed as:
[tex]\mathbf{H_2O_{(l)} \to H_{2(g)} + \dfrac{1}{2}O_{2(g)}}[/tex]
At the left side; the balanced half equation can be expressed as:
[tex]\mathbf{2H_2O _{(l)} + 2e^- \to H_{2(g)} + 2OH^{-}_{(aq)}}[/tex]
At the right side, the balanced half equation can be expressed as:
[tex]\mathbf{4OH^-_{(aq)} \to O_{2(g)} + 2H_2O_{(l)} + 4e^-}[/tex]
Learn more about the electrolysis of water here:
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