$2H_2O_{(l)} \rightarrow O_{2(g)}+4H^+_{(aq)}+4e^-$.

You assign a value of $+1.23 \text{ V}$ to this half-reaction (the oxidation of water) but that is incorrect. $+1.23 \text{ V}$ is in fact the value for the reduction of water, not the oxidation of water. So for:

$2H_2O_{(l)} \rightarrow O_{2(g)}+4H^+_{(aq)}+4e^-$, the half-potential is actually $-1.23 \text{ V}$.

One can understand this also as follows: the cell potential is calculated as $E=E_{Ox}+E_{Red}$ with $E_{Ox}$ the half-potential of the oxidation reaction and $E_{Red}$ the half-potential of the reduction reaction $(2H^+_{(aq)}+2e^- \rightarrow H_{2(g)})$.

So in our case we have $E=-1.23 \text{ V} + 0.0 \text{ V}=-1.23 \text{ V}$. A negative value conventionally in electrochemistry means that these reactions will not spontaneously occur. Well, that's precisely what we observe: water does not auto-oxidise!

So now we know that in order to make these reactions proceed, we need to apply at least $+1.23 \text{ V}$ (disregarding any overpotentials that might arise) across the electrodes.

The second thing to understand is that the half-potentials of half-cells are always relative to the half-potential of:

$2H^+_{(aq)}+2e^- \rightarrow H_{2(g)}$

Purely conventionally, the half-potential of a platinum/hydrogen half-cell has been assigned the value of $0.00 \text{ V}$. The half-potential of some half-reduction, say:

$O + n e^- \rightarrow R^{n-}$.

... is then determined relative to the platinum/hydrogen half-cell.

This means that in reality we never know the 'real' value of a half-cell, we can only know the potential difference between two electrodes (in a closed circuit).

Charge is added to each electrode in equal amounts. The actual potential on each electrode for this cell is (1.23V/2)=0.615V. The positive charge electrode is +0.615V and the negative charge electrode is -0.615V. The electrodes are is series, so the potentials add to 1.23V. Breaking the charge down further: the positive charge electrode has (+0.3075 charge and -0.3075 ionic charge from HSO−4 (or SO2−4) and the negative charge electrode has (-0.3075 charge and +0.3075 ionic charge from H+)

... is therefore a misinterpretation, albeit quite a smart one.

$2H_2O_{(l)} \rightarrow O_{2(g)}+4H^+_{(aq)}+4e^-$.

You assign a value of $+1.23 \text{ V}$ to this half-reaction (the oxidation of water) but that is incorrect. $+1.23 \text{ V}$ is in fact the value for the reduction of water, not the oxidation of water. So for:

$2H_2O_{(l)} \rightarrow O_{2(g)}+4H^+_{(aq)}+4e^-$, the half-potential is actually $-1.23 \text{ V}$.

One can understand this also as follows: the cell potential is calculated as $E=E_{Ox}+E_{Red}$ with $E_{Ox}$ the half-potential of the oxidation reaction and $E_{Red}$ the half-potential of the reduction reaction $(2H^+_{(aq)}+2e^- \rightarrow H_{2(g)})$.

So in our case we have $E=-1.23 \text{ V} + 0.0 \text{ V}=-1.23 \text{ V}$. A negative value conventionally in electrochemistry means that these reactions will not spontaneously occur. Well, that's precisely what we observe: water does not auto-oxidise!

So now we know that in order to make these reactions proceed, we need to apply at least $+1.23 \text{ V}$ (disregarding any overpotentials that might arise) across the electrodes.

The second thing to understand is that the half-potentials of half-cells are always relative to the half-potential of:

$2H^+_{(aq)}+2e^- \rightarrow H_{2(g)}$

Purely conventionally, the half-potential of a platinum/hydrogen half-cell has been assigned the value of $0.00 \text{ V}$. The half-potential of some half-reduction, say:

$O + n e^- \rightarrow R^{n-}$.

... is then determined relative to the platinum/hydrogen half-cell.

This means that in reality we never know the 'real' value of a half-cell, we can only know the potential difference between the two electrodes (in a closed circuit).

Charge is added to each electrode in equal amounts. The actual potential on each electrode for this cell is (1.23V/2)=0.615V. The positive charge electrode is +0.615V and the negative charge electrode is -0.615V. The electrodes are is series, so the potentials add to 1.23V. Breaking the charge down further: the positive charge electrode has (+0.3075 charge and -0.3075 ionic charge from HSO−4 (or SO2−4) and the negative charge electrode has (-0.3075 charge and +0.3075 ionic charge from H+)

... is therefore a misinterpretation, albeit quite a smart one.