Answer:
![5H_2O+C_5H_(10)O_5\rightarrow 5CO_2+20H^++20e^-](https://img.qammunity.org/2021/formulas/chemistry/college/861ckkmibbnou68r05ywtqkzb2wavma5om.png)
Step-by-step explanation:
Hello.
In this case, when ribose (C₅H₁₀O₅) yields carbon dioxide (CO₂) we write:
![C_5H_(10)O_5\rightarrow CO_2](https://img.qammunity.org/2021/formulas/chemistry/college/9gno61jo0i74x3bkgfacxbc0zjfbd0nr44.png)
Which needs to be balanced by adding water and hydrogen ions:
![5H_2O+C_5H_(10)O_5\rightarrow 5CO_2+20H^++20e^-](https://img.qammunity.org/2021/formulas/chemistry/college/861ckkmibbnou68r05ywtqkzb2wavma5om.png)
You can also see that there are 20 transferred electrons, since the carbon atoms in the ribose have 0 as their oxidation state and the carbon atoms in the carbon dioxide have +4 as the oxidation state, thus, each carbon transfers 4 electrons, a five carbon atoms transfer 20 electrons overall.
In such a way, since the carbon is increasing its oxidation state, such half reaction is an oxidation half reaction.
Best regards.