Question

Write the two half-reactions that make up the following balanced redox reaction. $3{\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4+2{\mathrm{HAsO}}_2\to 6{\mathrm{CO}}_2+2\mathrm{As}+4{\mathrm{H}}_2\mathrm{O}$

Short answer

The two balanced half-reactions that make up the given redox reaction are: Oxidation half-reaction: $$5{\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4\to 10{\mathrm{CO}}_2+10e^{-}$$ Reduction half-reaction: $$2{\mathrm{HAsO}}_2+10e^{-}\to 2\mathrm{As}+2{\mathrm{H}}_2\mathrm{O}$$

Step-by-step solution

Identify the species that undergo changes in oxidation states

First, we need to identify the species that undergo changes in their oxidation states during the reaction. In this case, we can see that Carbon (C) in ${\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4$ is oxidized to Carbon (C) in ${\mathrm{CO}}_2$ since the oxidation state changes from +3 to +4, and Arsenic (As) in ${\mathrm{HAsO}}_2$ is reduced to elemental Arsenic $\mathrm{As}$ since the oxidation state changes from +5 to 0.

Write the oxidation half-reaction

Next, let's write the oxidation half-reaction. To do this, we need to account for the balanced number of electrons that are lost or gained during the process. We know that ${\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4$ is converted into ${\mathrm{CO}}_2$. Notice that each Carbon atom loses one electron, making a total of two lost electrons per each ${\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4$ molecule, so the oxidation half-reaction is: $${\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4\to 2{\mathrm{CO}}_2+2e^{-}$$

Write the reduction half-reaction

Now, let's write the reduction half-reaction. We know that ${\mathrm{HAsO}}_2$ is converted into $\mathrm{As}$. Notice that each Arsenic atom gains five electrons during the process. Thus, the reduction half-reaction is: $${\mathrm{HAsO}}_2+5e^{-}\to \mathrm{As}+{\mathrm{H}}_2\mathrm{O}$$

Balance the half-reactions

To balance the half-reactions, we need to make sure that the total number of electrons lost in the oxidation process equals the total number of electrons gained in the reduction process. In our case, we find the lowest common multiple of the electrons in both half-reactions, which is 10. Then, we multiply each half-reaction by the appropriate factor to balance the electrons: Oxidation half-reaction x 5: $$5{\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4\to 10{\mathrm{CO}}_2+10e^{-}$$ Reduction half-reaction x 2: $$2{\mathrm{HAsO}}_2+10e^{-}\to 2\mathrm{As}+2{\mathrm{H}}_2\mathrm{O}$$

Finalize the half-reactions

Finally, we present the two balanced half-reactions that make up the given redox reaction: Oxidation half-reaction: $$5{\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4\to 10{\mathrm{CO}}_2+10e^{-}$$ Reduction half-reaction: $$2{\mathrm{HAsO}}_2+10e^{-}\to 2\mathrm{As}+2{\mathrm{H}}_2\mathrm{O}$$

Key concepts

Oxidation State

Oxidation states, also known as oxidation numbers, are an essential concept in understanding redox reactions. They help us to track the transfer of electrons between atoms in a chemical reaction. When you look at a molecule, the oxidation state provides a number to each atom that represents its charge after assuming all its bonds are ionic. This becomes particularly handy in identifying which elements are oxidized and which are reduced in a reaction.

For instance, in the reaction given in the exercise, the carbon atom in $${\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4$$ is initially at an oxidation state of +3. When it is turned into ${\mathrm{CO}}_2$, it becomes +4. This indicates a loss of electrons, which means oxidation has occurred. Meanwhile, arsenic in ${\mathrm{HAsO}}_2$ from an oxidation state of +5 is reduced to 0 in elemental arsenic form $\mathrm{As}$, indicating a gain of electrons and thus reduction. Understanding these changes in oxidation states is crucial as they highlight the exchange of electrons between reactants.

To determine the oxidation state of an atom in a molecule:

• Assign the oxidation number of -2 for oxygen.
• Assign +1 for hydrogen.
• Calculate the total oxidation number for the entire molecule must equal to its charge.

With these simple rules, one can compute the oxidation states of the involved atoms and keep tabs on electron movements.

Half-Reaction

In redox reactions, it's important to view the process in halves: oxidation and reduction. This splitting into half-reactions is helpful for balancing redox reactions and understanding the transfer of electrons. By identifying these two separate processes, it becomes clear which elements lose or gain electrons.

For example, consider the reaction between ${\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4$ and ${\mathrm{HAsO}}_2$. The oxidation half-reaction focuses on one component of the reaction losing electrons. Here, carbon in ${\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4$ loses electrons as it turns into ${\mathrm{CO}}_2$:$${\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4\to 2{\mathrm{CO}}_2+2e^{-}$$
Each molecule loses two electrons, and through understanding this sequence, students learn to write oxidation half-reactions.

On the flip side, the reduction half-reaction deals with elements gaining electrons. In our exercise, arsenic $\mathrm{As}$ gains electrons to reduce its oxidation state:$${\mathrm{HAsO}}_2+5e^{-}\to \mathrm{As}+{\mathrm{H}}_2\mathrm{O}$$Each molecule gains five electrons. Breaking down the overall redox reaction into these two halves lets us see exactly what's happening on an atomic level and helps balance the reaction equations.

To write half-reactions:

• Identify the species experiencing a change in oxidation state.
• Determine the number of electrons lost or gained based on their oxidation state change.
• Write separate equations for oxidation and reduction processes.

Balancing Equations

Balancing chemical equations is critical in chemistry as it ensures that the law of conservation of mass holds true. When dealing with redox reactions, we use half-reactions to make this task more manageable. The challenge is to ensure that the number of atoms and the charge are balanced across the process.

In the given problem, we determine the oxidation and reduction half-reactions individually. Once identified, it's also necessary to make sure the number of electrons lost in the oxidation reaction equals the number gained in the reduction reaction.

To achieve this:

• Calculate the least common multiple (LCM) of the number of electrons transferred in each half-reaction.
• Multiply each half-reaction by a factor that would equalize the number of electrons involved in each process. For example, multiplying the oxidation half-reaction by 5 and the reduction half-reaction by 2 makes both involve 10 electrons.

Combining these steps for the given redox reaction results in:

• Oxidation: $5{\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4\to 10{\mathrm{CO}}_2+10e^{-}$
• Reduction: $2{\mathrm{HAsO}}_2+10e^{-}\to 2\mathrm{As}+2{\mathrm{H}}_2\mathrm{O}$

These balanced half-reactions can be combined to give a fully balanced redox equation where both mass and charge are conserved.