Question

The reaction ${\mathrm{H}}_2{\mathrm{SeO}}_3(aq)+6{\mathrm{I}}^{-}(aq)+4{\mathrm{H}}^{+}(aq)$ $⟶\mathrm{Se}(s)+2{\mathrm{I}}_3^{-}(aq)+3{\mathrm{H}}_2\mathrm{O}(l)$ was studied at $0^{\circ }\mathrm{C}$, and the following data were obtained:

Short answer

The balanced redox reaction is ${\mathrm{H}}_2{\mathrm{SeO}}_3(aq)+6{\mathrm{I}}^{-}(aq)+4{\mathrm{H}}^{+}(aq)\to \mathrm{Se}(s)+2{\mathrm{I}}_3^{-}(aq)+3{\mathrm{H}}_2\mathrm{O}(l)$.

Step-by-step solution

Determine the half-reactions

First, we have to identify the species that are being oxidized and reduced in the given reaction. In this case, Se (selenium) from H2SeO3 is being reduced to Se(s), and I⁻(aq) is being oxidized to I₃⁻(aq). We can now write the unbalanced half-reactions: Oxidation half-reaction: $${\mathrm{I}}^{-}(aq)\to {\mathrm{I}}_3^{-}(aq)$$ Reduction half-reaction: $${\mathrm{H}}_2{\mathrm{SeO}}_3(aq)\to \mathrm{Se}(s)$$

Balance the half-reactions

Now we will balance the oxidation and reduction half-reactions: For the oxidation half-reaction: 2I⁻ → I₃⁻ + I⁻, manage charges by multiplying the iodine ion by 6. Thus, we get: $6{\mathrm{I}}^{-}(aq)\to 2{\mathrm{I}}_3^{-}(aq)$ For the reduction half-reaction: To balance the oxygens, we add 3H2O(l) as a product: ${\mathrm{H}}_2{\mathrm{SeO}}_3(aq)\to \mathrm{Se}(s)+3{\mathrm{H}}_2\mathrm{O}(l)$ To balance hydrogen, we add 6H⁺(aq) on the left side: ${\mathrm{H}}_2{\mathrm{SeO}}_3(aq)+6{\mathrm{H}}^{+}(aq)\to \mathrm{Se}(s)+3{\mathrm{H}}_2\mathrm{O}(l)$ Now, we subtract 2H⁺(aq) from both sides to get the balanced reduction half-reaction: ${\mathrm{H}}_2{\mathrm{SeO}}_3(aq)+4{\mathrm{H}}^{+}(aq)\to \mathrm{Se}(s)+3{\mathrm{H}}_2\mathrm{O}(l)$

Combine the balanced half-reactions

Finally, we can combine the balanced oxidation and reduction half-reactions to get the balanced overall reaction: $6{\mathrm{I}}^{-}(aq)\to 2{\mathrm{I}}_3^{-}(aq)$ (Balanced Oxidation Half-Reaction) ${\mathrm{H}}_2{\mathrm{SeO}}_3(aq)+4{\mathrm{H}}^{+}(aq)\to \mathrm{Se}(s)+3{\mathrm{H}}_2\mathrm{O}(l)$ (Balanced Reduction Half-Reaction) ${\mathrm{H}}_2{\mathrm{SeO}}_3(aq)+6{\mathrm{I}}^{-}(aq)+4{\mathrm{H}}^{+}(aq)\to \mathrm{Se}(s)+2{\mathrm{I}}_3^{-}(aq)+3{\mathrm{H}}_2\mathrm{O}(l)$ (Balanced Overall Reaction) We have now successfully broken down and balanced the given redox reaction.

Key concepts

Oxidation Half-Reaction

Understanding oxidation half-reactions is a key part of mastering redox reactions. In a redox process, a substance loses electrons during oxidation. This can sometimes be confusing, so let's explore what actually happens with the oxygen half-reaction in the provided chemical equation.
In the given exercise, iodine (${\mathrm{I}}^{-}$(aq)) is the species being oxidized. This means that the iodide ions are losing electrons as they transform into tri-iodide ions (${\mathrm{I}}_3^{-}$(aq)). The general half-reaction is represented as follows:

• Unbalanced oxidation half-reaction: $${\mathrm{I}}^{-}(aq)\to {\mathrm{I}}_3^{-}(aq)$$
• Balanced oxidation half-reaction: $$6{\mathrm{I}}^{-}(aq)\to 2{\mathrm{I}}_3^{-}(aq)$$

To balance an oxidation half-reaction like this, ensure that both the number of atoms and the charges are equal on both sides. This involves balancing the ions and the electrons in the equation, making sure that charge conservation is respected. In this particular reaction, iodine ions are multiplied and adjusted to ensure the balance of charges and moles is maintained.

Reduction Half-Reaction

In contrast to oxidation, the reduction half-reaction involves the gain of electrons. Here, a substance decreases its oxidation state by gaining electrons. In our exercise, selenium (${\mathrm{H}}_2{\mathrm{SeO}}_3$(aq)) is being reduced to elemental selenium ($\mathrm{Se}$(s)).The reduction half-reaction for this process is represented as follows:

• Unbalanced reduction half-reaction: $${\mathrm{H}}_2{\mathrm{SeO}}_3(aq)\to \mathrm{Se}(s)$$
• Balanced reduction half-reaction: $${\mathrm{H}}_2{\mathrm{SeO}}_3(aq)+4{\mathrm{H}}^{+}(aq)\to \mathrm{Se}(s)+3{\mathrm{H}}_2\mathrm{O}(l)$$

Balancing this half-reaction involves ensuring the same number of hydrogen and oxygen atoms on both sides. Initially, water molecules (${\mathrm{H}}_2\mathrm{O}(l)$) are added to the product side to balance the oxygen. Then, hydrogen ions (${\mathrm{H}}^{+}(aq)$) are added to the reactant side to maintain hydrogen balance. This step ensures that the charges and the atoms are balanced, adhering to the law of conservation of mass and charge.

Balancing Chemical Equations

Balancing chemical equations is an essential skill for working with chemical reactions. When handling redox equations, the process involves ensuring the number of atoms and the charge balance is consistent on both sides of the equation. Let's break it down further with the given exercise:
Initially, we need to separately balance each half-reaction for both mass and charge.
Here's a simplified procedure to follow:

• **Identify the oxidation and reduction components**: Determine which species is being oxidized and which is being reduced as done in previous sections.
• **Write the half-reactions**: Formulate the separate reactions showing oxidation and reduction.
• **Balance each half-reaction**: Add electrons, water, and ${\mathrm{H}}^{+}(aq)$ as necessary.
• **Equalize electron transfer**: Multiply the half-reactions by appropriate coefficients so that the number of electrons lost and gained are equal.
• **Sum up the half-reactions**: Add them together to get the overall balanced equation.

In the exercise, combining the two balanced half-reactions yields:$${\mathrm{H}}_2{\mathrm{SeO}}_3(aq)+6{\mathrm{I}}^{-}(aq)+4{\mathrm{H}}^{+}(aq)\to \mathrm{Se}(s)+2{\mathrm{I}}_3^{-}(aq)+3{\mathrm{H}}_2\mathrm{O}(l)$$This represents the balance of the chemical equation, showing that all atoms and charges are accounted for and consistent on both sides.