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Chapter 21: Problem 5

Oxalic acid is often used to remove rust stains. What properties of oxalic acid allow it to do this?

Short Answer

Expert verified
Oxalic acid effectively removes rust stains due to its chelating property, which allows it to form soluble iron oxalate complexes with iron ions in rust. Additionally, its function as a weak reducing agent enables it to reduce iron(III) oxide back to iron(II) oxide, further aiding in rust removal.

Step by step solution

01

Properties of Oxalic Acid

Oxalic acid, also known as ethanedioic acid, is an organic compound with the molecular formula C2H2O4. Its structure consists of two carboxyl groups (COOH) attached to a central carbon atom. It is readily soluble in water and acts as a weak acid.
02

Rust Formation

Rust forms when iron or its alloys (such as steel) come in contact with oxygen and moisture. In this process, the iron is oxidized, creating hydrated iron(III) oxide (Fe2O3. xH2O). This hydrated oxide is a reddish-brown color, which we recognize as rust.
03

Chelating Property of Oxalic Acid

Oxalic acid is a dicarboxylic acid, which means it can form chelate complexes with metallic ions. Chelating agents can form stable complexes with metal ions, and in this case, oxalic acid binds with the iron ions in rust (Fe3+), forming a soluble complex. This soluble iron oxalate complex then dissolves in water, effectively removing the rust stain.
04

Reduction of Iron Oxide

Oxalic acid is a weak reducing agent and can reduce iron(III) oxide (Fe2O3) in rust to form iron(II) oxide (FeO) and water. This reduces the rust back to a less oxidized form of iron. The reaction is as follows: \[ Fe_2O_3 + 3H_2C_2O_4 鉄� 2FeC_2O_4 + 3H_2O + 3CO_2\] In conclusion, the properties of oxalic acid that enable it to remove rust stains effectively are its ability to chelate with iron ions in rust, forming a soluble iron oxalate complex, and its function as a weak reducing agent, reducing iron(III) oxide back to iron(II) oxide.

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Most popular questions from this chapter

Use standard reduction potentials to calculate \(\mathscr{E}^{\circ}, \Delta G^{\circ}\), and \(K\) (at \(298 \mathrm{~K}\) ) for the reaction that is used in production of gold: $$2 \mathrm{Au}(\mathrm{CN})_{2}^{-}(a q)+\mathrm{Zn}(s) \longrightarrow 2 \mathrm{Au}(s)+\mathrm{Zn}(\mathrm{CN})_{4}^{2-}(a q)$$ The relevant half-reactions are $$\begin{aligned}\mathrm{Au}(\mathrm{CN})_{2}^{-}+\mathrm{e}^{-} \longrightarrow \mathrm{Au}+2 \mathrm{CN}^{-} & \mathscr{C}^{\circ} &=-0.60 \mathrm{~V} \\ \mathrm{Zn}(\mathrm{CN})_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}+4 \mathrm{CN}^{-} & \mathscr{E}^{\circ} &=-1.26 \mathrm{~V} \end{aligned}$$

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