Question

Chromium has been investigated as a coating for steel cans. The thickness of the chromium film is determined by dissolving a sample of a can in acid and oxidizing the resulting \(\mathrm{Cr}^{3+}\) to \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) with the peroxydisulfate ion: $$\begin{aligned} \mathrm{S}_{2} \mathrm{O}_{\mathrm{x}}^{2-}(a q)+\mathrm{Cr}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \\ &+\mathrm{SO}_{4}^{2-}(a q)+\mathrm{H}^{+}(a q)(\text { Unbalanced }) \end{aligned} $$ After removal of unreacted \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-},\) an excess of ferrous ammonium sulfate \(\left[\mathrm{Fe}\left(\mathrm{NH}_{4}\right)_{2}\left(\mathrm{SO}_{4}\right)_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\right]\) is added, reacting with \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) produced from the first reaction. The unreacted \(\mathrm{Fe}^{2+}\) from the excess ferrous ammonium sulfate is titrated with a separate \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) solution. The reaction is: $$\begin{array}{c}{\mathrm{H}^{+}(a q)+\mathrm{Fe}^{2+}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \longrightarrow \mathrm{Fe}^{3+}(a q)} \\\ {+\mathrm{Cr}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(I) \text { (Unbalanced) }}\end{array}$$ a. Write balanced chemical equations for the two reactions. b. In one analysis, a \(40.0-\mathrm{cm}^{2}\) sample of a chromium-plated can was treated according to this procedure. After dissolution and removal of excess \(\mathrm{S}_{2} \mathrm{O}_{8}^{2}-, 3.000 \mathrm{g}\) of \(\operatorname{Fe}\left(\mathrm{NH}_{4}\right)_{2}\left(\mathrm{SO}_{4}\right)_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) was added. It took 8.58 \(\mathrm{mL}\) of 0.0520 \(\mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) solution to completely react with the excess \(\mathrm{Fe}^{2+}\) . Calculate the thickness of the chromium film on the can. (The density of chromium is 7.19 \(\mathrm{g} / \mathrm{cm}^{3} .\) )

Short answer

The thickness of the chromium film on the can is \(2.995 * 10^{-4}\) cm.

Step-by-step solution

Balancing the chemical equations

For the first reaction, we have to balance the following equation: $$\mathrm{S}_{2}\mathrm{O}_{\mathrm{x}}^{2-}(aq)+\mathrm{Cr}^{3+}(aq)+\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}(aq)+\mathrm{SO}_{4}^{2-}(aq)+\mathrm{H}^{+}(aq)$$ Balanced equation: $$2\mathrm{S}_{2} \mathrm{O}_{8}^{2-}(aq) + 3\mathrm{Cr}^{3+}(aq) + 8\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow 2\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}(aq) + 4\mathrm{SO}_{4}^{2-}(aq) + 24\mathrm{H}^{+}(aq)$$ For the second reaction, we have to balance the following equation: $$\mathrm{H}^{+}(aq)+\mathrm{Fe}^{2+}(aq)+\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}(aq) \longrightarrow \mathrm{Fe}^{3+}(aq)+\mathrm{Cr}^{3+}(aq)+\mathrm{H}_{2}\mathrm{O}(l)$$ Balanced equation: $$14\mathrm{H}^{+}(aq) + 6\mathrm{Fe}^{2+}(aq) + \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}(aq) \longrightarrow 6\mathrm{Fe}^{3+}(aq) + 2\mathrm{Cr}^{3+}(aq) + 7\mathrm{H}_{2}\mathrm{O}(l)$$

Calculate moles of \(\mathrm{Fe}^{2+}\) in the excess ferrous ammonium sulfate

We are given that 3.000 g of ferrous ammonium sulfate was added to the sample. We need to find the number of moles of \(\mathrm{Fe}^{2+}\) present in it. First, let's find the molar mass of the ferrous ammonium sulfate: Molar mass of \(\mathrm{Fe}\left(\mathrm{NH}_{4}\right)_{2}\left(\mathrm{SO}_{4}\right)_{2} \cdot 6\mathrm{H}_{2}\mathrm{O}\) = 1(55.85) + 2(14.01 + 4(1.01)) + 2(32.07 + 4(16.00)) + 12(1.01) + 6(18.02) = 392.14 g/mol Now, we can calculate the moles of ferrous ammonium sulfate: Moles = mass / molar mass = 3.000 g / 392.14 g/mol = 7.654 * 10^{-3} mol Each mole of ferrous ammonium sulfate gives 1 mole of \(\mathrm{Fe}^{2+}\), so there are 7.654 * 10^{-3} mol of \(\mathrm{Fe}^{2+}\) in the excess ferrous ammonium sulfate.

Calculate moles of unreacted \(\mathrm{Fe}^{2+}\) and reacted \(\mathrm{Fe}^{2+}\)

It took 8.58 mL of 0.0520 M \(\mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}\) solution to completely react with the excess \(\mathrm{Fe}^{2+}\). Let's find the moles of \(\mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}\) used: Moles of \(\mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}\) = volume * concentration = 8.58 mL * (0.0520 mol/L) * (1 L / 1000 mL) = 4.4616*10^{-4} mol Now, looking at the balanced equation of the second reaction, 1 mole of \(\mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}\) reacts with 6 moles of \(\mathrm{Fe}^{2+}\). So, we will now find the moles of \(\mathrm{Fe}^{2+}\) that reacted with the \(\mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}\): Moles of unreacted \(\mathrm{Fe}^{2+}\) = (6 * moles of \(\mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}\)) = 6 * 4.4616*10^{-4} mol = 2.67696*10^{-3} mol Next, we need to find the moles of \(\mathrm{Fe}^{2+}\) that reacted with \(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}\) in the first reaction: Moles of reacted \(\mathrm{Fe}^{2+}\) = moles (total \(\mathrm{Fe}^{2+}\)) - moles (unreacted \(\mathrm{Fe}^{2+}\)) = 7.654 * 10^{-3} mol - 2.67696 * 10^{-3} mol = 4.97704 * 10^{-3} mol

Calculate moles of \(\mathrm{Cr}^{3+}\) from the first reaction

We now need to find out the moles of \(\mathrm{Cr}^{3+}\) that must have reacted in the first reaction. From the balanced equation, for every 6 moles of \(\mathrm{Fe}^{2+}\) that reacts, 2 moles of \(\mathrm{Cr}^{3+}\) must have reacted: Moles of \(\mathrm{Cr}^{3+}\) = (2 * moles of reacted \(\mathrm{Fe}^{2+}\)) / 6 = (2 * 4.97704 * 10^{-3}) / 6 = 1.65835 * 10^{-3} mol

Calculate mass and volume of chromium

Now, let's calculate the mass of chromium using its molar mass (Molar mass of \(\mathrm{Cr}\) = 51.996 g/mol): Mass of \(\mathrm{Cr}\) = moles of \(\mathrm{Cr}^{3+}\) * molar mass of \(\mathrm{Cr}\) = 1.65835 * 10^{-3} mol * 51.996 g/mol = 0.086156 g Next, let's calculate the volume of chromium using its density (Density of \(\mathrm{Cr}\) = 7.19 g/cm³): Volume of \(\mathrm{Cr}\) = Mass of \(\mathrm{Cr}\) / Density of \(\mathrm{Cr}\) = 0.086156 g / 7.19 g/cm³ = 0.01198 cm³

Calculate the thickness of the chromium film

We are given that the area of the sample is 40.0 cm². Now, to find the thickness of the film, we will divide the volume of chromium by the area: Thickness of chromium film = volume / area = 0.01198 cm³ / 40.0 cm² = 2.995 * 10^{-4} cm The thickness of the chromium film on the can is 2.995 * 10^{-4} cm.

Key concepts

Oxidation-Reduction

In chemistry, oxidation-reduction reactions are also known as redox reactions. These are processes in which electrons are transferred between substances. Oxidation refers to the loss of electrons, while reduction refers to the gain of electrons. In the context of the problem, the ^Cr3+ ions are oxidized to ^Cr₂O₇2- by the peroxydisulfate ion, where the oxidation state of chromium increases, indicating a loss of electrons. Understanding these concepts is crucial since different reactions in this process involve these exchanges. For instance, ^S₂O₈2- acts as the oxidizing agent by taking in electrons and reducing its oxidation number in the reaction process.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They show the reactants transformed into products, often with coefficients indicating the number of molecules or atoms involved. Balancing these equations is essential to accurately reflect the law of conservation of mass. In the exercise, two equations are given, both initially unbalanced. It's important to balance them by ensuring equal numbers of each type of atom on both sides of the equation. An unbalanced equation is misleading as it doesn't accurately represent the stoichiometry of a chemical reaction.

Here, balancing these equations allows us to understand how many molecules of each reactant contribute to forming the products. This is vital for computations, like determining how much chromium reacts and accordingly how much of it forms the coating.

Chromium Coating

Chromium coating provides an anti-corrosive and protective layer over materials like steel cans. This coating process involves transforming chromium into a more durable and typically reactive form like ^Cr₂O₇2-. Understanding how much chromium is needed for an effective coating involves dissolving the coating in a controlled chemical reaction. By performing these chemical analyses, such as those in the exercise, you can calculate the thickness of the chromium layer, ensuring the correct amount has been applied. This not only assesses the effectiveness but also optimizes resource use for large-scale applications.

Titration

Titration is a laboratory method used primarily in chemistry to determine the concentration of a known reactant. It involves adding a titrant of known concentration to a solution to determine the concentration of an unknown solution. In this exercise, after deriving ^Cr₂O₇2-, the excess ^Fe2+ ions are titrated using a ^{K₂Cr₂O₇} solution. Titration helps in determining exactly how many of the ^Fe2+ ions had reacted with the dichromate ( ^Cr₂O₇2-) to produce measurable amounts of chromate ( ^Cr3+). This method, being precise in measurement, is key to finding the initial concentration of various reactants, which further helps in the computation of desired parameters, like film thickness in the exercise.

Stoichiometry

Stoichiometry refers to the measurement and calculation of the quantitative relationships of the reactants and products in a chemical reaction. It's a central part of balancing chemical equations. By using stoichiometry, you can calculate how much of each reactant is needed to form a desired amount of product. In the exercise, stoichiometric calculations involve determining the amount of unreacted and reacted ^Fe2+ ions. It allows us to figure out from the balanced equations how many moles of a ^Cr3+ ion are required to react with a certain number of ^Fe2+ ions. By extending this concept, you can use given data inputs like concentration and volume to predict the amount of chromium coating produced, ensuring precise application according to necessary specifications.