Chapter 4, Problem 155CP

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 Cr³⁺ to Cr₂O₇²⁻ with the peroxydisulfate ion:

S₂O₈²⁻(aq) + Cr³⁺(aq) + H₂O(l) → Cr₂O₇²⁻(aq) + SO₄²⁻(aq) + H⁺(aq) (Unbalanced)

After removal of unreacted S₂O₈²⁻ an excess of ferrous ammonium sulfate [Fe(NH₄)₂(SO₄)₂·6H₂O] is added, reacting with Cr₂O₇²⁻ produced from the first reaction. The unreacted Fe²⁺ from the excess ferrous ammonium sulfate is titrated with a separate K₂Cr₂O₇ solution. The reaction is:

H⁺(aq) + Fe²⁺(aq) + Cr₂O₇²⁻(aq) → Fe³⁺(aq) + Cr³⁺(aq) + H₂O(l) (Unbalanced)

1. a. Write balanced chemical equations for the two reactions.
2. b. In one analysis, a 40.0-cm² sample of a chromium-plated can was treated according to this procedure. After dissolution and removal of excess S₂O₈²⁻, 3.000 g of Fe(NH₄)₂(SO₄)₂·6H₂O was added. It took 8.58 mL of 0.0520 M K₂Cr₂O₇ solution to completely react with the excess Fe²⁺. Calculate the thickness of the chromium film on the can. (The density of chromium is 7.19 g/cm³)

Interpretation Introduction

Interpretation:

The balanced equation for the given reactions and the thickness of chromium film on the can are needed to be determined if the $\text{8}\text{.58mL}$ of $\text{0}\text{.0520M}$ ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ is completely react with $\text{3}\text{.0g}$ of ${\text{Fe(NH}}_{\text{4}}{\text{)}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{2}}{\text{.6H}}_{\text{2}}\text{O}$.

Concept introduction:

Steps followed to balance a redox (reduction-oxidation) reaction occurs in acidic medium,

• Redox reactions can be separated into two half-reactions, such as oxidation reaction and reduction reaction.
• Balancing of oxygen and hydrogen are carried out by adding water and ions in the half-reactions, and balance the number of electrons then make the number of electrons in oxidation and reduction reactions equal by multiplying by integers.
• Club the both half-reactions.
• Mole ratio between the reactants of a reaction are depends upon the coefficients of reactants in a balanced chemical equation.
• Equation for number of grams of a substance from its number of moles is,

$\text{Number of moles}\text{×}\text{Molecular}\text{mass in grams}\text{=}\text{Number}\text{of}\text{grams}$

• Equation for volume from density is,

  $\text{Volume}\text{=}\frac{\text{Mass}}{\text{Density}}$

• Equation for finding thickness from volume and area is,

$\text{thickness}\text{=}\frac{\text{volume}}{\text{area}}$

Answer to Problem 155CP

The balanced equations for the given reactions are,

${\text{3S}}_{\text{2}}{\text{O}}_{\text{8}}{}^{\text{2-}}\text{(aq)}+{\text{2Cr}}^{\text{3+}}\text{(aq)}\text{+}{\text{7H}}_2\text{O}\text{(aq)}\to {\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2-}}\text{(aq)}\text{+}{\text{14H}}^{+}\text{(aq)}+{\text{6SO}}_{\text{4}}{}^{\text{2-}}\text{(aq)}$

${\text{6Fe}}^{\text{2+}}\text{(aq)}+{\text{14H}}^{+}+{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2-}}\text{(aq)}\to {\text{2Cr}}^{\text{3+}}\text{(aq)}\text{+}{\text{7H}}_{\text{2}}\text{O}\text{+}6{\text{Fe}}^{\text{3+}}\text{(aq)}$

Explanation of Solution

To determine: The balanced equations for the given reactions.

The chemical equations for the given titration is,

${\text{S}}_{\text{2}}{\text{O}}_{\text{8}}{}^{\text{2-}}\text{(aq)}\text{+}{\text{Cr}}^{\text{3+}}\text{(aq)}\text{+}{\text{H}}_{\text{2}}\text{O}\text{(l)}\to {\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2-}}\text{(aq)}\text{+}{\text{SO}}_{\text{4}}{}^{\text{2-}}\text{(aq)}\text{+}{\text{H}}^{\text{+}}\text{(aq)}$

${\text{H}}^{+}\text{+}{\text{Fe}}^{\text{2+}}\text{(aq)}\text{+}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2-}}\text{(aq)}\to {\text{Fe}}^{\text{3+}}\text{(aq)}\text{+}{\text{Cr}}^{3+}\text{(aq)}\text{+}{\text{H}}_2\text{O}\text{(l)}$

Steps (I, II and III) followed for balance the redox equations,

Balance the reaction,

${\text{S}}_{\text{2}}{\text{O}}_{\text{8}}{}^{\text{2-}}\text{(aq)}\text{+}{\text{Cr}}^{\text{3+}}\text{(aq)}\text{+}{\text{H}}_{\text{2}}\text{O}\text{(l)}\to {\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2-}}\text{(aq)}\text{+}{\text{SO}}_{\text{4}}{}^{\text{2-}}\text{(aq)}\text{+}{\text{H}}^{\text{+}}\text{(aq)}$

I- Separate the redox reaction into two, such as oxidation and reduction.

Therefore,

Oxidation reaction is,

${\text{Cr}}^{\text{3+}}\text{(aq)}\to {\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2-}}\text{(aq)}$

Reduction reaction is,

${\text{S}}_{\text{2}}{\text{O}}_{\text{8}}{}^{\text{2-}}\text{(aq)}\to {\text{SO}}_{\text{4}}{}^{\text{2-}}\text{(aq)}$

II- Balance the atoms and electrons in each half-reactions,

Balancing of oxygen and hydrogen are carried out by adding water and ions in the half-reactions

${\text{2Cr}}^{\text{3+}}\text{(aq)}\text{+}{\text{7H}}_2\text{O}\text{(aq)}\to {\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2-}}\text{(aq)}\text{+}{\text{14H}}^{+}\text{(aq)}+{\text{6e}}^{-}$

${\text{S}}_{\text{2}}{\text{O}}_{\text{8}}{}^{\text{2-}}\text{(aq)}+{\text{2e}}^{-}\to 2{\text{SO}}_{\text{4}}{}^{\text{2-}}\text{(aq)}$

Equalizing the numbers of electrons in oxidation and reduction reactions,

${\text{3S}}_{\text{2}}{\text{O}}_{\text{8}}{}^{\text{2-}}\text{(aq)}+{\text{6e}}^{-}\to {\text{6SO}}_{\text{4}}{}^{\text{2-}}\text{(aq)}$

III- club the both half-reactions into a single equation,

${\text{3S}}_{\text{2}}{\text{O}}_{\text{8}}{}^{\text{2-}}\text{(aq)}+{\text{6e}}^{-}+{\text{2Cr}}^{\text{3+}}\text{(aq)}\text{+}{\text{7H}}_2\text{O}\text{(aq)}\to {\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2-}}\text{(aq)}\text{+}{\text{14H}}^{+}\text{(aq)}+{\text{6e}}^{-}+{\text{6SO}}_{\text{4}}{}^{\text{2-}}\text{(aq)}$

The same number of electrons and molecules in the both sides will cancel each other,

So, the balanced equation for the given equation is,

${\text{3S}}_{\text{2}}{\text{O}}_{\text{8}}{}^{\text{2-}}\text{(aq)}+{\text{2Cr}}^{\text{3+}}\text{(aq)}\text{+}{\text{7H}}_2\text{O}\text{(aq)}\to {\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2-}}\text{(aq)}\text{+}{\text{14H}}^{+}\text{(aq)}+{\text{6SO}}_{\text{4}}{}^{\text{2-}}\text{(aq)}$

Balance the reaction,

${\text{H}}^{+}\text{+}{\text{Fe}}^{\text{2+}}\text{(aq)}\text{+}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2-}}\text{(aq)}\to {\text{Fe}}^{\text{3+}}\text{(aq)}\text{+}{\text{Cr}}^{3+}\text{(aq)}\text{+}{\text{H}}_2\text{O}\text{(l)}$

I- Separate the redox reaction into two, such as oxidation and reduction.

Therefore,

Oxidation reaction is,

${\text{Fe}}^{\text{2+}}\text{(aq)}\to {\text{Fe}}^{\text{3+}}\text{(aq)}$

Reduction reaction is,

${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2-}}\text{(aq)}\to {\text{Cr}}^{\text{3+}}\text{(aq)}$

II- Balance the atoms and electrons in each half-reactions,

Balancing of oxygen and hydrogen are carried out by adding water and ions in the half-reactions

${\text{Fe}}^{\text{2+}}\text{(aq)}\to {\text{Fe}}^{\text{3+}}\text{(aq)}\text{+}{\text{1e}}^{-}$

${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2-}}\text{(aq)}\text{+}{\text{14H}}^{+}\text{(aq)}+{\text{6e}}^{-}\to {\text{2Cr}}^{\text{3+}}\text{(aq)}\text{+}{\text{7H}}_2\text{O}\text{(aq)}$

Equalizing the numbers of electrons in oxidation and reduction reactions,

${\text{6Fe}}^{\text{2+}}\text{(aq)}\to {\text{6Fe}}^{\text{3+}}\text{(aq)}\text{+}{\text{6e}}^{-}$

III- club the both half-reactions into a single equation,

${\text{6Fe}}^{\text{2+}}\text{(aq)}\text{+}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2-}}\text{(aq)}\text{+}{\text{14H}}^{+}\text{(aq)}+{\text{6e}}^{-}\to {\text{2Cr}}^{\text{3+}}\text{(aq)}\text{+}{\text{7H}}_2\text{O}\text{(aq)}\text{+}{\text{6Fe}}^{\text{3+}}\text{(aq)}\text{+}{\text{6e}}^{-}$

The same number of electrons and molecules in the both sides will cancel each other,

So, the balanced equation for the given equation is,

${\text{6Fe}}^{\text{2+}}\text{(aq)}+{\text{14H}}^{+}+{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2-}}\text{(aq)}\to {\text{2Cr}}^{\text{3+}}\text{(aq)}\text{+}{\text{7H}}_{\text{2}}\text{O}\text{+}6{\text{Fe}}^{\text{3+}}\text{(aq)}$

Conclusion

The molarity of the permanganate solution is determined by using the given data’s for the titration.

Interpretation Introduction

Interpretation:

The balanced equation for the given reactions and the thickness of chromium film on the can are needed to be determined if the $\text{8}\text{.58mL}$ of $\text{0}\text{.0520M}$ ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ is completely react with $\text{3}\text{.0g}$ of ${\text{Fe(NH}}_{\text{4}}{\text{)}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{2}}{\text{.6H}}_{\text{2}}\text{O}$.

Concept introduction:

Steps followed to balance a redox (reduction-oxidation) reaction occurs in acidic medium,

• Redox reactions can be separated into two half-reactions, such as oxidation reaction and reduction reaction.
• Balancing of oxygen and hydrogen are carried out by adding water and ions in the half-reactions, and balance the number of electrons then make the number of electrons in oxidation and reduction reactions equal by multiplying by integers.
• Club the both half-reactions.
• Mole ratio between the reactants of a reaction are depends upon the coefficients of reactants in a balanced chemical equation.
• Equation for number of grams of a substance from its number of moles is,

$\text{Number of moles}\text{×}\text{Molecular}\text{mass in grams}\text{=}\text{Number}\text{of}\text{grams}$

• Equation for volume from density is,

  $\text{Volume}\text{=}\frac{\text{Mass}}{\text{Density}}$

• Equation for finding thickness from volume and area is,

$\text{thickness}\text{=}\frac{\text{volume}}{\text{area}}$

Answer to Problem 155CP

The thickness of chromium film is $\text{4}\text{.6105}\text{×}{\text{10}}^{\text{-4}}\text{cm}$.

Explanation of Solution

To determine: The molarity of the permanganate if the $\text{28}\text{.97mL}$ of permanganate is required to titrate $\text{0}\text{.1058g}$ of oxalic acid.

The mass of ${\text{Fe(NH}}_{\text{4}}{\text{)}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{2}}{\text{.6H}}_{\text{2}}\text{O}$ reacted in the titration is given as $\text{3}\text{.0g}$.

Equation for number of moles of a substance is,

$\text{number}\text{of}\text{moles}\text{=}\frac{\text{given}\text{mass}}{\text{molar}\text{mass}}$

Therefore,

The number of moles of ${\text{Fe(NH}}_{\text{4}}{\text{)}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{2}}{\text{.6H}}_{\text{2}}\text{O}$ is,

$\text{number}\text{of}\text{moles}\text{=}\frac{\text{3}\text{.0g}}{\text{392}\text{.13g}}=\text{0}\text{.7651}\text{×}{\text{10}}^{\text{-2}}\text{mol}$

There is 1 mole of ${\text{Fe}}^{\text{2+}}$ ion in 1 mole of ${\text{Fe(NH}}_{\text{4}}{\text{)}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{2}}{\text{.6H}}_{\text{2}}\text{O}$. So the number of mole of ${\text{Fe}}^{\text{2+}}$ is $\text{0}\text{.7651}\text{×}{\text{10}}^{\text{-2}}\text{mol}$(equals to the number of moles of ${\text{Fe(NH}}_{\text{4}}{\text{)}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{2}}{\text{.6H}}_{\text{2}}\text{O}$).

The balanced equation of the reaction of ${\text{Fe}}^{\text{2+}}$ ions and $\text{8}\text{.58mL}$ of $\text{0}\text{.0520M}$ ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ is,

${\text{6Fe}}^{\text{2+}}\text{(aq)}+{\text{14H}}^{+}+{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2-}}\text{(aq)}\to {\text{2Cr}}^{\text{3+}}\text{(aq)}\text{+}{\text{7H}}_{\text{2}}\text{O}\text{+}6{\text{Fe}}^{\text{3+}}\text{(aq)}$

The mole ratio between ${\text{Fe}}^{\text{2+}}$ and ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2-}}$ ions in the balanced equation is $\text{6:1}\text{or}\text{1}\text{:}\frac{\text{1}}{\text{6}}$.

Therefore,

The number of moles of ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2-}}$ ionsreacted with $\text{0}\text{.7651}\text{×}{\text{10}}^{\text{-2}}\text{mol}$ of ${\text{Fe}}^{\text{2+}}$ ionsis,

$\frac{\text{1}}{\text{6}}\text{×}\text{0}\text{.7651}\text{×}{\text{10}}^{\text{-2}}\text{mol}\text{=}\text{0}\text{.1275}\text{×}{\text{10}}^{\text{-2}}\text{mol}$

The balanced equation for dissolution of chromium is,

${\text{3S}}_{\text{2}}{\text{O}}_{\text{8}}{}^{\text{2-}}\text{(aq)}+{\text{2Cr}}^{\text{3+}}\text{(aq)}\text{+}{\text{7H}}_2\text{O}\text{(aq)}\to {\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2-}}\text{(aq)}\text{+}{\text{14H}}^{+}\text{(aq)}+{\text{6SO}}_{\text{4}}{}^{\text{2-}}\text{(aq)}$

The mole ratio between ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2-}}$ and ${\text{Cr}}^{\text{2+}}$ ions in the balanced equation is $\text{1:2}$.

Therefore,

The number of moles chromium in the can is,

$\text{2}\times \text{0}\text{.1275}\text{×}{\text{10}}^{\text{-2}}\text{mol}\text{=}\text{0}\text{.255}\text{×}{\text{10}}^{\text{-2}}\text{mol}$

Equation for number of grams of a substance from its number of moles is,

$\text{Number of moles}\text{×}\text{Molecular}\text{mass in grams}\text{=}\text{Number}\text{of}\text{grams}$

Therefore,

The mass of chromium in the can is,

$\text{0}\text{.255}\text{×}{\text{10}}^{\text{-2}}\text{mol}\text{×}\text{51}\text{.9961g}\text{=}\text{0}\text{.1326g}$

The density of chromium is given as $\text{7}{\text{.19g/cm}}^{\text{3}}$.

Equation for volume from density is,

$\text{Volume}\text{=}\frac{\text{Mass}}{\text{Density}}$

Therefore,

$\text{Volume}\text{=}\frac{\text{0}\text{.1326g}}{\text{7}{\text{.19g/cm}}^{\text{3}}}\text{=}\text{0}{\text{.01844cm}}^{\text{3}}$

The area of the chromium plate is given as ${\text{40cm}}^{\text{2}}$.

Equation for finding thickness from volume and area is,

$\text{thickness}\text{=}\frac{\text{volume}}{\text{area}}$

Therefore,

$\text{thickness}\text{=}\frac{\text{0}{\text{.01844cm}}^{\text{3}}}{{\text{40cm}}^{\text{2}}}\text{=}\text{4}\text{.6105}\text{×}{\text{10}}^{\text{-4}}\text{cm}$

Conclusion

The molarity of the permanganate solution is determined by using the given data’s for the titration.