Chapter 8.14 - Redox Reactions as the Basis for Titrations

In the previous section, we completed a discussion on balancing of redox reactions. In this section, we will see redox titrations.

• Let us first see some basics about acid-base titration. It can be explained by using the reaction between HCl and NaOH as an example.
• We want to find the concentration of a sample of NaOH.
   ♦ We are given a sample of HCl. The concentration of this HCl sample is given.
• We can use this HCl to find the concentration of the NaOH sample.
• The procedure can be written in 8 steps:

1. Take a known volume (V₁) of NaOH in a Erlenmeyer flask (also known as conical flask or titration flask)
• Add a few drops of Phenolphthalein indicator.
• When this indicator is added, the NaOH solution becomes pink. The pink color indicates the presence of OH^- ions. We know that, those OH^- ions come from the NaOH.
2. To this solution, add the HCl slowly from a burette.
• Shake the flask gently while adding the HCl.
• Continuous shaking will ensure proper interaction between the reactants. The H⁺ ions from the burette will slowly consume the OH^- ions in the flask
• So the number of OH^- ions in the flask will slowly decrease. Due to this decrease in OH^- ions, the pink color becomes paler and paler.
3. Now reduce speed and add the HCl drop by drop.
• Add the drops until the pale pink color completely become colorless at the addition of a single drop of HCl.
• Colorless solution indicates that, no more OH^- ions are present in the flask. That means, all the OH^- ions in the flask are used up.
4. From the readings in the burette, we can easily find the volume (V₂) of the HCl transfered to the flask.
• So we can write: V₂ mL of HCl is required to neutralize V₁ mL of NaOH
5. The balanced equation for the reaction is:
HCl + NaOH → NaCl + H₂O
• The ionic equation is:
H⁺ + Cl^- + Na⁺ + OH^- → Na⁺ + Cl^- + H₂O
• Na⁺ and Cl^- do not take part in the reaction. They are spectator ions. So the net ionic equation is:
H⁺ + OH^- → H₂O
• From the stoichiometric coefficients, it is clear that, one mole of HCl can neutralize one mole of NaOH.
6. So our next task is to find how many moles of HCl were transferred into the flask. It can be done in 2 steps:
(i) Let the concentration of HCl be M₂ moles per liter. (Remember that, we are taking HCl of known concentration. So M₂ is known)
• That means, one liter of that HCl will contain M₂ moles of HCl
• That means, 1 mL of that HCl will contain $\mathbf\small{\rm{\frac{M_2}{1000}}}$ moles of HCl
(ii) We transfered V₂ mL of HCl into the flask.
• That means, we transferred $\mathbf\small{\rm{V_2 \times\frac{M_2}{1000}}}$ moles into the flask.
7. Since the stoichiometric coefficients are thar same, it is clear that, the same $\mathbf\small{\rm{V_2 \times\frac{M_2}{1000}}}$ moles of NaOH was present in the flask.
• Remember that, we took V₁ mL of NaOH for doing the titration.
• That means, V₁ mL of that NaOH contains $\mathbf\small{\rm{V_2 \times\frac{M_2}{1000}}}$ moles of NaOH
• That means, 1 mL of that NaOH will contain $\mathbf\small{\rm{\frac{V_2}{V_1} \times\frac{M_2}{1000}}}$ moles of NaOH
• That means 1000 mL of that NaOH will contain:
$\mathbf\small{\rm{\left(\frac{V_2}{V_1} \times\frac{M_2}{1000}\times 1000\right)}}$ = $\mathbf\small{\rm{\frac{M_2 V_2}{V_1}}}$ moles of NaOH
8. But number of moles per 1000 mL (1 liter) is molarity. (It can also be considered as concentration). So if M₁ is the molarity of NaOH, we can write: $\mathbf\small{\rm{M_1=\frac{M_2 V_2}{V_1}}}$
• Thus, using titration, we calculated the concentration of the given sample of NaOH

---

• In the above example, one mole of HCl can neutralize one mole of NaOH. So this example does not teach us the 'importance of stoichiometric coefficients'. We will see another example. It can be written in 6 steps:
1. Assume that, H₂SO₄ is used in the burette. Let V₂ mL of H₂SO₄ be transferred into the flask. If molarity of that H₂SO₄ is M₂, how many moles of H₂SO₄ is transferred into the flask?
• As before, we get: $\mathbf\small{\rm{V_2 \times\frac{M_2}{1000}}}$ mols of H₂SO₄
This is shown in fig.8.7(a) below:

Fig.8.7

2. Now, let the volume of NaOH in the flask be V₁ mL
• If M1 is the molarity of that NaOH, how many moles of NaOH are present in the flask?
• As before, we get: $\mathbf\small{\rm{V_1 \times\frac{M_1}{1000}}}$ mols of NaOH
This is also shown in fig.8.7(a) above.
• That means, $\mathbf\small{\rm{V_1 \times\frac{M_1}{1000}}}$ moles of NaOH is neutralized by $\mathbf\small{\rm{V_2 \times\frac{M_2}{1000}}}$ moles of H₂SO₄
3. Now consider the balanced equation:
2NaOH + H₂SO₄ → Na₂SO₄ + H₂O
• We see that: one mole of H₂SO₄ can neutralize two moles of NaOH
• That means, two mole of NaOH require one mole of H₂SO₄
◼ So we can divide all the NaOH moles into groups. Each group will contain exactly two moles of NaOH. Two such groups are shown in fig.8.7(b). The groups are indicated by dotted circles.
• Number of groups available = (Number of moles ÷ 2) = $\mathbf\small{\rm{\frac{M_1V_1}{2000}}}$
4. Equality of two numbers:
    ♦ The number of groups
    ♦ must be exactly equal to
    ♦ The number of moles of H₂SO₄
• Then only complete neutralization can take place
5. Equating the two numbers, we get:
$\mathbf\small{\rm{\frac{M_1 V_1}{2000}=\frac{M_2 V_2}{1000}}}$
Thus we get: $\mathbf\small{\rm{M_1=\frac{2M_2 V_2}{V_1}}}$
5. Previously, when it was HCl in the burette, we obtained: $\mathbf\small{\rm{M_1=\frac{M_2 V_2}{V_1}}}$
    ♦ M₁ is the unknown concentration in the flask.
    ♦ M₂ is the known concentration (moles per liter) in the burette.
    ♦ V₁ is the known volume (mL) taken in the flask.
    ♦ V₂ is the calculated volume (mL) from burette, required for neutralization.
6. So it is clear that, stoichiometric coefficients are very important while doing titration problems.

---

• In the above discussion, we saw how unknown concentrations can be determined using titration. The reactions that we saw, were neutralization reactions between acids and bases.
• We are able to use titration successfully because, there is a clearly visible end point. The end point is the point at which no more OH^- ions remain in the flask. This point is clearly visible because, the pink color completely vanishes with the addition of a single drop.
• If end point is clearly visible, redox reactions can also be used just like neutralization reactions. Then we will be able to find unknown concentrations of oxidants or reductants. Let us see some examples:

Example 1:
This can be written in 11 steps:
1. Consider the redox reaction:
$\mathbf\small{\rm{\overset{+7}{Mn}\,\overset{-2}{O_4^{-}}\,(aq)
+5\,\overset{+2}{Fe^{2+}}\,(aq)+8\,H^{+}\,(aq)\rightarrow
\overset{+2}{Mn^{2+}}\,(aq)+5\,\overset{+3}{Fe^{3+}}\,(aq)+4H_2O\,(l)}}$
• Fe is being oxidized from an oxidation state of +2 to an oxidation state of +3
• Mn is being reduced from an oxidation state of +7 to an oxidation state of +2
2. We have a solution containing Fe²⁺ ions. The concentration (M₁) of Fe2+ in that solution is not known. Our task is to determine M₁.
3. We take a small quantity (about 20 mL) of that solution in the Erlenmeyer flask. Let this quantity be V₁
• As before, the number of moles of Fe²⁺ ions in this solution will be: $\mathbf\small{\rm{V_1 \times\frac{M_1}{1000}}}$ mols
This is shown in fig.8.8(a) below:

Fig.8.8

4. We take potassium permanganate solution in the burette. This solution will supply the necessary MnO₄^- ions required to carry out the redox reaction. The concentration of this solution is known. Let it be M₂
• As before, the number of moles of MnO₄^- ions in this solution will be: $\mathbf\small{\rm{V_2 \times\frac{M_2}{1000}}}$ mols
This is shown in fig.8.8(a) above.
5. We add the permanganate solution drop by drop into the flask. Each drop has an intense purple color.
• But when the drop reaches the flask, it reacts with the Fe²⁺ and gets reduced to Mn²⁺
• The Mn²⁺ is colorless. So, when we shake the flask after adding each drop, the solution becomes colorless.
6. We continue adding the drops. When the drops are being added, the Fe²⁺ are being used up. As long as Fe²⁺ ions are available, the incoming MnO₄^- ions will be turned into Mn²⁺ ions
• So, as long as Fe²⁺ ions are available in the flask, the solution in the flask will become colorless when the flask is shaken.
7. A stage will be reached when all the Fe²⁺ ions are used up. Then the MnO₄^- ions in the next incoming drop cannot be reduced to Mn²⁺. Those new MnO₄^- ions will remain as such. This will give a tinge of pink color (small amount of pink color) to the solution. This pink tinge will be a lasting tinge because, it will not become colorless even after shaking the flask.
8. The appearance of this lasting pink tinge is the equivalence point.
• Equivalence point is the point at which the number of moles of MnO₄^- required to oxidize all the Fe²⁺ ions is available in the flask.
9. So, when the pink tinge appears, we stop adding the drops. We can now begin the calculations. It can be written in 5 steps:
(i) Every one liter in the burette will contain M₂ moles of MnO₄^- ions.
• So one mL in the burette will contain $\mathbf\small{\rm{\frac{M_2}{1000}}}$ moles of MnO₄^- ions
(ii) If V₂ (mL) is the volume used from the burette, we can write:
• $\mathbf\small{\rm{V_2 \times\frac{M_2}{1000}}}$ moles of MnO₄^- reached the flask.
• This is shown in fig.8.8(a) above.
(iii) Similarly, if M₁ is the unknown concentration of Fe²⁺ ions in the flask, the number of those ions in the flask will be $\mathbf\small{\rm{V_1 \times\frac{M_1}{1000}}}$
This is shown in fig.8.8(a) above:
V₁ is the known volume of the Fe²⁺ solution taken in the flask
(iv) So now we have the number of moles of each item in the flask:
    ♦ There are $\mathbf\small{\rm{\frac{M_1V_1}{1000}}}$ moles of Fe²⁺ ions in the flask.
    ♦ There are $\mathbf\small{\rm{\frac{M_2V_2}{1000}}}$ moles of MnO₄^- ions in the flask.
(v) Five moles of Fe²⁺ ions require one mole of MnO₄^- ions
• So we can divide all the Fe²⁺ moles into groups. Each group will contain exactly five moles of Fe²⁺. Two such groups are shown in fig.8.8(b). The groups are indicated by dotted circles.
• Number of groups available = (Number of moles ÷ 5) = $\mathbf\small{\rm{\frac{M_1V_1}{5000}}}$
10. Equality of two numbers:
    ♦ The number of groups
    ♦ must be exactly equal to
    ♦ The number of moles of MnO₄^-
• Then only complete oxidation and reduction can take place
11. Equating the two numbers, we get:
$\mathbf\small{\rm{\frac{M_1 V_1}{5000}=\frac{M_2 V_2}{1000}}}$
• Thus we get: $\mathbf\small{\rm{M_1=\frac{5M_2 V_2}{V_1}}}$
    ♦ M₁ is the unknown concentration in the flask.
    ♦ M₂ is the known concentration (moles per liter) in the burette.
    ♦ V₁ is the known volume (mL) taken in the flask.
    ♦ V₂ is the calculated volume (mL) from burette, required for neutralization.

---

Example 2:
• In the example 1 above, the MnO₄^- ions coming from the burette, are colored.
• The loss of color after adding each drop indicates that, reaction is proceeding.
• That means, as long as the added drops lose color, we can be sure about the presence of Fe²⁺ ions in the flask.
◼  But what if the solutions (solution coming from the burette and the solution in the flask) are both colorless ?
• In such a situation, we use another technique. It can be explained using Cr₂O₇^2- an example. It can be written in 5 steps:
1. The Cr₂O₇^2- solution is taken in the burette. It is added drop by drop into the flask.
2. Diphenylamine is added as indicator to the reagent in the flask.
3. The reagent present in the flask consumes the incoming Cr₂O₇^2- ions. But there is no visual indication of this consumption because, both solutions are colorless.
4. Since there is no visual indication, we find it difficult to identify the point at which the consumption of incoming Cr₂O₇^2- ions stops.
• But the indicator diphenylamine comes to our help. It can be explained in 2 steps:
(i) Consider the point at which the consumption of incoming Cr₂O₇^2- ions stops. This is the equivalence point.
(ii) The next incoming Cr₂O₇^2- ions will not be consumed. That means, the Cr₂O₇^2- ions will remain as such in the solution.
• This newly added Cr₂O₇^2- ions will oxidize the diphenylamine indicator to produce an intense blue color.
5. So the appearance of the intense blue color is the end point. At this point, we stop adding Cr₂O₇^2- ions from the burette.

---

Example 3:
In this example, we will see another type of end point. It can be written in steps:
1. We have a solution containing Cu²⁺ ions. We want to find the concentration of Cu²⁺ ions in that solution.
2. We take a small known quantity (about 20 mL) of that solution in a flask. To that, we add an excess quantity of KI solution.
3. The I^- ions from the KI solution will react with the Cu²⁺ ions according to the equation:
2Cu²⁺ (aq) + 4I^- (aq) gives Cu₂I₂ (s) + I₂ (aq)
• Excess amount of KI solution is used to ensure that, all Cu²⁺ in the solution are used up.
4. If we can find the number of moles of I₂ in the resulting solution, we will be able to calculate the number of moles of Cu²⁺ ions. For this, we use the coefficients in the balanced equation.
• For example:
   ♦ If 1 mole of I₂ is present in the resulting solution, 2 moles of Cu²⁺ ions was present in the original solution.       
   ♦ If 1.5 moles of I₂ is present in the resulting solution, 3 moles of Cu²⁺ ions was present in the original solution.
3. So our next aim is to find the number of moles of I₂ in the resulting solution. For that, we use titration. We add a solution of thiosulphate ions (S₂O₃^2-), from the burette. (We can take sodium thiosulphate solution in the burette. It will give enough supply of thiosulphate ions)
• The thiosulphate ions react with I₂ according to the equation:
I₂ (aq) + 2S₂O₃^2- (aq) gives 2I^- (aq) + S₄O₆^2- (aq)
4. We add the thiosulphate ions drop by drop. We want the equivalence point at which all the I₂ molecules are consumed.
• The solution is initially brown in color. This is due to the presence of I₂ in solution.
• As we continue to add the thiosulphate ions, the number of I₂ molecules gradually decreases. Due to this decrease, the color gradually changes from brown to yellow.
5. As the equivalence point approach, the color become very very pale yellow.
• When all I₂ are consumed, the solution becomes colorless.
6. We want the point at which all the I₂ are consumed. But the transition from 'very very pale yellow' to complete colorless is not easily identifiable.
• If we miss the point, we will be adding excess thiosulphate ions from the burette. This will lead to error.
7. To obtain an easily visible transition, we add a few drops of starch to the solution in the flask. This starch is added when the color becomes straw yellow.
• The I₂ will get attached to the starch molecules to form a complex. This complex is not a chemical product. That is., I₂ does not react chemically with starch.
• However, due to the formation of the complex, the color of the solution becomes dark blue.
8. Then we continue to add the thiosulphate ions drop by drop. The I₂ which are attached to the starch, will detach and react with the thiosulphate ions.
• Thus The number of I₂ molecules again begin to decrease.
• The dark blue color begins to fade because, the number of complex molecules are decreasing.
• At a particular drop, the blue color completely becomes colorless. This is the end point. At this point no more I₂ molecules are available in the solution.
9. From the readings in the burette, we can write the volume of thiosulphate solution used up to react with all the I₂ molecules.

---

In the next section, we will see redox reactions and electrode processes.


Previous

Contents

Next

Copyright©2021 Higher secondary chemistry.blogspot.com