how to solve redox titration problems

First, you would write the balanced equation for the reaction.

The skeleton ionic equation is

#"MnO"_4^(-) + "C"_2"O"_4^(2-) → "Mn"^(2+) + "CO"_2#

Then you would probably balance this by the ion electron method to get the balanced net ionic equation

#"2MnO"_4^(-) + "5C"_2"O"_4^(2-) + "16H"^+ → "2Mn"^(2+) + "10CO"_2 + "8H"_2"O"#

and perhaps put the spectator ions back in to get the molecular equation

#"2KMnO"_4 + "5Na"_2"C"_2"O"_4 + "8H"_2"SO"_4→ "2MnSO"_4 + "10CO"_2 +"K"_2"SO"_4 + "5Na"_2"SO"_4 + "8H"_2"O" #

From here on, the calculation is the same as for an acid-base titration.

Step 1. Calculate the moles of titrant.

#"Moles of KMnO"_4 = 0.3101 cancel("g Na₂C₂O₄") × (1 cancel("mol Na₂C₂O₄"))/(134.00 cancel("g Na₂C₂O₄")) × ("2 mol KMnO"_4)/(5 cancel("mol Na₂C₂O₄")) = 9.2567 × 10^-4"mol KMnO"_4#

Step 2. Calculate the molarity of the titrant.

#"Molarity" = (9.2567 × 10^-4"mol")/"0.024 90 L" = "0.037 17 mol/L"#

Related questions

• How do you do acid base titration calculations?
• How do you use titration calculations to find pH?
• What is a redox titration and what is it used for?
• Why is titration used when standardizing a solution?
• Is titration suitable for sodium nitrate?
• How can I calculate the titration of a weak acid and a strong base?
• How can I make back titration calculations?
• How does titration affect molarity?
• How does the endpoint of a titration differ from the equivalence point?
• How does pH change during titration?

Impact of this question

Redox Titration Questions

Titration, also known as titrimetry, is a chemical qualitative analysis technique for determining the concentration of an analyte in a mixture. Titration is an important technique in analytical chemistry, and it is also known as volumetric analysis.

Redox Titration in Chemistry Questions with Solutions

Q1. Which of the following are redox titrations?

a.) Iodometric titration

b.) Acid-base titration

c.) Complexometric titration

d.) All of the above

Correct Answer – (d.) All of the above

Q2. The ____ can detect the endpoint of a titration involving the reaction of MnO 4 – to Mn 2+ .

a.) Use of oH probe

b.) Disappearance of the pink colour of the MnO 4

c.) Addition of an indicator like methyl orange

d.) Use of red or blue litmus

Correct Answer – (b.) Disappearance of the pink colour of the MnO 4

Q3. Due to which of the following reasons can potassium permanganate not be prepared as a primary standard?

a.) It has a relatively low molar mass.

b.) It only reacts in an acidic solution.

c.) Its solutions are volatile.

d.) It is not highly soluble in water.

Correct Answer – (c.) Its solutions are volatile.

Q4. Excess KI reacts with CuSO 4 solution, and then Na 2 S 2 O 3 solution is added to it. Which of the statements is incorrect for this reaction?

a.) Na 2 S 2 O 3 is oxidised

b.) CuI 2 is formed

c.) Cu 2 I 2 is formed

d.) Evolved I 2 is reduced

Correct Answer – (b.) CuI 2 is formed.

Q5. Dichromate ion reduces to ______ ion in an acidified solution of potassium dichromate (VI).

a.) Chromate (V) ions

b.) Chromium (III) ions

c.) Chromium (II) ions

d.) Chromium (IV) ions

Correct Answer – (b.) Chromium (III) ions

Q6. What is the principle of redox titrations?

Answer. The oxidation-reduction titrations work on the principle that the oxidation process involves electron loss while the reduction process involves electron gain.

Q7. How is the endpoint of a redox titration determined?

Answer. When we add a redox indicator to the titrand, the indicator imparts a colour based on the potential of the solution. As the potential of the solution changes due to the addition of titrant, the indicator changes oxidation state and colour, signalling the endpoint.

Q8. What is the redox titration curve?

Answer. To evaluate a redox titration, first, determine the shape of its titration curve. The titration curve in an acid-base titration or a complexation titration shows how the concentration of H 3 O + (as pH) or M n+ (as pM) changes as the titrant is added. It is more convenient to monitor the potential of the titration reaction rather than the concentration of one species during a redox titration.

Q9. What are the types of redox titrations?

Answer. Redox titrations are named after the titrant used, and they are:

• A bromine (Br 2 ) titrant is used in bromometry.
• Cerimetry makes use of cerium(IV) salts.
• Potassium dichromate is used in dichrometry.
• Iodine is used in iodometry (I 2 ).
• Potassium permanganate is used in permanganometry.

Q10. What is the principle of redox titration?

Answer. If the following conditions are met, a redox reaction can be used as the basis for a titration:

• The redox reaction must be rapid and nearly complete (less than 99% success rate is unacceptable).
• The endpoint must be quantifiable or identifiable using a colour indicator or potentiometry.
• Stoichiometric electron exchange means that the redox systems (oxidising and reducing agents) must be adequate and equal.

Q11. Give the general sequence for calculations in redox titration.

Answer. The general sequence for calculating redox titration is as follows:

• Write down the half equations for the oxidant and reductant.
• Determine the overall equation.
• Determine the amount of manganate(VII) or dichromate(VI) used.
• From the overall redox equation, calculate the ratio of moles of oxidant to moles of reductant.
• Determine the number of moles in the reductant sample solution.
• Determine the number of moles of reductant in the original solution.
• Determine either the concentration of the original solution or the percentage of reductant in a known quantity of the sample.

Q12. In dilute sulfuric acid, an iron tablet weighing 0.960 g was dissolved. To reach the endpoint, an average titre of 28.50 cm 3 of 0.0180 mol dm -3 potassium manganate(VII) solution was required. What is the iron percentage of the tablet in terms of mass?

Answer. The equation of the reaction is:

MnO 4 – (aq) + 8H + (aq) + 5Fe 2+ → Mn 2+ (aq) + 5Fe 3+ (aq) + 4H 2 O (l)

Number of moles of MnO 4 – = 0.0180 × 0.0285 = 0.000513 = 5.13 × 10 –4 moles.

According to the balanced equation, the molar ratio of permanganate and ferrous ions is 1:5

Therefore, moles of iron (II) = 5 × 5.13 × 10 –4 = 2.565 × 10 –3 moles

Mass of iron (II) = 56 × 2.565 × 10 –3 = 0.14364 g

% by mass of iron = 90.14364/0.960) × 100 = 15%

Q13. A 32.15 mL sample of MoO 4 2– (aq) solution was passed through a Jones reductor (a zinc powder column) to convert all of the MoO 4 2– (aq) to Mo3+ (aq). For the reaction given, the filtrate required 20.85 mL of 0.09550 M KMnO 4 – (aq).

MnO 4 ¯(aq) + Mo 3+ (aq) → Mn 2+ (aq) + MoO 2 2+ (aq)

Balance this equation, and determine the concentration of the original MoO 4 2– (aq) solution.

Answer. Half reactions:

5e – + 8H + +MnO 4 – → Mn 2+ + 4H 2 O

2H 2 O + Mo 3+ → MoO 2 2+ + 4H + + 3e –.

The balanced equation is:

4H + + 3MnO 4 – 5Mo 3+ → 3Mn 2+ + 5MoO 2 2+ + 2H 2 O

Moles of permanganate required:

0.09550 mol L –1 × 0.02085 L = 0.001991175 mol

Moles of molybdate required:

According to the balanced equation, the molar ratio of permanganate and molybdate is 3:5.

Therefore, 0.001991175 × (5/3) = 0.003318625 mol molybdate.

Concentration:

0.003318625 mol/0.03215 L = 0.1032 M

Q14. A flask contains 0.2640 g of sodium oxalate, which requires 30.74 mL of potassium permanganate (from a buret) to titrate and turn pink (the endpoint).

This reaction’s equation is:

5Na 2 C 2 O 4 (aq) + 2KMnO 4 (aq) + 8H 2 SO 4 (aq) → 2MnSO 4 (aq) + K 2 SO 4 (aq) + 5Na 2 SO 4 (aq) + 10CO 2 (g) + 8H 2 O(l)

a.) How many moles of sodium oxalate are there in the flask?

b.) How many moles of potassium permanganate were titrated into the flask to reach the endpoint?

c.) What is the molecular weight of potassium permanganate?

The number of moles of sodium oxalate in the flask is 0.00197 mol.

b.) According to the balanced equation, the molar ratio of oxalate to permanganate is 5:2.

= 0.001970149 mol oxalate ×(⅖) = 0.00078806 mol permanganate

The number of moles of potassium permanganate = 0.0007881 mol.

c.) The molecular weight of potassium permanganate = 0.00078806 mol / 0.03074 L = 0.02564 mol/L

Q15. Potassium dichromate is used to titrate a sample containing an unknown percentage of iron. To reduce all of the iron to Fe 2+ ions, the sample is dissolved in an H 3 PO 4 /H 2 SO 4 mixture. The solution is then titrated with 0.01625 M K 2 Cr 2 O 7 , resulting in the formation of Fe 3+ and Cr 3+ ions in an acidic solution. For 1.2765 g of sample, the titration requires 32.26 mL of K 2 Cr 2 O 7 .

(a) Using the half-reaction method, balance the net ionic equation.

(b) Determine the iron % in the sample.

(c) Does the sample consist of ferrous iodate, ferrous phosphate or ferrous acetate?

a.) 6Fe 2+ + Cr 2 O 7 2– + 14H + → 6Fe 3+ + 2Cr 3+ + 7H 2 O

b.) Determination of Fe(II) in solution:

(0.01625 mol/L) × (0.03226 L) = 0.000524225 mol dichromate

Since the ratio of ferrous ion to dichromate is 6:1

0.000524225 × (6/1) = 0.00314535 mol Fe

0.00314535 mol Fe(II) × 55.845 g/mol = 0.175652 g

% of iron in the sample:

(0.175652 g/1.2765 g) × 100 = 13.76 %

c.) Calculating the % composition of ferrous iodate, ferrous phosphate, or ferrous acetate will help in determining the sample.

Ferrous iodate = Fe(IO 3 ) 2

%Fe = (55.845 / 405.67) = 13.77 %

Ferrous phosphate = Fe 3 (PO 4 ) 2

%Fe = (55.845 / 357.48) = 15.62 %

Ferrous acetate Fe(C 2 H 3 O 2 ) 2

%Fe = (55.845 /173.93) = 32. 11%

Hence, the sample consists of ferrous iodate.

Practice Questions on Redox Titration

Q1. For redox titrations, KMnO 4 is acidified by:

c.) H 2 SO 4

Q2. Which of the following is the colour change when acidified potassium dichromate (VI) is used as an oxidising agent?

a.) Orange to red

b.) Orange to green

c.) Yellow to red

d.) Yellow to green

Q3. What are redox indicators?

Q4. 0.500 g of zinc powder reduced a 25.50 mL acidified solution of 0.200 M AO 2 + . What is metal A’s final oxidation number?

Q5. Titration with a solution containing a known concentration of S 2 O 3 2– (aq) can be used to determine the amount of I 3 – (aq) in a solution (thiosulfate ion). The balanced equation is used to make the determination:

I 3 ¯(aq) + 2S 2 O 3 2– (aq) → 3I – (aq) + S 4 O 6 2¯ (aq)

Calculate the molarity of I 3 ¯ (aq) in the solution, given that it takes 36.40 mL of 0.3300 M Na 2 S 2 O 3 (aq) to titrate the I 3 – (aq) in a 15.00 mL sample.

Click the PDF to check the answers for Practice Questions. Download PDF

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9.4: Redox Titrations

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Analytical titrations using redox reactions were introduced shortly after the development of acid–base titrimetry. The earliest Redox titration took advantage of the oxidizing power of chlorine. In 1787, Claude Berthollet introduced a method for the quantitative analysis of chlorine water (a mixture of Cl 2 , HCl, and HOCl) based on its ability to oxidize indigo, a dye that is colorless in its oxidized state. In 1814, Joseph Gay-Lussac developed a similar method for determining chlorine in bleaching powder. In both methods the end point is a change in color. Before the equivalence point the solution is colorless due to the oxidation of indigo. After the equivalence point, however, unreacted indigo imparts a permanent color to the solution.

The number of redox titrimetric methods increased in the mid-1800s with the introduction of MnO 4 – , Cr 2 O 7 2– , and I 2 as oxidizing titrants, and of Fe 2 + and S 2 O 3 2– as reducing titrants. Even with the availability of these new titrants, redox titrimetry was slow to develop due to the lack of suitable indicators. A titrant can serve as its own indicator if its oxidized and reduced forms differ significantly in color. For example, the intensely purple MnO 4 – ion serves as its own indicator since its reduced form, Mn 2 + , is almost colorless. Other titrants require a separate indicator. The first such indicator, diphenylamine, was introduced in the 1920s. Other redox indicators soon followed, increasing the applicability of redox titrimetry.

9.4.1 Redox Titration Curves

To evaluate a redox titration we need to know the shape of its titration curve. In an acid–base titration or a complexation titration, the titration curve shows how the concentration of H 3 O + (as pH) or M n + (as pM) changes as we add titrant. For a redox titration it is convenient to monitor the titration reaction’s potential instead of the concentration of one species.

You may recall from Chapter 6 that the Nernst equation relates a solution’s potential to the concentrations of reactants and products participating in the redox reaction. Consider, for example, a titration in which a titrand in a reduced state, A red , reacts with a titrant in an oxidized state, B ox .

\[A_\textrm{red}+B_\textrm{ox} \rightleftharpoons B_\textrm{red}+A_\textrm{ox}\]

where A ox is the titrand’s oxidized form, and B red is the titrant’s reduced form. The reaction’s potential, E rxn , is the difference between the reduction potentials for each half-reaction.

\[E_\textrm{rxn}=E_{B_\mathrm{\Large ox}/B_\mathrm{\Large red}}-E_{A_\mathrm{\Large ox}/A_\mathrm{\Large red}}\]

After each addition of titrant the reaction between the titrand and the titrant reaches a state of equilibrium. Because the potential at equilibrium is zero, the titrand’s and the titrant’s reduction potentials are identical.

\[E_{B_\mathrm{\Large ox}/B_\mathrm{\Large red}}=E_{A_\mathrm{\Large ox}/A_\mathrm{\Large red}}\]

This is an important observation because we can use either half-reaction to monitor the titration’s progress.

Before the equivalence point the titration mixture consists of appreciable quantities of the titrand’s oxidized and reduced forms. The concentration of unreacted titrant, however, is very small. The potential, therefore, is easier to calculate if we use the Nernst equation for the titrand’s half-reaction

\[E_\textrm{rxn}= E^o_{A_\mathrm{\Large ox}/A_\mathrm{\Large red}}-\dfrac{RT}{nF}\ln\dfrac{[A_\textrm{red}]}{[A_\textrm{ox}]}\]

Although the Nernst equation is written in terms of the half-reaction’s standard state potential, a matrix-dependent formal potential often is used in its place. See Appendix 13 for the standard state potentials and formal potentials for selected half-reactions.

After the equivalence point it is easier to calculate the potential using the Nernst equation for the titrant’s half-reaction.

\[E_\textrm{rxn}= E^o_{B_\mathrm{\Large ox}/B_\mathrm{\Large red}}-\dfrac{RT}{nF}\ln\dfrac{[B_\textrm{red}]}{[B_\textrm{ox}]}\]

Calculating the Titration Curve

Let’s calculate the titration curve for the titration of 50.0 mL of 0.100 M Fe 2 + with 0.100 M Ce 4 + in a matrix of 1 M HClO 4 . The reaction in this case is

\[\textrm{Fe}^{2+}(aq)+\textrm{Ce}^{4+}(aq)\rightleftharpoons \textrm{Ce}^{3+}(aq)+\textrm{Fe}^{3+}(aq)\tag{9.15}\]

In 1 M HClO 4 , the formal potential for the reduction of Fe 3 + to Fe 2 + is +0.767 V, and the formal potential for the reduction of Ce 4 + to Ce 3 + is +1.70 V.

Because the equilibrium constant for reaction 9.15 is very large—it is approximately 6 × 10 15 —we may assume that the analyte and titrant react completely.

Step 1: Calculate the volume of titrant needed to reach the equivalence point.

The first task is to calculate the volume of Ce 4 + needed to reach the titration’s equivalence point. From the reaction’s stoichiometry we know that

\[\textrm{moles Fe}^{2+}=\textrm{moles Ce}^{4+}\]

\[M_\textrm{Fe}\times V_\textrm{Fe} = M_\textrm{Ce}\times V_\textrm{Ce}\]

Solving for the volume of Ce 4 + gives the equivalence point volume as

\[V_\textrm{eq} = V_\textrm{Ce} = \dfrac{M_\textrm{Fe}V_\textrm{Fe}}{M_\textrm{Ce}}=\dfrac{\textrm{(0.100 M)(50.0 mL)}}{\textrm{(0.100 M)}}=\textrm{50.0 mL}\]

Step 2: Calculate the potential before the equivalence point by determining the concentrations of the titrand’s oxidized and reduced forms, and using the Nernst equation for the titrand’s reduction half-reaction.

Before the equivalence point, the concentration of unreacted Fe 2 + and the concentration of Fe 3 + are easy to calculate. For this reason we find the potential using the Nernst equation for the Fe 3 + /Fe 2 + half-reaction.

\[E = E^o_\mathrm{\large Fe^{3+}/Fe^{2+}} - \dfrac{RT}{nF}\log\dfrac{[\mathrm{Fe^{2+}}]}{[\mathrm{Fe^{3+}}]}=+0.767\textrm V - 0.05916\log\dfrac{[\mathrm{Fe^{2+}}]}{[\mathrm{Fe^{3+}}]}\tag{9.16}\]

For example, the concentrations of Fe 2 + and Fe 3 + after adding 10.0 mL of titrant are

\[\begin{align} [\textrm{Fe}^{2+}]&=\dfrac{\textrm{initial moles Fe}^{2+} - \textrm{moles Ce}^{4+}\textrm{ added}}{\textrm{total volume}}=\dfrac{M_\textrm{Fe}V_\textrm{Fe} - M_\textrm{Ce}V_\textrm{Ce}}{V_\textrm{Fe}+V_\textrm{Ce}}\\ &\mathrm{= \dfrac{(0.100\;M)(50.0\;mL)-(0.100\;M)(10.0\;mL)}{50.0\;mL+10.0\;mL} = 6.67\times10^{-2}\;M} \end{align}\]

\[\begin{align} [\mathrm{Fe^{3+}}]&=\mathrm{\dfrac{moles\;Ce^{4+}\;added}{total\;volume}}=\dfrac{M_\textrm{Ce}V_\textrm{Ce}}{V_\textrm{Fe} + V_\textrm{Ce}}\\ &=\dfrac{\textrm{(0.100 M)(10.0 mL)}}{\textrm{50.0 mL + 10.0 mL}}=1.67\times10^{-2}\textrm{ M} \end{align}\]

Substituting these concentrations into equation 9.16 gives a potential of

\[E = +0.767\textrm{ V} - 0.05916 \log\dfrac{6.67\times10^{-2}\textrm{ M}}{1.67\times10^{-2}\textrm{ M}}=+0.731\textrm{ V}\]

Step 3: Calculate the potential after the equivalence point by determining the concentrations of the titrant’s oxidized and reduced forms, and using the Nernst equation for the titrant’s reduction half-reaction.

After the equivalence point, the concentration of Ce 3 + and the concentration of excess Ce 4 + are easy to calculate. For this reason we find the potential using the Nernst equation for the Ce 4 + /Ce 3 + half-reaction.

\[E=E^o_\mathrm{\large{Ce^{4+}/Ce^{3+}}}-\dfrac{RT}{nF}\log\mathrm{\dfrac{[Ce^{3+}]}{[Ce^{4+}]}}=+ 1.70\textrm{ V} - 0.05916 \log\mathrm{\dfrac{[Ce^{3+}]}{[Ce^{4+}]}}\tag{9.17}\]

For example, after adding 60.0 mL of titrant, the concentrations of Ce 3 + and Ce 4 + are

\[\begin{align} [\textrm{Ce}^{3+}]&={\dfrac{\textrm{initial moles Fe}^{2+}}{\textrm{total volume}}}=\dfrac{M_\textrm{Fe}V_\textrm{Fe}}{V_\textrm{Fe}+V_\textrm{Ce}}\\ &=\dfrac{\textrm{(0.100 M)(50.0 mL)}}{\textrm{50.0 mL + 60.0 mL}}=4.55\times10^{-3}\textrm{ M} \end{align}\]

\[\begin{align} [\textrm{Ce}^{4+}]&=\dfrac{\textrm{moles Ce}^{4+}\textrm{ added} - \textrm{initial moles Fe}^{2+}}{\textrm{total volume}}=\dfrac{M_\textrm{Ce}V_\textrm{Ce}-M_\textrm{Fe}V_\textrm{Fe}}{V_\textrm{Fe}+V_\textrm{Ce}}\\ &=\dfrac{\textrm{(0.100 M)(60.0 mL)}-\textrm{(0.100 M)(50.0 mL)}}{\textrm{50.0 mL + 60.0 mL}}=9.09\times10^{-3}\textrm{ M} \end{align}\]

Substituting these concentrations into equation 9.17 gives a potential of

\[E=+1.70\textrm{ V}-0.05916\log\dfrac{4.55\times10^{-2}\textrm{ M}}{9.09\times10^{-3}\textrm{ M}}=+1.66\textrm{ V}\]

Step 4: Calculate the potential at the equivalence point.

At the titration’s equivalence point, the potential, E eq , in equation 9.16 and equation 9.17 are identical. Adding the equations together to gives

\[2E_\textrm{eq}= E^o_\mathrm{\large Fe^{3+}/Fe^{2+}}+E^o_\mathrm{\large Ce^{4+}/Ce^{3+}}-0.05916\log\dfrac{\mathrm{[{Fe}^{2+}][Ce^{3+}]}}{\mathrm{[Fe^{3+}][Ce^{4+}]}}\]

Because [Fe 2 + ] = [Ce 4 + ] and [Ce 3+ ] = [Fe 3 + ] at the equivalence point, the log term has a value of zero and the equivalence point’s potential is

\[E_\textrm{eq}=\dfrac{E^o_\mathrm{\large Fe^{3+}/Fe^{2+}} + E^o_\mathrm{\large Ce^{4+}/Ce^{3+}}}{2}=\dfrac{\textrm{0.767 V + 1.70 V}}{2}=1.23\textrm{ V}\]

Additional results for this titration curve are shown in Table 9.15 and Figure 9.36.

Figure 9.36 Titration curve for the titration of 50.0 mL of 0.100 M Fe 2 + with 0.100 M Ce 4 + . The red points correspond to the data in Table 9.15. The blue line shows the complete titration curve.

Practice Exercise 9.17

Calculate the titration curve for the titration of 50.0 mL of 0.0500 M Sn 2 + with 0.100 M Tl 3 + . Both the titrand and the titrant are 1.0 M in HCl. The titration reaction is

\[\textrm{Sn}^{2+}(aq)+\textrm{Tl}^{3+}(aq)\rightarrow \textrm{Sn}^{4+}(aq)+\textrm{Tl}^+(aq)\]

Click here to review your answer to this exercise.