Redox Titration

Chemistry indeed was my fave in my early days of college. There was a lot new to learn and experience every time we did a practical back in those old labs. We learned the difference between acids and bases, what is titration, what are compounds, its equations, calculations and so on. Every solution and every theory seems never-ending in the books of chem! There is always a room for some mix 'n' match, mumbo-jumbo of chemicals and excogitations of unfathomable possibilities.

Defining Redox Titration
Two things in a redox titration are essential to help define the word alone. One is the titrant and the other is substance participating in the reaction. This two things are formerly important to commence a reaction. Well there are other components equally important as well, but they are introduced when the reaction just begins. Technically, this titration is also called an oxidation-reduction titration. This means that the redox titration is in between titrant and an analyst. Determining process is done by using a potentiometer and or a redox indicator.

The process is too simple for any sort of insurrection to come up. It's the titration of an oxidizing agent by a reducing agent and contrariwise. In general there are various types of redox titration where in there compulsorily has to be a solution passing through a reducing agent. In most of the cases its starch is used as an indicator. There in a blue starch-iodine composite is formed only when there is a presence of an excess amount of iodine. This is one of the useful redox indicator stating the termination of the chemical reaction.

Why is this Titration Used?
Well the whole purpose of using this titration method is simple. All you have to do is discover the quantity of the reacting substance in the reaction. This will lead you to calculate all the chemical reactions and help you come to an appropriate deduction. For performing a redox reaction you need to bring together an acid and a base. An acid-base reaction caters much better results and is known to be more suited in the domain of titration.

Redox Titration Problems

Problem #1: The amount of Fe²⁺ (aq) in an FeSO₄ (aq) solution can be determined by a titration having a solution containing a known concentration of Ce⁴⁺ (aq). Thus the determination is based on this reaction:

Reaction: Fe²⁺ (aq) + Ce⁴⁺ (aq) -----> Fe³⁺ (aq) + Ce³⁺ (aq). It is given that it requires 37.50 mL of 0.09650 M Ce⁴⁺ (aq) to oxide the Fe²⁺ (aq) in a 35.00 mL sample to Fe³⁺ (aq). Now calculate the molarity of Fe²⁺ (aq) and also the number of milligrams of iron in the given sample.

Solution
#1 Determine the amount of Ce(IV) which reacts:
(0.09650 mol/L) (0.03750 L) = 0.00361875 mol

#2 Determine the amount of iron(II) ion which reacts:
There is a 1:1 molar ratio in between Fe(II) and Ce(IV), thus;
0.00361875 mol of Fe(II) reacts.

#3 Determine the molarity by using the molarity formula:
0.00361875 mol/ 0.03500 L = 0.1034 M

#4 Determine milligrams:
0.00361875 mol times 55.845 b/mol= 0.2021 g = 202.1 mg

Problem #2: This problem has a rock sample which is to be assayed for a tin content by an oxidation-reduction titration with I₃^- (aq). A 10.00 g of rock sample is crushed and dissolved in sulfuric acid which is then passed over a reducing agent so that the whole tin is in the form of Sn²⁺. The Sn²⁺ (aq) is now completely oxidized by 34.60 mL of a 0.5560 M solution of NaI₃. Thus the determination is based on this reaction:
I₃^- (aq) + Sn²⁺ (aq) -----> Sn⁴⁺ (aq) + 3I₃^- (aq). Now calculate the amount of tin in the sample and also its mass percentage in the rock.

Solution
#1 Determine the amount of triiodide used:
(0.5560 mol/L) (0.03460 L) = 0.0192376 mol of I₃^- reacted.

#2 Determine the Sn(II) that reacted:
There is a 1:1 molar ratio between triiodide and the stannous ion. Therefore;
0.0192376 mol of Sn(II) reacted.

#3 Calculate grams of tin(II) ion:
0.0192376 mol times 118.710 g/mol = 2.284 g

#4 Calculate Mass Percentage
2.284 g/ 10.00 g = 22.84%

Problem #3: Sodium oxalate is dissolved in the flask which measures 0.2640 g and yet requires 30.74 mL of potassium permanganate from a buret in order to titrate the composition. The composition is going to turn pink till it reaches the end. Thus the determination is based on this reaction:

5Na₂C₂O₄ (aq) + 2KMnO₄ + 8H₂SO₄ (aq) -----> 2MnSO₄(aq) + K₂SO₄(aq) + 5Na₂SO₄(aq) + 10 CO₂(g) + 8H₂O(I)

a) Determine how many moles of sodium oxalate are present in the flask?
b) Determine how many moles of potassium permanganate have been titrated into the flask to reach an end point?
c) What is the molarity of potassium permanganate?

Solution
For (a): 0.2640 g/134.00 g/mol = 0.001970149 mol (to four sig figs) which would be;
0.001970 mol

For (b): From the given balanced equation, the oxalate permanganate molar ratio is 5:2
(0.001970149 mol oxalate) x (2 mol permanganate/5 mol oxalate) = 0.00078806 mol permanganate (to four sig figs) which is 0.0007881 mol

For (c): 0.00078806 mol/ 0.03074 L = 0.02564 M (to four sig figs)

All you need to do is apprehend the determining methods correctly and you are good to go! Carefully, read the stated problem, take a look at the given balanced equation and proceed with the lab procedures that's all. In many cases there is a table provided to fill up all the details, calculations and solutions to keep a better record of your experiments. Make sure you enjoy doing so! (smiley)