Problem 86
Question
In the standardization of \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) using \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) by iodometry, the equivalent weight of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) is (a) (molecular weight)/2 (b) (molecular weight) \(/ 6\) (c) (molecular weight \() / 3\) (d) same as molecular weight.
Step-by-Step Solution
Verified Answer
(b) (molecular weight) / 6
1Step 1: Identify the Redox Reaction
In this reaction,
K2Cr2O7 acts as an oxidizing agent and Na2S2O3 acts as a reducing agent. In acidic conditions,
K2Cr2O7 will get reduced to
Cr3+ ions, and iodide ions from KI will be oxidized to iodine. This iodine is then titrated with Na2S2O3.
2Step 2: Determine Change in Oxidation State
The equation for the reduction of Cr2O7^{2-} to 2Cr^{3+} in acidic medium requires 6 electrons:\[Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O\]
3Step 3: Calculate Equivalent Weight
The equivalent weight of a species is calculated by dividing its molecular weight by the number of electrons lost per formula unit in the redox reaction. Since
Cr2O7^{2-} gains 6 electrons per molecule,
K2Cr2O7 has to divide its molecular weight by 6.
Key Concepts
Equivalent WeightOxidizing AgentsReducing Agents
Equivalent Weight
Equivalent weight is a crucial concept in redox reactions. It refers to the mass of a substance that reacts with or supplies one mole of electrons in a chemical reaction. This concept helps in understanding how different substances participate in reactions. For example, in the exercise you're dealing with, you have to find the equivalent weight of potassium dichromate, \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\).
A simple way to calculate equivalent weight is to take the molecular weight of that substance and divide it by the number of electrons transferred per molecule in the redox process. So, if you have a reaction where a substance loses or gains a certain number of electrons, the equivalent weight helps to decide how much of that substance is needed to interact with one mole of electrons.
In the specific example given in your exercise, \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\) gains 6 electrons when it is reduced to \(\mathrm{Cr}^{3+}\). Therefore, you divide the molecular weight by 6 to get the equivalent weight.
A simple way to calculate equivalent weight is to take the molecular weight of that substance and divide it by the number of electrons transferred per molecule in the redox process. So, if you have a reaction where a substance loses or gains a certain number of electrons, the equivalent weight helps to decide how much of that substance is needed to interact with one mole of electrons.
- Molecular weight of the substance
- Number of electrons transferred
In the specific example given in your exercise, \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\) gains 6 electrons when it is reduced to \(\mathrm{Cr}^{3+}\). Therefore, you divide the molecular weight by 6 to get the equivalent weight.
Oxidizing Agents
An oxidizing agent is a substance that causes oxidation by keeping and causing another substance to lose electrons. This means the oxidizing agent itself gets reduced in the process. In redox reactions, identifying the oxidizing agent is vital for setting up the reaction equation and understanding the electron flow.
Potassium dichromate, \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\), acts as an oxidizing agent in your exercise scenario. When reduced, its chromium goes from an oxidation state of +6 to +3. This reduction is balanced by the oxidizing agent gaining electrons, specifically 6 electrons here. So, an oxidizing agent can be seen as driving force behind the oxidization process because it readily accepts the electrons.
Understanding which compound acts as the oxidizing agent helps you to determine the course and balance of the redox reaction, which is vital when calculating things like equivalent weight.
Potassium dichromate, \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\), acts as an oxidizing agent in your exercise scenario. When reduced, its chromium goes from an oxidation state of +6 to +3. This reduction is balanced by the oxidizing agent gaining electrons, specifically 6 electrons here. So, an oxidizing agent can be seen as driving force behind the oxidization process because it readily accepts the electrons.
- Accepts electrons and gets reduced
- Drives the oxidation of other substances
Understanding which compound acts as the oxidizing agent helps you to determine the course and balance of the redox reaction, which is vital when calculating things like equivalent weight.
Reducing Agents
In contrast to oxidizing agents, reducing agents are substances that donate electrons in redox reactions. By donating electrons, a reducing agent itself becomes oxidized. This is how it facilitates the reduction of another substance.
In your exercise, sodium thiosulfate, \(\mathrm{Na}_2 \mathrm{S}_2 \mathrm{O}_3\), acts as the reducing agent. During the redox reaction, it provides electrons that are used to convert iodine into iodide, which is an essential step in the iodometry process employed in the given exercise.
Identifying the reducing agent is crucial to fully understand how the redox reaction is balanced, as well as to know which substances are working together in the reaction. It also helps in discovering how much of a reducing agent is needed under the given conditions, especially when calculating equivalent weight.
In your exercise, sodium thiosulfate, \(\mathrm{Na}_2 \mathrm{S}_2 \mathrm{O}_3\), acts as the reducing agent. During the redox reaction, it provides electrons that are used to convert iodine into iodide, which is an essential step in the iodometry process employed in the given exercise.
- Donates electrons to another substance
- Gets oxidized as part of the reaction
Identifying the reducing agent is crucial to fully understand how the redox reaction is balanced, as well as to know which substances are working together in the reaction. It also helps in discovering how much of a reducing agent is needed under the given conditions, especially when calculating equivalent weight.
Other exercises in this chapter
Problem 83
In the process of extraction of gold, roasted gold ore \(+\mathrm{CN}^{-}+\mathrm{H}_{2} \mathrm{O} \stackrel{\mathrm{O}_{2}}{\longrightarrow}[\mathrm{X}]+[\mat
View solution Problem 85
Anhydrous ferric chloride is prepared by (a) heating hydrated ferric chloride at a high temperature in a stream of air (b) heating metallic iron in a stream of
View solution Problem 87
Which of the following statement is correct with reference to the ferrous and ferric ions? 1\. \(\mathrm{Fe}^{3+}\) gives brown colour with potassium ferricyani
View solution Problem 88
Which of the following statement is correct when a mixture of \(\mathrm{NaCl}\) and \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) is gently warmed with conc
View solution