A \(1 \mathrm{M}\) solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is electrolyzed. Select right statement with products at anode and cathode respectively : Given : $$ \begin{aligned} 2 \mathrm{SO}_{4}^{2-} & \longrightarrow \mathrm{S}_{2} \mathrm{O}_{8}^{2-}+2 e^{-} ; E^{\circ}=-2.01 \mathrm{~V} \\ \mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 2 \mathrm{H}^{+}(a q)+1 / 2 \mathrm{O}_{2}(g)+2 e^{-} ; E^{\circ}=-1.23 \mathrm{~V} \end{aligned} $$ (a) concentration of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) remain constant; \(\mathrm{H}_{2}, \mathrm{O}_{2}\) (b) concentration of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) increases; \(\mathrm{O}_{2}, \mathrm{H}_{2}\) (c) concentration of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) decreases; \(\mathrm{O}_{2}, \mathrm{H}_{2}\) (d) concentration of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) remains constant; \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}, \mathrm{H}_{2}\)

Short Answer

Expert verified
The correct statement is (a) concentration of H\(_2\)SO\(_4\) remain constant; H\(_2\), O\(_2\).

Step by step solution

01

Identify Reaction Occurring at Cathode

The cathode is the electrode where reduction occurs. In this case, the likely reduction reaction is the production of hydrogen gas via the reduction of H\(^{+}\) ions to H\(_2\). The reaction is \(2 \text{H}^{+} + 2 e^{-} \rightarrow \text{H}_2\).
02

Identify Reaction Occurring at Anode

At the anode, oxidation occurs. We have two possible oxidation reactions. The given standard electrode potentials (E\(^{\text{o}}\)) for both \(2 \text{SO}_4^{2-} \rightarrow \text{S}_2 \text{O}_8^{2-} + 2 e^{-}\) and \(\text{H}_2\text{O}(l) \rightarrow 2 \text{H}^{+} + \frac{1}{2} \text{O}_2(g) + 2 e^{-}\) are -2.01 V and -1.23 V respectively. The reaction with the lower (more positive) potential will occur preferentially, which is the oxidation of water to O\(_2\).
03

Determine Changes in Concentration

Since water is being oxidized at the anode, the sulfate ions are not consumed during the electrolysis and thus, the concentration of H\(_2\)SO\(_4\) remains constant.
04

Choose the Correct Option

Based on the anode and cathode reactions and the concentration change, the correct statement is (a) concentration of H\(_2\)SO\(_4\) remain constant; H\(_2\), O\(_2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Electrode Potentials
The concept of standard electrode potentials, often represented as E°, is crucial in understanding the behavior of cells in electrochemistry. These potentials are measured under standard conditions, which typically involve solutions at 1 molar concentration, a pressure of 1 atmosphere, and a temperature of 25°C (298 K).

For instance, the electrolysis of a 1M solution of H2SO4 involves comparing the standard electrode potentials of possible reactions to predict which will occur at the anode and cathode. A higher (more positive) E° value indicates a greater tendency to gain electrons and undergo reduction, while a lower (more negative) E° suggests a propensity for oxidation.

In the example problem, we examine the E° values of two potential oxidation reactions at the anode. The reaction producing O2 from water has a less negative E° than the reaction forming S2O82- from sulfate ions, signifying that water oxidation is the favorable process under standard conditions. Understanding these potentials helps in predicting the direction and extent of chemical reactions during electrolysis.
Electrochemical Series
An important tool in predicting the outcomes of electrochemical reactions is the electrochemical series. This series is essentially a list of elements and some compounds organized according to their standard electrode potentials.

This list allows chemists to quickly determine which species are more likely to be oxidized or reduced during electrolysis. The more positive the electrode potential, the greater the element's ability to be reduced. Conversely, a more negative potential indicates a greater tendency to lose electrons and be oxidized.

When considering multiple species in a solution, the electrochemical series can help determine the dominant reaction at both the anode and cathode. For instance, in the given exercise, the series can help us understand why O2 is produced at the anode instead of S2O82-, because O2 has a less negative standard electrode potential.
Redox Reactions for JEE Chemistry
Redox reactions are a class of chemical reactions involving the transfer of electrons between two species. In the context of JEE Chemistry, understanding redox reactions is essential for analyzing electrochemical processes. The term 'redox' comes from the two concepts of reduction and oxidation.

Reduction refers to the gain of electrons, while oxidation involves the loss of electrons. These reactions always occur together; when one species undergoes oxidation, another must be reduced. This is well illustrated in the exercise, where water is oxidized to oxygen gas, while hydrogen ions are reduced to hydrogen gas.

To master redox reactions for JEE Chemistry, students need to be able to identify the oxidation states of elements in compounds, write half-reaction equations, and balance these equations. This competence is critical for predicting the products of electrolysis and assessing how the concentrations of reactants change during the reaction.
Chemical Changes During Electrolysis
Electrolysis is a process that uses electrical energy to drive non-spontaneous chemical reactions. During electrolysis, chemical changes occur at the electrodes submerged in an electrolytic solution. At the cathode, cations, which are positively charged ions, gain electrons in a process called reduction. At the anode, anions, which are negatively charged ions, lose electrons in a process known as oxidation.

In the provided exercise, the chemical changes involve the reduction of hydrogen ions to hydrogen gas at the cathode and the oxidation of water molecules to oxygen gas and hydrogen ions at the anode. As a result, understanding the nature of these changes is crucial when analyzing the overall process and the resulting changes in concentration.

For instance, since the sulfate ions are not involved in the oxidation process taking place at the anode, the concentration of H2SO4 does not change, an important conclusion that comes directly from understanding the chemical changes during electrolysis. Students can better predict such outcomes by thoroughly understanding these principles.

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Most popular questions from this chapter

Given the listed standard electrode potentials, what is \(E^{\circ}\) for the cell : \(4 \mathrm{BiO}^{+}(a q)+3 \mathrm{~N}_{2} \mathrm{H}_{5}^{+}(a q) \longrightarrow 4 \mathrm{Bi}(s)+3 \mathrm{~N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)+7 \mathrm{H}^{+}(a q)\) \(\mathrm{N}_{2}(g)+5 \mathrm{H}^{+}(a q)+4 e^{-} \longrightarrow \mathrm{N}_{2} \mathrm{H}_{5}^{+}(a q), E^{\circ}=-0.23 \mathrm{~V}\) \(\mathrm{BiO}^{+}(a q)+2 \mathrm{H}^{+}(a q)+3 e^{-} \longrightarrow \mathrm{Bi}(s)+\mathrm{H}_{2} \mathrm{O}(l), \quad E^{\circ}=+0.32 \mathrm{~V}\) (a) \(+0.55\) (b) \(+0.34\) (c) \(+1.88\) (d) \(+0.09\)

The relation among conductance \((G)\), specific conductance \((\kappa)\) and cell constant \((l / A)\) is : (a) \(G=\kappa \frac{l}{A}\) (b) \(G=\kappa \frac{A}{l}\) (c) \(G \kappa=\frac{l}{A}\) (d) \(G=\kappa A l\) \(\therefore\)

The conductivity of a strong electrolyte : (a) Increases on dilution (b) Decreases on dilution (c) Does not change with dilution (d) Depends upon density of electrolytes

For the electrochemical cell \(\operatorname{Pt}(s) \mid \begin{aligned}&\mathrm{H}_{2}(g)\left|\mathrm{H}^{+}(1 M) \| \mathrm{Cu}^{2+}(1 M)\right| \mathrm{Cu}(s), \text { which one of the } \\\&1 \text { atm }\end{aligned}\) following statements is true ? (a) \(\mathrm{H}_{2}\) liberated at anode and \(\mathrm{Cu}\) is deposite at cathode. (b) \(\mathrm{H}_{2}\) liberated at cathode and \(\mathrm{Cu}\) is deposite at anode. (c) Oxidation occurs at cathode. (d) Reduction occurs at anode.

How many grams of \(\mathrm{Cr}\) are deposited in the electrolysis of solution of \(\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3}\) in the same time that it takes to deposit \(0.54 \mathrm{~g}\) of \(\mathrm{Ag}\) in a silver coulometer arranged in series with the \(\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3}\) cell? \(\quad\) (Atomic weight : \(\mathrm{Cr}=52.0 ; \mathrm{Ag}=108\) ) (a) \(0.0866\) (b) \(0.0288\) (c) \(0.173\) (d) \(0.220\)

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