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Assertion (A): The electrolysis of NaCl ...

Assertion `(A):` The electrolysis of `NaCl` solution gives `H_(2)(g)` at cathode and `Cl_(2)(g)` at anode.
Reason `(R) : Cl_(2)` has higher oxidation potential than `H_(2)O`

A

If both `(A)` and `(R)` are correct, and `(R)` is the correct explanation of `(A)`.

B

If both `(A)` and `(R)` are correct, but `(R)` is not the correct explanation of `(A)`.

C

If `(A)` is correct, but `(R)` is incorrect.

D

If `(A)` is incorrect, `(R)` is correct.

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The correct Answer is:
To solve the question regarding the electrolysis of NaCl solution and the assertion and reason provided, we can break it down into a step-by-step analysis. ### Step 1: Understanding the Electrolysis Process Electrolysis involves the breakdown of a compound in solution using an electric current. In the case of NaCl (sodium chloride) dissolved in water, the solution contains the following ions: - Sodium ions (Na⁺) - Chloride ions (Cl⁻) - Hydrogen ions (H⁺) - Hydroxide ions (OH⁻) ### Step 2: Identifying the Cathode Reaction At the cathode, reduction occurs. The species that can be reduced are H⁺ and Na⁺. The reduction potential for H⁺ to H₂ is higher than that for Na⁺ to Na. Therefore, H⁺ will be reduced to form hydrogen gas (H₂): \[ 2H^+ + 2e^- \rightarrow H_2(g) \] Thus, hydrogen gas is produced at the cathode. ### Step 3: Identifying the Anode Reaction At the anode, oxidation occurs. The species that can be oxidized are Cl⁻ and OH⁻. Chloride ions (Cl⁻) can be oxidized to form chlorine gas (Cl₂), while hydroxide ions (OH⁻) can lead to the formation of oxygen gas (O₂). However, Cl⁻ has a higher tendency to be oxidized compared to OH⁻ under standard conditions, leading to the formation of chlorine gas: \[ 2Cl^- \rightarrow Cl_2(g) + 2e^- \] Thus, chlorine gas is produced at the anode. ### Step 4: Evaluating the Assertion and Reason - **Assertion (A)**: The electrolysis of NaCl solution gives H₂(g) at the cathode and Cl₂(g) at the anode. This statement is **correct** based on the reactions identified. - **Reason (R)**: Cl₂ has a higher oxidation potential than H₂O. This statement is **incorrect**. In fact, Cl₂ has a lower oxidation potential compared to the oxidation of water. ### Conclusion The assertion is correct, while the reason is incorrect. Therefore, the correct answer is: **Assertion is correct, reason is incorrect.** ### Final Answer **Option C: Assertion is correct, Reason is incorrect.** ---

To solve the question regarding the electrolysis of NaCl solution and the assertion and reason provided, we can break it down into a step-by-step analysis. ### Step 1: Understanding the Electrolysis Process Electrolysis involves the breakdown of a compound in solution using an electric current. In the case of NaCl (sodium chloride) dissolved in water, the solution contains the following ions: - Sodium ions (Na⁺) - Chloride ions (Cl⁻) - Hydrogen ions (H⁺) - Hydroxide ions (OH⁻) ...
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The electrolysis of a solution resulted in the formation of H_2 (g) at the cathode and Cl_(2) (g) at anode. The solution is :

Assertion(A) Electrolysis of NaCl solution gives chlorine at anode instead of O_(2) . Reason(R) Formation of oxygen at anode requires over voltage.

Assertion (A): The enthalpy of formation of H_(2)O(l) is greater than that of H_(2)O(g) . Reason (R ) : Enthalpy change is negative for the condensation reaction H_(2)O(g) rarr H_(2)O(l) .

The passage of current through a solution of certain electrolyte results in the evolution of H_2 (g) at cathode and Cl_2 (g) at anode. The electrolytic solution is :

Assertion (A): Magnesium can be obtained by the electronlysis of aqueous solution of MgCl_(2) . Reason (R ): The electrode potential of Mg^(2+) is much higher than H^(o+) .

(a). Explain why electrolysis of an aqueous solution of NaCl gives H_(2) at cathode and Cl_(2) at anode. Given E_(Na^(+)//Na)^(@)=-2.71V,E_(H_(2)O//H_(2)^(@)=-0.83V E_(Cl_(2)//2Cl^(-))^(@)=+1.36V,E_(2H^(+)//(1)/(2)O_(2)//H_(2)O)^(@)=+1.23V (b). The resistance of a conductivity cell when filled with 0.05 M solution of an electrolyte X is 100Omega at 40^(@)C . the same conductivity cell filled with 0.01 M solution of electrolyte Y has a resistance of 50Omega . The conductivity of 0.05M solution of electrolyte X is 1.0xx10^(-4)scm^(-1) calculate (i). Cell constant (ii). conductivity of 0.01 M Y solution (iii). Molar conductivity of 0.01 M Y solution.

Assertion:In electrolysis of aqueous NaCl the product obtained is H_2 gas. Reason:Gases are liberted faster than the metals.

A : 2 F^(-) + Cl_(2) to 2 Cl^(-) + F_(2) is a reaction having Delta G^0 = -ve . R : Cl_(2) is better oxidizing agent than F_(2) .

Assertion (A) Decomposition of H_(2)O_(2) is a disproportionation reaction. Reason (R ) H_(2)O_(2) molecule simultaneously undergoes oxidation and reduction.

Assertion (A) Chlorine reacts more rapidly with H_(2) in comparison to D_(2) . Reason (R ) D -CL bond is stronger in comparison to H - Cl bond .

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