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Question: (١٢) من جهود الأقطاب الذي: $X^{2+} + 2e^{-} \longrightarrow X \quad , E^{\circ} = -0.41 V$ $Y \lon...

(١٢) من جهود الأقطاب الذي:

X2++2e−⟶X,E∘=−0.41VX^{2+} + 2e^{-} \longrightarrow X \quad , E^{\circ} = -0.41 V

Y⟶Y2++2e−,E∘=−0.23VY \longrightarrow Y^{2+} + 2e^{-} \quad , E^{\circ} = -0.23 V

أي الاختيارات التالية يمثل التفاعل التالي؟

X+Y2+⟶X2++YX + Y^{2+} \longrightarrow X^{2+} + Y

A

غير تلقائي ، وكتلة القطب (X) تقل.

B

غير تلقائي ، وكتلة القطب (Y) تقل.

C

تلقائي ، وكتلة القطب (X) تقل.

D

تلقائي ، وكتلة القطب (Y) تقل.

Answer

تلقائي ، وكتلة القطب (X) تقل.

Explanation

Solution

The problem asks us to determine the spontaneity of a given redox reaction and the change in mass of the electrodes, based on the provided standard electrode potentials.

1. Identify the half-reactions and their standard potentials: We are given:

  • Reduction of X: X2++2e−⟶X,EX2+/X∘=−0.41VX^{2+} + 2e^{-} \longrightarrow X \quad , E^{\circ}_{X^{2+}/X} = -0.41 V (This is the standard reduction potential for X).
  • Oxidation of Y: Y⟶Y2++2e−,EY/Y2+∘=−0.23VY \longrightarrow Y^{2+} + 2e^{-} \quad , E^{\circ}_{Y/Y^{2+}} = -0.23 V (This is the standard oxidation potential for Y).

2. Analyze the target reaction: The reaction to be analyzed is: X+Y2+⟶X2++YX + Y^{2+} \longrightarrow X^{2+} + Y

Let's break this down into oxidation and reduction half-reactions:

  • Oxidation half-reaction: X⟶X2++2e−X \longrightarrow X^{2+} + 2e^{-}
  • Reduction half-reaction: Y2++2e−⟶YY^{2+} + 2e^{-} \longrightarrow Y

3. Determine the standard potentials for the half-reactions in the target reaction:

  • For the oxidation of X (X⟶X2++2e−X \longrightarrow X^{2+} + 2e^{-}): The standard oxidation potential is the negative of the standard reduction potential. Eox∘(X/X2+)=−Ered∘(X2+/X)=−(−0.41V)=+0.41VE^{\circ}_{ox}(X/X^{2+}) = -E^{\circ}_{red}(X^{2+}/X) = -(-0.41 V) = +0.41 V.
  • For the reduction of Y (Y2++2e−⟶YY^{2+} + 2e^{-} \longrightarrow Y): The standard reduction potential is the negative of the standard oxidation potential given for Y. Ered∘(Y2+/Y)=−Eox∘(Y/Y2+)=−(−0.23V)=+0.23VE^{\circ}_{red}(Y^{2+}/Y) = -E^{\circ}_{ox}(Y/Y^{2+}) = -(-0.23 V) = +0.23 V.

4. Calculate the standard cell potential (Ecell∘E^{\circ}_{cell}): The standard cell potential is the sum of the standard oxidation potential and the standard reduction potential for the reaction. Ecell∘=Eoxidation∘+Ereduction∘E^{\circ}_{cell} = E^{\circ}_{oxidation} + E^{\circ}_{reduction} Ecell∘=Eox∘(X/X2+)+Ered∘(Y2+/Y)E^{\circ}_{cell} = E^{\circ}_{ox}(X/X^{2+}) + E^{\circ}_{red}(Y^{2+}/Y) Ecell∘=(+0.41V)+(+0.23V)E^{\circ}_{cell} = (+0.41 V) + (+0.23 V) Ecell∘=+0.64VE^{\circ}_{cell} = +0.64 V

5. Determine the spontaneity of the reaction: Since Ecell∘E^{\circ}_{cell} is positive (+0.64V>0+0.64 V > 0), the reaction is spontaneous.

6. Analyze the mass change of the electrodes:

  • Oxidation of X: X⟶X2++2e−X \longrightarrow X^{2+} + 2e^{-} In this process, solid X is converted into X2+X^{2+} ions in the solution. This means the mass of the solid X electrode will decrease.
  • Reduction of Y: Y2++2e−⟶YY^{2+} + 2e^{-} \longrightarrow Y In this process, Y2+Y^{2+} ions from the solution are deposited as solid Y. This means the mass of the solid Y electrode will increase.

Therefore, the correct statement is that the reaction is spontaneous, and the mass of electrode X decreases.