Write balanced equations and solubility product expressions for the following compounds cubr zn c2o4 agcro4

1 answer:

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The solubility product expressions for the CuBr, ZnC₂O₄ and Ag₂CrO₄ will be represented as [Cu⁺][Br⁻], [Zn²⁺][C₂O₄²⁻] and [Ag⁺]²[CrO₄²⁻] respectively.

<h3>What is solubility product expressions?</h3>

The solubility product expression or constant is denoted by Ksp, and it is the equilibrium constant for a solid substance dissolving in an aqueous solution.

Balanced chemical reaction of CuBr is CuBr → Cu⁺ + Br⁻ and value of Ksp will be written as Ksp = [Cu⁺][Br⁻].
Balanced chemical reaction of ZnC₂O₄ is ZnC₂O₄ → Zn²⁺ + C₂O₄²⁻ and value of Ksp will be written as Ksp = [Zn²⁺][C₂O₄²⁻].
Balanced chemical reaction of Ag₂CrO₄ is AgCrO₄ → 2Ag⁺ + CrO₄²⁻ and value of Ksp will be written as Ksp = [Ag⁺]²[CrO₄²⁻].

Hence balanced equations with solubility product expressions will be shown above.

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Given the E0 values of the following two half-reactions: Zn  Zn2+ + 2e- E0 = 0.763 volt Fe  Fe2+ + 2e- E0 = 0.441 volt a) Writ

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Answer:

For a: The balanced chemical equation is written below.

For b: The corrosion of iron pipe will take place in the presence of zinc.

For c: Zinc will not protect iron pipe from corrosion.

Explanation:

For a:

The given half reaction follows:

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, zinc will undergo reduction reaction will get reduced.

Oxidation half reaction: $Zn\rightarrow Zn^{2+}+2e^-;E^o_{Zn^{2+}/Zn}=-0.763V$

Reduction half reaction: $Fe+2e^-\rightarrow Fe;E^o_{Fe^{2+}/Fe}=-0.441V$

The balanced chemical equation follows:

$Fe+Zn^{2+}\rightarrow Fe^{2+}+Zn$

For b:

For a reaction to be spontaneous (thermodynamically feasible) , the standard electrode potential must be positive.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

Calculating the $E^o_{cell}$ using above equation, we get:

$E^o_{cell}=-0.441-(-0.763)=0.322V$

As, the EMF is coming out to be positive, the reaction will be thermodynamically feasible and corrosion of iron pipe will take place in the presence of zinc.