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evablogger [386]
3 years ago
14

A buffer can be prepared by mixing two solutions. Determine if each of the following mixtures will result in a buffer solution o

r not.1) Mixing 100.0 mL of 0.1 M HF with 100.0 mL of 0.05 M mol KF [ Select ] ["Yes, it will result in a buffer solution.", "No, it will not result in a buffer solution."] 2) Mixing 100.0 mL of 0.1 M NH3 with 100.0 mL of 0.1 M NH4Br [ Select ] ["Yes, it will result in a buffer solution.", "No, it will not result in a buffer solution."] 3) Mixing 100.0 mL of 0.1 M HCN with 100.0 mL of 0.05 M KOH [ Select ] ["No, it will not result in a buffer solution.", "Yes, it will result in a buffer solution."] 4) Mixing 100.0 mL of 0.1 M HCl with 100.0 mL of 0.1 M KCl [ Select ] ["Yes, it will result in a buffer solution.", "No, it will not result in a buffer solution."] 5) Mixing 100.0 mL of 0.1 M HCN with 100.0 mL of 0.1 M KOH [ Select ] ["Yes, it will result in a buffer solution.", "No, it will not result in a buffer solution."]
Chemistry
1 answer:
GenaCL600 [577]3 years ago
6 0

Answer:

The answers are in the explanation

Explanation:

A buffer is the mixture of a weak acid with its conjugate base or vice versa. Thus:

<em>1)</em> Mixing 100.0 mL of 0.1 M HF with 100.0 mL of 0.05 M mol KF. <em>Will </em>result in a buffer because HF is a weak acid and KF is its conjugate base.

<em>2)</em> Mixing 100.0 mL of 0.1 M NH₃ with 100.0 mL of 0.1 M NH₄Br. <em>Will not </em>result in a buffer because NH₃ is a strong base.

<em>3) </em>Mixing 100.0 mL of 0.1 M HCN with 100.0 mL of 0.05 M KOH. <em>Will </em>result in a buffer because HCN is a weak acid and its reaction with KOH will produce CN⁻ that is its conjugate base.

<em>4)</em> Mixing 100.0 mL of 0.1 M HCl with 100.0 mL of 0.1 M KCl <em>Will not </em>result in a buffer because HCl is a strong acid.

<em>5)</em> Mixing 100.0 mL of 0.1 M HCN with 100.0 mL of 0.1 M KOH <em>Will not </em>result in a buffer because each HCN will react with KOH producing CN⁻, that means that you will have just CN⁻ (Conjugate base) without HCN (Weak acid).

I hope it helps!

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Answer:

The E°cell for the galvanic cell is 1.56 V.

Explanation:

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The gain of electrons is called reduction and the loss of electrons oxidation. That is to say, there is oxidation whenever an atom or group of atoms loses electrons (or increases its positive charges) and in the reduction an atom or group of atoms gains electrons, increasing its negative charges or decreasing the positive ones.

The species that supplies electrons is the reducing agent (that is, it is that species that oxidizes, yielding electrons and increasing its positive charge, or decreasing the negative one causing the reduction of the other species) and the one that gains them is the oxidizing agent ( that is, it is that species that is reduced, capturing electrons and increasing its negative charge, or decreasing its positive charge, causing oxidation of the other species).

The galvanic cell works as follows: In the anodic half-cell oxidations occur, while in the cathodic half-cell reductions occur. The anode electrode, conducts the electrons that are released in the oxidation reaction, to the metallic conductors. These electrical conductors conduct the electrons and carry them to the cathode electrode; the electrons thus enter the cathode half-cell and the reduction takes place in it.

To determine the oxidizing and reducing agent you must first know the reduction potentials. For this you consult the list of standard reduction potentials. In this list you can see that the semi-reactions that occur with their corresponding potentials are:

Ag⁺ + e⁻ ⇒ Ag E°= 0.80 V

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The species that has the greatest potential for reduction will be the species that will be reduced, that is, it will be the oxidizing agent. In this case, it will be the experience corresponding to silver (Ag). Therefore, to obtain the redox reaction, the half-reaction corresponding to zinc (Zn) must be reversed to be an oxidation, keeping its E ° value constant. Then:

Reduction: Ag⁺ + e⁻ ⇒ Ag E°= 0.80 V

Oxidation: Zn ⇒ Zn²⁺ + 2 e⁻ E° -0.76 V

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Or what is the same<em> E°cell=Ecathode - Eanode </em>because the reduction always occurs in the cathode and oxidation in the anode.

E°cell=0.80 V - (-0.76) V

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Then <u><em>the E°cell for the galvanic cell is 1.56 V.</em></u>

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