On dissout dans 250 mL d'eau 10 g de nitrate d'argent et 20 g de cyanure de potassium KCN, pu

On dissout dans 250 mL d'eau 10 g de nitrate d'argent et...

On dissout dans $250 \mathrm{~mL}$ d'eau $10 \mathrm{~g}$ de nitrate d'argent et $20 \mathrm{~g}$ de cyanure de potassium KCN, puis on utilise une partie de cette solution pour constituer, avec un électrode d'argent métallique, une demi-pile. Quelle est la f.é.m. de la pile formée par l'association de cette demi-pile et d'une électrode normale à hydrogène ? Quelle réaction s'y produit lorsqu'elle débite du courant? L'électrode d'argent est-elle son anode ou sa cathode? Rappel : les ions cyanure $\mathrm{CN}^{-}$forment avec les ions $\mathrm{Ag}^{+}$un complexe $\left[\mathrm{Ag}(\mathrm{CN})_2\right]^{-}$.

On dissout dans $250 \mathrm{~mL}$ d'eau $10 \mathrm{~g}$ de nitrate d'argent et $20 \mathrm{~g}$ de cyanure de potassium KCN, puis on utilise une partie de cette solution pour constituer, avec un électrode d'argent métallique, une demi-pile.
Quelle est la f.é.m. de la pile formée par l'association de cette demi-pile et d'une électrode normale à hydrogène ? Quelle réaction s'y produit lorsqu'elle débite du courant? L'électrode d'argent est-elle son anode ou sa cathode? Rappel : les ions cyanure $\mathrm{CN}^{-}$forment avec les ions $\mathrm{Ag}^{+}$un complexe $\left[\mathrm{Ag}(\mathrm{CN})_2\right]^{-}$.

Key Concepts

Electrochemical Cells

This concept involves the construction and operation of cells that convert chemical energy into electrical energy. In such cells, two half?cells are connected via an external circuit and a salt bridge, where one half-cell undergoes oxidation (anode) and the other undergoes reduction (cathode). Understanding the role of each electrode and the flow of electrons between them is essential in determining the overall cell electromotive force (f.e.m.) and predicting the direction of the redox reaction.

Standard Hydrogen Electrode

The standard hydrogen electrode (SHE) serves as the universal reference for measuring electrode potentials. It is assigned a potential of 0.00 V under standard conditions, and other electrode potentials are measured relative to it. In galvanic cells, pairing a half-cell with the SHE allows for the determination of the cell’s overall electromotive force.

Redox Reactions

These are chemical reactions involving the transfer of electrons between species, with one species being oxidized (losing electrons) and another reduced (gaining electrons). In the context of electrochemical cells, identifying which species is oxidized or reduced, and at which electrode each process takes place, is fundamental for determining the cell’s operation, including current production and electrode designation as an anode or cathode.

Complex Ion Formation

Complex ion formation occurs when central metal ions, such as Ag?, bind with ligands like CN? to form stable complexes (such as [Ag(CN)?]?). This process reduces the concentration of free metal ions in solution, which can significantly affect the redox equilibrium and the measured electrode potential. Recognizing the formation of complex ions is crucial in predicting the behavior of metal electrodes in solution.

Nernst Equation

The Nernst equation links the electrode potential with the concentrations (or activities) of the reactants and products. It provides a quantitative method to calculate the actual electrode potential under non?standard conditions by incorporating concentration terms. This is particularly important when complexation or dilution affects ion concentrations in the solution.

Transcript

00:01 To answer this multiple part question, we first need to understand the electrochemical cell that has been provided.

00:08 The electrochemical cell that has been provided shows a picture of the standard silver chloride reference electrode with a one molar chloride concentration.

00:25 This is serving as the cathode because the positive lead of the volt meter is attached to it.

00:34 Then the negative lead is attached to a silver -silver ion electrode, well half -cell, with a silver concentration of 1 times 10 to negative 3 that we are going to end up changing with the addition of particular reagents.

00:58 So when we solve for the initial reading of the volt -meal, based on how the leaves are attached, we need to do cathode minus anode.

01:11 Well, we know the cathode potential because that is simply the silver, silver chloride reference electrode potential given to us in problem 88 at 0 .223 volts.

01:27 To solve for the half -cell potential of the anode, we need to use the nernsd equation where we take the half cell potential under standard conditions minus 0 .0916 over n and will be one as silver goes from silver plus to silver solid.

01:49 Multiply by the log of in the half reaction we have silver ion going to silver metal.

01:57 So a product is a solid so there's no concentration for that.

02:01 We just put one.

02:02 Divided by the concentration of the silver ion, which is given to us at 1 .00 times 10 to negative 3.

02:10 So this then is the half -cell potential for the anode.

02:15 The entire cell potential will be the cathode potential, 0 .223, minus the anode potential, 0 .6225.

02:27 So we end up getting a negative .40440 volts.

02:32 So they really should have, negative voltage really doesn't make much sense.

02:36 They should have switched the leads.

02:38 But i have noticed this author does a lot of little additional confusing details.

02:45 And here's an example.

02:49 Next, it says that we add 10 milliliters of 0 .01 molar potassium chromate.

03:00 After adding the potassium chromate, you need to recognize how this will affect the silver concentration.

03:09 Well, the silver can react with the chromate according to ksp and produce the solid silver chromate.

03:22 Ksp for this process is 1 .1 times 10 to the negative 12.

03:28 So we can solve for the new silver concentration if we know the chromate concentration.

03:39 So we'll assume that if as many moles of chromate is added, as there are silver, which is the case, because we are adding 10 milliliters at 0 .01 molar, which is 1 times 10 to the negative 4 moles chromate, to a 100 milliliter solution at 1 times 10 to negative 3 molar, which is also 1 times 10 to the negative 4 moles.

04:08 Moles you can get by taking the volume in liters multiplied by the concentration...

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