Consider the standard galvanic cell based on the following...

Consider the standard galvanic cell based on the following halfreactions: $$\begin{array}{r}\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu} \\ \mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}\end{array}$$ The electrodes in this cell are $\mathrm{Ag}(s)$ and $\mathrm{Cu}(s)$. Does the cell potential increase, decrease, or remain the same when the following changes occur to the standard cell? a. $\operatorname{CuSO}_{4}(s)$ is added to the copper half-cell compartment (assume no volume change). b. $\mathrm{NH}_{3}(a q)$ is added to the copper half-cell compartment. [Hint: $\mathrm{Cu}^{2+}$ reacts with $\mathrm{NH}_{3}$ to form $\left.\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) .\right]$ c. $\mathrm{NaCl}(s)$ is added to the silver half-cell compartment. [Hint: $\mathrm{Ag}^{+}$ reacts with $\mathrm{Cl}^{-}$ to form $\left.\mathrm{AgCl}(s) .\right]$ d. Water is added to both half-cell compartments until the volume of solution is doubled. e. The silver electrode is replaced with a platinum electrode. $$\mathrm{Pt}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt} \quad \mathscr{E}^{\circ}=1.19 \mathrm{~V}$$

Consider the standard galvanic cell based on the following halfreactions:
$$\begin{array}{r}\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu} \\
\mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}\end{array}$$
The electrodes in this cell are $\mathrm{Ag}(s)$ and $\mathrm{Cu}(s)$. Does the cell potential increase, decrease, or remain the same when the following changes occur to the standard cell?
a. $\operatorname{CuSO}_{4}(s)$ is added to the copper half-cell compartment (assume no volume change).
b. $\mathrm{NH}_{3}(a q)$ is added to the copper half-cell compartment. [Hint:
$\mathrm{Cu}^{2+}$ reacts with $\mathrm{NH}_{3}$ to form $\left.\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) .\right]$
c. $\mathrm{NaCl}(s)$ is added to the silver half-cell compartment. [Hint:
$\mathrm{Ag}^{+}$ reacts with $\mathrm{Cl}^{-}$ to form $\left.\mathrm{AgCl}(s) .\right]$
d. Water is added to both half-cell compartments until the volume of solution is doubled.
e. The silver electrode is replaced with a platinum electrode.
$$\mathrm{Pt}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt} \quad \mathscr{E}^{\circ}=1.19 \mathrm{~V}$$

Key Concepts

Dilution Effects on Cell Potential

Diluting the solution changes the concentration of the ionic species involved in the redox reactions. According to the Nernst Equation, these changes can adjust the electrode potentials, even if the overall reaction remains the same, highlighting the impact of solution concentration on cell voltage.

Role of Inert Electrodes

Inert electrodes, such as platinum, serve as conductive surfaces without chemically participating in the redox reaction. Their use helps in accurately measuring the potential of a redox couple in solution since the electrode itself does not alter the solution's chemistry.

Complex Ion Formation

When a ligand binds to a metal ion to form a complex, it reduces the free ion concentration in solution. This change influences the equilibrium of the redox reaction according to the Nernst equation, potentially altering the cell potential.

Precipitation Reactions

The formation of an insoluble compound from its ions in solution decreases the concentration of the participating ions. This reduction can shift the reaction equilibrium in the respective half-cell and, as a result, modify the measured electrode potential.

Nernst Equation

The Nernst Equation quantitatively relates the electrode potential to the concentrations (activities) of the redox species. It allows us to calculate how deviations from standard conditions, such as changes from added reagents or dilution, will modify the cell potential.

Standard Electrode Potential

This is the voltage associated with a half-cell under standard conditions (usually 1 M concentration, 1 atm pressure, and 25°C). It represents the inherent tendency of a chemical species to be reduced and forms the baseline for comparing redox reactions in electrochemical cells.

Le Chatelier's Principle

This principle states that a system at equilibrium will adjust to counteract any imposed change. In electrochemical cells, perturbations like adding a reagent that reacts with one of the ions will disturb the equilibrium and shift the concentrations of redox-active species, thereby affecting the electrode potential.

Transcript

00:01 We are asked to consider these two equations, and we've got several things that happens.

00:05 And we're going to, i'm going to go ahead, and before we get started on this, do my reaction.

00:11 This is a galvanic cell, so it will be spontaneous.

00:20 So let's figure out which one we need where.

00:25 I'm going to go look at my cell potentials here.

00:31 So this one is 0 .34.

00:35 0 .34 this is 0 .80 so i need this to be positive so i need to reverse this one so it'll be cu and cu 2 plus plus 2 e minus and negative 0 .34 and here i will have oops so my for my cell this will will be 0 .46 volts and i'll have cu plus, there we go.

01:59 So now for a, we are told that we add copper sulfate.

02:12 And what's my question? will cell potential increase, decrease, or stay the same? and this is the question we're going to answer for each one of these, i think five that we have to do so let's take a look at here my e for my cell will equal this number minus 0 .0592 over n times the log of q and q will be equal to my cu2 plus over my ag2 plus excuse me ag plus squared so if if cu2 plus increases, then q will increase and the e cell will decrease.

03:54 So my e for my cell will decrease.

04:00 Okay, let's look at our next one.

04:09 What is my next scenario? e and h3 is added to the cu half cell.

04:33 And we're reminded that cu2 plus reacts with nh3 to form the complex, which will have sort of the opposite factor here.

05:03 So my q, which is equal to my copper 2 plus again over my ag, my cu2 plus will decrease.

05:24 Q will decrease.

05:33 Thus, my e for my cell will increase for b.

05:46 Now let's do c...

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