A galvanic cell is based on the following half-reactions:...

A galvanic cell is based on the following half-reactions: $$\begin{aligned}\mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s) & & \mathscr{E}^{\circ}=0.80 \mathrm{~V} \\\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) & & \mathscr{6}^{\circ}=0.34 \mathrm{~V}\end{aligned}$$ In this cell, the silver compartment contains a silver electrode and excess $\mathrm{AgCl}(s)\left(K_{\mathrm{sp}}=1.6 \times 10^{-10}\right)$, and the copper compartment contains a copper electrode and $\left[\mathrm{Cu}^{2+}\right]=2.0 M$. a. Calculate the potential for this cell at $25^{\circ} \mathrm{C}$. b. Assuming $1.0 \mathrm{~L}$ of $2.0 \mathrm{M} \mathrm{Cu}^{2+}$ in the copper compartment, calculate the moles of $\mathrm{NH}_{3}$ that would have to be added to give a cell potential of $0.52 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$ (assume no volume change on addition of $\mathrm{NH}_{3}$ ). $$\mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons{\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q)} \quad K=1.0 \times 10^{13}$$

A galvanic cell is based on the following half-reactions:
$$\begin{aligned}\mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s) & & \mathscr{E}^{\circ}=0.80 \mathrm{~V} \\\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) & & \mathscr{6}^{\circ}=0.34 \mathrm{~V}\end{aligned}$$
In this cell, the silver compartment contains a silver electrode and excess $\mathrm{AgCl}(s)\left(K_{\mathrm{sp}}=1.6 \times 10^{-10}\right)$, and the copper compartment contains a copper electrode and $\left[\mathrm{Cu}^{2+}\right]=2.0 M$.
a. Calculate the potential for this cell at $25^{\circ} \mathrm{C}$.
b. Assuming $1.0 \mathrm{~L}$ of $2.0 \mathrm{M} \mathrm{Cu}^{2+}$ in the copper compartment, calculate the moles of $\mathrm{NH}_{3}$ that would have to be added to give a cell potential of $0.52 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$ (assume no volume change on addition of $\mathrm{NH}_{3}$ ).
$$\mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons{\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q)} \quad K=1.0 \times 10^{13}$$

Key Concepts

Complex Ion Formation Equilibrium

Complex ion formation equilibrium involves the reversible binding of metal ions with ligands, forming complex ions. The associated equilibrium constant can be very large, indicating that most metal ions form complexes under the given conditions. This phenomenon is crucial for controlling the effective concentration of free metal ions in solution, thereby influencing various equilibria and impacting cell potential calculations.

Galvanic Cell

A galvanic cell is an electrochemical cell that converts chemical energy into electrical energy through spontaneous redox reactions. It consists of two distinct half-cells where oxidation and reduction occur, connected by an external circuit and a salt bridge, allowing electrons to flow from the anode to the cathode.

Standard Electrode Potentials

Standard electrode potentials quantify the inherent tendency of a chemical species to be reduced under standard conditions. These values, expressed in volts, allow for the prediction of the direction of electron flow in redox reactions and enable the calculation of overall cell potential when combining half-reactions.

Nernst Equation

The Nernst equation is a fundamental tool in electrochemistry that relates the electrode potential to the concentrations or activities of the reactants and products involved in a redox reaction. This equation is critical for determining cell potentials under non-standard conditions and for understanding how changes in concentration affect the driving force of the reaction.

Solubility Product (Ksp)

The solubility product, Ksp, is the equilibrium constant for the dissolution of a sparingly soluble salt. It represents the maximum product of the ion concentrations at equilibrium and plays an essential role in predicting whether precipitation occurs, which directly affects the concentration of ions in solution and impacts electrochemical potentials.

Transcript

00:01 We are asked to consider a cell based on these half reactions.

00:04 So the silver compartment, we're told our silver compartment contains a silver electrode and excess silver chloride, which has a ksp of 1 .6 times 10 to the minus 10th.

00:39 And it's got the copper compartment.

00:41 It contains a copper electrode and a copper 2 ion concentration of 2 .0 molar.

01:05 And we're asked to a, calculate the cell potential.

01:12 Okay, so we're going to use the following equation.

01:23 Equation this will equal this plus or minus excuse me minus 0 .0591 over n times the log of q.

01:46 Okay first let's do e for our cell.

01:52 That will equal we've got our silver which we're going to multiply by two and my e will equal 0 .80 volts and my copper will be here this one will equal negative 0 .34.

02:31 So my e for my cell will equal 0 .46 volts.

02:41 E for my cell equals 0 .46 volts.

02:48 There's one part of our answer.

02:53 Okay, my q will be equal to my copper concentration over my silver concentration, which we will have to square.

03:09 So first let's figure out our silver.

03:20 And i know that my silver chloride will dissociate into silver and chloride.

03:36 That my initial will be zero and zero.

03:40 This will increase by x and x and it will have x and x for my equilibrium concentrations.

03:53 And my ksp will be equal to silver times chloride, whoops, which will equal x squared.

04:07 So 1 .6 times 10 to the minus 10 equals x squared.

04:13 X will equal something times 10 to the fifth.

04:24 1 .265 times 10 to the minus fifth molar.

04:32 That'll be my silver concentration.

04:40 So now let's put all of our values into our problem here.

04:45 So e for my cell will equal 0 .46 minus 0 .0591 divided by 2 because n equals 2 from our balanced chemical equation times the log of of 0, that was 2 .0, 2, not .2, 2 .0 divided by 1 .265 times 10 to the minus 5th squared.

05:24 Good heavens.

05:27 So 0 .46 minus parentheses 0 .0591 divided by 2.

05:39 Over here i'm going to add one more parentheses, okay, times and then i'm going to do the log of 2 divided by parentheses 1 .265 times 10 to the minus 5th close x squared close close.

06:05 I think i got my parentheses right.

06:07 Right.

06:07 So i'm getting my e for my cell is equal to 0 .162 volts.

06:20 I can only go to two sig figs here.

06:22 So e for my cell will be equal to 0 .1 to 6 volts...

Please give Ace some feedback