Sketch the galvanic cells based on the following...

Sketch the galvanic cells based on the following half-reactions. Show the direction of electron flow, show the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced equation, and determine $\mathscr{E}^{\circ}$ for the galvanic cells. Assume that all concentrations are $1.0 M$ and that all partial pressures are $1.0 \mathrm{~atm}$. a. $\mathrm{Cl}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-} \quad \mathscr{E}^{\circ}=1.36 \mathrm{~V}$ $\mathrm{Br}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Br}^{-} \quad 8^{\circ}=1.09 \mathrm{~V}$ b. $\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}$ $\mathscr{b}^{\circ}=1.51 \mathrm{~V}$ $\mathrm{IO}_{4}^{-}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{IO}_{3}^{-}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E ^ { \circ }}=1.60 \mathrm{~V}$

Sketch the galvanic cells based on the following half-reactions. Show the direction of electron flow, show the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced equation, and determine $\mathscr{E}^{\circ}$ for the galvanic cells. Assume that all concentrations are $1.0 M$ and that all partial pressures are $1.0 \mathrm{~atm}$.
a. $\mathrm{Cl}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-} \quad \mathscr{E}^{\circ}=1.36 \mathrm{~V}$
$\mathrm{Br}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Br}^{-} \quad 8^{\circ}=1.09 \mathrm{~V}$
b. $\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}$
$\mathscr{b}^{\circ}=1.51 \mathrm{~V}$
$\mathrm{IO}_{4}^{-}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{IO}_{3}^{-}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E ^ { \circ }}=1.60 \mathrm{~V}$

Key Concepts

Balancing Redox Half-Reactions

Balancing redox half-reactions is essential for constructing the overall cell reaction. This involves ensuring that both mass and charge are balanced, which may require the appropriate combination and scaling of the half-reactions so that the electrons lost in oxidation equal the electrons gained in reduction.

Overall Cell Reaction and Calculation of E°

The overall cell reaction is obtained by combining the balanced oxidation and reduction half-reactions, ensuring that electrons cancel out. The standard cell potential (E°cell) is calculated by subtracting the standard reduction potential of the anode from that of the cathode. This calculation helps determine whether the cell reaction is spontaneous under standard conditions and provides a quantitative measure of the cell’s driving force.

Electron Flow in Galvanic Cells

In a galvanic cell, electrons generated by the oxidation reaction at the anode travel through an external circuit to the cathode, where reduction occurs. The direction of electron flow is from the anode (site of oxidation) to the cathode (site of reduction), and this movement of electrons is what generates electrical current in the cell.

Salt Bridge

A salt bridge is used in galvanic cells to connect the two half-cells without allowing the direct mixing of their solutions. Its primary role is to maintain electrical neutrality in the cell by allowing ions to migrate between the half-cells. Typically, anions move toward the anode to counterbalance the loss of negative charge, while cations move toward the cathode to replenish the positive charge.

Standard Reduction Potentials

Standard reduction potentials (E°) are measured under standard conditions and indicate the tendency of a chemical species to be reduced. In galvanic cells, the half-reaction with the higher reduction potential typically occurs at the cathode because it is more favorable for reduction. The difference between the cathode and anode potentials gives the overall cell potential, which quantifies the driving force behind the cell's redox reaction.

Redox Reactions

Redox reactions involve the transfer of electrons between chemical species, with one species undergoing oxidation (losing electrons) and the other undergoing reduction (gaining electrons). In the context of galvanic cells, identifying which half-reaction is oxidation and which is reduction is critical to determine the direction of electron flow, and subsequently, the designation of the anode (site of oxidation) and the cathode (site of reduction).

Galvanic Cells

A galvanic cell is an electrochemical cell that derives electrical energy from spontaneous redox reactions occurring within the cell. It consists of two half-cells, each containing an electrode and an electrolyte, where oxidation and reduction reactions occur separately. These two half-cells are connected by a conductive wire for electron flow and a salt bridge or porous barrier to maintain charge balance by allowing ion migration.

Transcript

00:01 So in this question, we're going to sketch the gavannic cells using the half reactions given.

00:10 And we want the overall balance equation, and we're going to determine the cell potential.

00:17 And all of these are in the standard states.

00:20 So for the first one, we're going to look at the reduction potentials.

00:29 The higher one is the species that is more easily reduced.

00:34 So we know that chlorine must be the one that's reduced.

00:37 And then we're going to flip the other half reaction to make it oxidation.

00:42 And then we'll flip the reduction potential.

00:47 And you see that these two reactions, we already have the same number of electrons in both.

00:57 And also the elements are already balanced for both of them.

01:00 So we're just going to add them together.

01:02 I will get this overall reaction.

01:06 And then to get the cell potential, we're going to just do ecathode minus potential of the anode.

01:16 So basically we just add these two values together.

01:20 Remember, we negated the reduction potential, and we're just going to add the negative to the higher reduction potential.

01:36 So then for the next one, we're going to compare the, two reduction potentials.

01:47 So we're going to see that the one that's reduced is going to be this ion, the i .o4 negative, and the one that is oxidized is the manganese ion.

02:06 And in our sketches, the reduction occurs at the cathode, while oxidation occurs with the anode...

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