Consider the galvanic cell based on the following...

Consider the galvanic cell based on the following halfreactions: $$\begin{array}{ll}{\mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}} & {\mathscr{E}^{\circ}=-0.76 \mathrm{V}} \\ {\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}} & {\mathscr{E}^{\circ}=-0.44 \mathrm{V}}\end{array}$$ a. Determine the overall cell reaction and calculate $\mathscr{E}_{\text { cell }}$ b. Calculate $\Delta G^{\circ}$ and $K$ for the cell reaction at $25^{\circ} \mathrm{C}$ c. Calculate $\mathscr{E}_{\text { cell }}$ at $25^{\circ} \mathrm{C}$ when $\left[\mathrm{Zn}^{2+}\right]=0.10 M$ and $\left[\mathrm{Fe}^{2+}\right]=1.0 \times 10^{-5} \mathrm{M} .$

Consider the galvanic cell based on the following halfreactions:
$$\begin{array}{ll}{\mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}} & {\mathscr{E}^{\circ}=-0.76 \mathrm{V}} \\ {\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}} & {\mathscr{E}^{\circ}=-0.44 \mathrm{V}}\end{array}$$
a. Determine the overall cell reaction and calculate $\mathscr{E}_{\text { cell }}$
b. Calculate $\Delta G^{\circ}$ and $K$ for the cell reaction at $25^{\circ} \mathrm{C}$
c. Calculate $\mathscr{E}_{\text { cell }}$ at $25^{\circ} \mathrm{C}$ when $\left[\mathrm{Zn}^{2+}\right]=0.10 M$ and $\left[\mathrm{Fe}^{2+}\right]=1.0 \times 10^{-5} \mathrm{M} .$

Key Concepts

Nernst Equation

The Nernst equation is used to determine the cell potential under non-standard conditions by incorporating the effect of ion concentrations or partial pressures. It adjusts the standard cell potential based on the reaction quotient, providing a link between the measured cell potential and the concentrations of the reacting species, which is critical for real-world applications where conditions deviate from the standard state.

Equilibrium Constant in Electrochemical Reactions

The equilibrium constant (K) of an electrochemical reaction quantifies the ratio of product activities to reactant activities at equilibrium. It is related to the standard cell potential and Gibbs free energy by the expression ?G° = -RT ln K, connecting the thermodynamic favorability of the reaction to its equilibrium distribution of species.

Gibbs Free Energy and Its Relation to Cell Potential

Gibbs free energy change (?G°) is a thermodynamic quantity that determines the spontaneity of a reaction. In the context of electrochemistry, it is directly related to the cell potential through the equation ?G° = -nFE°cell, where n is the number of electrons transferred and F is Faraday’s constant. A negative value of ?G° indicates that the reaction is spontaneous under standard conditions.

Standard Reduction Potentials

Standard reduction potentials are measurements of the tendency of a chemical species to gain electrons and be reduced, measured under standard conditions (1 M concentration, 1 atm pressure, and 25°C). These potentials allow for the prediction of reaction spontaneity; comparing the potentials of different half-reactions indicates which species acts as the oxidizing agent and which as the reducing agent.

Galvanic Cell

A galvanic cell is an electrochemical cell that converts chemical energy into electrical energy through spontaneous redox reactions. It is composed of two half-cells, each containing an electrode in contact with an electrolyte, where oxidation occurs at the anode and reduction at the cathode. The cell’s overall performance is characterized by its cell potential, which is the voltage difference between the two electrodes.

Overall Cell Reaction

The overall cell reaction in a galvanic cell is obtained by combining the two half-reactions, ensuring that the electrons lost in oxidation equal the electrons gained in reduction. This reaction represents the net chemical change occurring in the cell and forms the basis for calculating cell potential and other thermodynamic quantities.

Transcript

00:01 In this problem, we are told that these two half reactions occur within a galvanic cell, and in part a, we want to determine what the overall cell reaction is, as well as the value for the overall cell potential.

00:13 We know that for a galvanic cell, the cell potential has to be positive.

00:18 Therefore, we have to change one of the half reactions so that the standard reduction potential that is larger between the two does not change.

00:30 And so we have to flip the one that is the least positive or the most negative.

00:36 And in this case, the most negative standard reduction potential between these two is one from the first half reaction.

00:43 So we have to flip that one.

00:45 So in part a, we are flipping that first reaction so that oxidation occurs.

00:51 We have zn going to zn 2 plus plus two electrons.

01:02 And again, for the standard reduction potentials, we now reverse that sign, so it is positive 0 .76.

01:12 And then the other reaction, since it is larger for its standard reduction potential, it will occur in the cathode, where reduction will proceed.

01:25 So we have fe2 plus plus two electrons going to fe solid.

01:34 And we keep that standard reduction potential of negative 0 .44.

01:41 So now we see that because we algebraically manipulated the two half reactions in this way, that the electrons can cancel, and we are left with a positive value for the cell potential, which is indicative of a galvanic cell.

01:55 So when we combine these two equations, we get the overall cell reaction, which is fe2 plus aqueous plus zn solid, goes to zn2 plus aqueous plus fe solid.

02:22 And now for the cell potential at standard conditions, we add these two reduction potentials together.

02:32 So that's 0 .76 minus 0 .44.

02:35 That comes out to 0 .32 volts.

02:41 So that is the answer for part a.

02:44 And now in part b, we want to determine first what the value is for the change in gives free energy, at standard conditions, delta g at standard conditions.

02:54 We know that that is equal to negative n, f, e...

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