Use tabulated electrode potentials to calculate ΔGKn^∘ for each reaction at 25^∘ C . a. 2

step 1 Okay, so these questions, we need to use this equation. So the delta g is equal to the delta g s

Use tabulated electrode potentials to calculate ΔGKn^∘ for...

Use tabulated electrode potentials to calculate $\Delta G_{\mathrm{Kn}}^{\circ}$ for each reaction at $25^{\circ} \mathrm{C} .$ $$ \begin{array}{l}{\text { a. } 2 \mathrm{Fe}^{3+}(a q)+3 \operatorname{Sn}(s) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{Sn}^{2+}(a q)} \\ {\text { b. } \mathrm{O}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{Cu}(s) \longrightarrow 4 \mathrm{OH}^{-}(a q)+2 \mathrm{Cu}^{2+}(a q)} \\ {\text { c. } \mathrm{Br}_{2}(l)+2 \mathrm{I}^{-(a q)} \longrightarrow 2 \mathrm{Br}^{-}(a q)+\mathrm{I}_{2}(s)}\end{array} $$

Use tabulated electrode potentials to calculate $\Delta G_{\mathrm{Kn}}^{\circ}$ for each reaction at $25^{\circ} \mathrm{C} .$
$$
\begin{array}{l}{\text { a. } 2 \mathrm{Fe}^{3+}(a q)+3 \operatorname{Sn}(s) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{Sn}^{2+}(a q)} \\ {\text { b. } \mathrm{O}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{Cu}(s) \longrightarrow 4 \mathrm{OH}^{-}(a q)+2 \mathrm{Cu}^{2+}(a q)} \\ {\text { c. } \mathrm{Br}_{2}(l)+2 \mathrm{I}^{-(a q)} \longrightarrow 2 \mathrm{Br}^{-}(a q)+\mathrm{I}_{2}(s)}\end{array}
$$

Key Concepts

Stoichiometry of Electron Transfer

The overall redox reaction must be balanced for both mass and charge, especially concerning the number of electrons transferred between oxidation and reduction half-reactions. The value of n in the ?G° equation (which is derived from the stoichiometric coefficients in the balanced overall reaction) is crucial, as it scales the impact of the cell potential on the free energy change.

Standard Electrode Potentials

Standard electrode potentials are tabulated values that indicate the intrinsic tendency of a species to be reduced, measured under standard conditions (typically 25°C, 1 M concentration, 1 atm pressure). These values are essential in predicting the direction of electron flow when two half-cells are connected in an electrochemical cell.

Half-Reactions and Their Reversals

Electrochemical reactions are broken down into half-reactions—one for reduction and one for oxidation. The tabulated standard reduction potentials pertain to the reduction direction, so if a reaction is written in the opposite (oxidation) direction, its electrode potential must be reversed in sign. This process is fundamental when combining half-reactions to calculate overall cell potential.

Calculation of Cell Potential (E°cell)

The overall standard cell potential is determined by taking the difference between the cathode and anode electrode potentials (E°cell = E°cathode ? E°anode). This value signifies the driving force for the redox reaction under standard conditions and is calculated after ensuring that each half-reaction is expressed in a consistent manner.

Relationship Between ?G° and Cell Potential

The change in standard Gibbs free energy (?G°) for an electrochemical reaction is linked to the standard cell potential by the equation ?G° = ?nFE°cell, where n represents the number of electrons transferred and F is Faraday's constant. This equation bridges the gap between thermodynamics and electrochemistry, indicating how the electrical potential drives the spontaneity of the reaction.

Transcript

00:04 Okay, so these questions, we need to use this equation.

00:11 So the delta g is equal to the delta g standard equals to minus n e standard times f.

00:21 F is the friday constant and e is the standard cell potential.

00:26 Okay, so the question now is transferred to how to solve, how to get the standard cell potential of each reactions.

00:39 First, let's look at the first reaction.

00:43 The first reaction is the iron three oxidized the team.

01:03 As we know, there are totally six electrons transferred.

01:10 So n here equals to six.

01:14 And e is standard now.

01:17 We should calculate it from the electrode potential table and we should calculate these two pairs.

01:34 Okay? right.

01:39 So we need to find it on the table.

01:42 So the fe 0 .036, which is minus 0 .36 minus the tins, minus 0 .14, minus 0 .36, minus 0 .14.

02:06 Okay, so it will be, the result will be, let me use my calculator, okay, it should be 0 .104 voltage.

02:33 And then we plug in this, this, into this equation.

02:43 And f, we use this value, okay? now we can get the result, which is the delta g will be minus 60 .21 k k .com or more.

03:18 Okay? now let's look at the second reaction...

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