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Given the following reduction half-reactions: $$ \mathrm{Fe}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+}(a q) $$ \(E_{\mathrm{red}}^{\circ}=+0.77 \mathrm{~V}\) $$ \mathrm{S}_{2} \mathrm{O}_{6}{ }^{2-}(a q)+4 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{SO}_{3}(a q) $$ \(E_{\mathrm{red}}^{\circ}=+0.60 \mathrm{~V}\) $$ \begin{array}{r} \mathrm{N}_{2} \mathrm{O}(g)+2 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \\ E_{\mathrm{red}}^{\circ}=-1.77 \mathrm{~V} \\ \mathrm{VO}_{2}^{+}(a q)+2 \mathrm{H}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{VO}^{2+}+\mathrm{H}_{2} \mathrm{O}(l) \\ E_{\mathrm{red}}^{\circ}=+1.00 \mathrm{~V} \end{array} $$ (a) Write balanced chemical equations for the oxidation of \(\mathrm{Fe}^{2+}(a q)\) by \(\mathrm{S}_{2} \mathrm{O}_{6}{ }^{2-}(a q),\) by \(\mathrm{N}_{2} \mathrm{O}(a q),\) and by \(\mathrm{VO}_{2}{ }^{+}(a q) .(\mathbf{b})\) Calculate \(\Delta G^{\circ}\) for each reaction at \(298 \mathrm{~K}\). (c) Calculate the equilibrium constant \(K\) for each reaction at \(298 \mathrm{~K}\).

Step by step solution

01

Oxidation half-reaction of Fe虏鈦�

We have the reduction half-reaction of Fe鲁鈦�: \[ \mathrm{Fe}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+}(a q) \] The oxidation half-reaction will be the reverse of this: \[ \mathrm{Fe}^{2+}(a q) \longrightarrow \mathrm{Fe}^{3+}(a q)+\mathrm{e}^{-} \] Step 2: Balance the redox reactions

02

Balanced redox reactions

The balanced redox reactions are: (i) With S鈧侽鈧喡测伝: \[ 2 \mathrm{Fe}^{2+}(aq) \longrightarrow 2 \mathrm{Fe}^{3+}(aq) + 2\mathrm{e}^- \] \[ \mathrm{S}_{2} \mathrm{O}_{6}^{2-}(a q)+4 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{SO}_{3}(a q) \] (ii) With N鈧侽: \[ 2 \mathrm{Fe}^{2+}(aq) \longrightarrow 2 \mathrm{Fe}^{3+}(aq) + 2\mathrm{e}^- \] \[ \mathrm{N}_{2} \mathrm{O}(g)+2 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \] (iii) With VO鈧傗伜: \[ \mathrm{Fe}^{2+}(aq) \longrightarrow \mathrm{Fe}^{3+}(aq) + \mathrm{e}^- \] \[ \mathrm{VO}_{2}^{+}(a q)+2 \mathrm{H}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{VO}^{2+}+\mathrm{H}_{2} \mathrm{O}(l) \] Step 3: Calculate the potentials for each redox reaction

03

Potentials for redox reactions

We need to subtract the reduction potentials: (i) E掳(Fe虏鈦�/Fe鲁鈦�) - E掳(S鈧侽鈧喡测伝/H鈧係O鈧�) = 0.77 - 0.60 = +0.17 V (ii) E掳(Fe虏鈦�/Fe鲁鈦�) - E掳(N鈧侽/N鈧�) = 0.77 - (-1.77) = +2.54 V (iii) E掳(Fe虏鈦�/Fe鲁鈦�) - E掳(VO鈧傗伜/VO虏鈦�) = 0.77 - 1.00 = -0.23 V Step 4: Calculate 螖G掳 for each reaction at 298 K

04

Gibbs free energy change

The relation between E掳 and 螖G掳 is: 螖G掳 = -nF 脳 E掳 (where F is Faraday's constant, about 96485 C/mol) (i) 螖G掳 = -2 脳 96485 脳 0.17 = -32784 J/mol (ii) 螖G掳 = -2 脳 96485 脳 2.54 = -490307 J/mol (iii) 螖G掳 = -1 脳 96485 脳 (-0.23) = +22192 J/mol Step 5: Calculate the equilibrium constant K for each reaction at 298 K

05

Equilibrium constant K

The relation between K and 螖G掳 is: 螖G掳 = -RT 脳 ln(K) (where R is the gas constant, about 8.314 J/(mol K)) (i) ln(K) = -(-32784)/(8.314 脳 298) = 1.315, K = e鹿路鲁鹿鈦� = 3.72 (ii) ln(K) = -(-490307)/(8.314 脳 298) = 19.6, K = e鹿鈦孤封伓 = 3.27 脳 10鈦� (iii) ln(K) = -22192/(8.314 脳 298) = -0.892, K = e^鈦烩伆路鈦糕伖虏 = 0.41

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Equations

Understanding how to balance chemical equations is fundamental to studying redox reactions in chemistry. A chemical equation represents a reaction where the reactant molecules are transformed into product molecules. In a redox reaction, one substance loses electrons (oxidation) while another gains electrons (reduction). In the exercise provided, the oxidation of \( \mathrm{Fe}^{2+}(aq) \) by various agents leads to different balanced chemical equations. For instance, the oxidation by \( \mathrm{S}_{2} \mathrm{O}_{6}^{2-}(aq) \) results in \( \mathrm{Fe}^{3+}(aq) \) and \( \mathrm{H}_{2} \mathrm{SO}_{3}(aq) \) as products.

The process of balancing these equations involves ensuring that the number of atoms for each element and the total charge is the same on both sides of the equation. This is important because it reflects the conservation of mass and charge, pivotal principles in chemistry.

Gibbs Free Energy

The Gibbs free energy (\( \Delta G^\circ \)), is an indicator of the spontaneity of a reaction at constant pressure and temperature. When \( \Delta G^\circ \) is negative, the reaction tends to occur without the addition of external energy, meaning it is spontaneous. Conversely, a positive value indicates that the reaction is non-spontaneous.

In the context of redox reactions, the Gibbs free energy can be calculated from the cell potentials using the formula \( \Delta G^\circ = -nFE^\circ \), where \( n \) is the number of moles of electrons exchanged, \( F \) is Faraday's constant, and \( E^\circ \) is the standard cell potential. The calculations in the solution result in negative values for reactions with \( \mathrm{S}_{2} \mathrm{O}_{6}^{2-}(aq) \) and \( \mathrm{N}_{2} \mathrm{O}(aq) \), indicating they are spontaneous, while the reaction with \( \mathrm{VO}_{2}^{+}(aq) \) has a positive Gibbs free energy, suggesting it is non-spontaneous under standard conditions.

Equilibrium Constant

The equilibrium constant (\( K \)) is a dimensionless value that expresses the ratio of products to reactants at equilibrium concentration for a reversible reaction. Its magnitude gives an insight into the position of equilibrium; a large \( K \) favors product formation, while a small \( K \) suggests reactants are predominating.

The equilibrium constant is connected to Gibbs free energy by the relationship \( \Delta G^\circ = -RT \ln(K) \), where \( R \) is the universal gas constant and \( T \) is the temperature in Kelvin. A negative \( \Delta G^\circ \) results in a positive natural logarithm of \( K \), which translates to a large equilibrium constant. Thus, reactions (i) and (ii) in the solution indicate that products are heavily favored under standard conditions. In contrast, reaction (iii)'s small \( K \) value implies a reaction that does not proceed to a great extent before reaching equilibrium.

Electrochemical Cell Potentials

The electrochemical cell potentials (\( E^\circ \)), also known as standard reduction potentials, are measures of the tendency of a chemical species to gain electrons and thereby be reduced. The more positive the reduction potential, the greater the species' affinity for electrons. In a redox reaction, the species with the higher reduction potential will undergo reduction, while the one with the lower potential will be oxidized.

In the provided exercise, different pairs of half-reactions combine to form complete redox reactions. The cell potential for each reaction is determined by subtracting the reduction potential of the oxidizing agent from the potential of the reducing agent. It's important to ensure that electrons are balanced when combining half-reactions. A positive cell potential indicates a spontaneous reaction under standard conditions, while a negative potential represents a non-spontaneous reaction. This concept is pivotal in predicting the direction of redox reactions and designing electrochemical cells such as batteries.