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For each of the following reactions, write a baFor each of the following reactions, write a balanced equation, calculate the standard emf, calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\), and calculate the equilibrium constant \(K\) at \(298 \mathrm{~K}\). (a) Aqueous iodide ion is oxidized to \(\mathrm{I}_{2}(s)\) by \(\mathrm{Hg}_{2}{ }^{2+}(a q) .\) (b) In acidic solution, copper(I) ion is oxidized to copper(II) ion by nitrate ion. (c) In basic solution, \(\mathrm{Cr}(\mathrm{OH})_{3}(s)\) is oxidized to \(\mathrm{CrO}_{4}^{2-}(a q)\) by \(\mathrm{ClO}^{-}(a q) .\)lanced equation, calculate the standard emf, calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\), and calculate the equilibrium constant \(K\) at \(298 \mathrm{~K}\). (a) Aqueous iodide ion is oxidized to \(\mathrm{I}_{2}(s)\) by \(\mathrm{Hg}_{2}{ }^{2+}(a q) .\) (b) In acidic solution, copper(I) ion is oxidized to copper(II) ion by nitrate ion. (c) In basic solution, \(\mathrm{Cr}(\mathrm{OH})_{3}(s)\) is oxidized to \(\mathrm{CrO}_{4}^{2-}(a q)\) by \(\mathrm{ClO}^{-}(a q) .\)

Step by step solution

01

Write the balanced redox equation

First we write the half-reactions: Oxidation: \(2I^-(aq) \rightarrow I_2(s) + 2e^-\) Reduction: \(Hg_2^{2+}(aq) + 2e^- \rightarrow 2Hg(l)\) Now, combine the two half-reactions to obtain the balanced redox equation: \(2I^-(aq) + Hg_2^{2+}(aq) \rightarrow I_2(s) + 2Hg(l)\)

02

Calculate the standard emf (E掳)

Using the standard electrode potential table, we can find the E掳 for each half-reaction: E掳(I鈦�/I鈧�) = +0.54 V E掳(Hg鈧偮测伜/Hg) = +0.85 V Next, calculate the overall E掳 for the reaction: E掳(cell) = E掳(Hg鈧偮测伜/Hg) - E掳(I鈦�/I鈧�) = +0.85 V - (+0.54 V) = +0.31 V

03

Calculate the standard Gibbs free energy change (螖G掳)

Since there are 2 electrons transferred in this reaction, n=2. Now, calculate 螖G掳 using 螖G掳 = -nFE掳: 螖G掳 = -(2 mol e鈦�)(96485 C/mol e鈦�)(0.31 V) = -59.4 kJ/mol

04

Calculate the equilibrium constant (K)

Now, use the relationship 螖G掳 = -RTlnK to calculate the equilibrium constant: \(-59.4 kJ/mol = -(8.314 J/mol*K)(298 K)lnK\) Solving for K, we get: K = 3.3 脳 10鹿虏 #(b) In acidic solution, copper(I) ion is oxidized to copper(II) ion by nitrate ion# Follow a similar process for steps 1-4:

05

Write the balanced redox equation

In acidic solution, we need to balance the half-reactions using H鈦� ions and H鈧侽: Oxidation: \(Cu^+(aq) \rightarrow Cu^{2+}(aq) + e^-\) Reduction: \(NO_3^-(aq) + 2H^+(aq) + e^- \rightarrow NO_2(g) + H_2O(l)\) Now, combine the two half-reactions to obtain the balanced redox equation: \(2Cu^+(aq) + 2NO_3^-(aq) + 4H^+(aq) \rightarrow 2Cu^{2+}(aq) + 2NO_2(g) + 2H_2O(l)\)

06

Calculate the standard emf (E掳)

Using the standard electrode potential table, we can find the E掳 for each half-reaction: E掳(Cu虏鈦�/Cu鈦�) = +0.15 V E掳(NO鈧冣伝/NO鈧�) = +0.96 V Next, calculate the overall E掳 for the reaction: E掳(cell) = E掳(NO鈧冣伝/NO鈧�) - E掳(Cu虏鈦�/Cu鈦�) = +0.96 V - (+0.15 V) = +0.81 V

07

Calculate the standard Gibbs free energy change (螖G掳)

Since there is 1 electron transferred in this reaction, n=1. Now, calculate 螖G掳 using 螖G掳 = -nFE掳: 螖G掳 = -(1 mol e鈦�)(96485 C/mol e鈦�)(0.81 V) = -78.2 kJ/mol

08

Calculate the equilibrium constant (K)

Now, use the relationship 螖G掳 = -RTlnK to calculate the equilibrium constant: \(-78.2 kJ/mol = -(8.314 J/mol*K)(298 K)lnK\) Solving for K, we get: K = 1.1 脳 10鹿鲁 #(c) In basic solution, Cr(OH)鈧�(s) is oxidized to CrO鈧劼测伝(aq) by ClO鈦�(aq)# Follow a similar process for steps 1-4:

09

Write the balanced redox equation

In basic solution, we need to balance the half-reactions using OH鈦� ions and H鈧侽: Oxidation: \(Cr(OH)_3(s) + 3OH^-(aq) \rightarrow CrO_4^{2-}(aq) + 2H_2O(l) + 3e^-\) Reduction: \(2ClO^-(aq) + e^- + 2H_2O(l) \rightarrow Cl_2(g) + 4OH^-(aq)\) Now, combine the two half-reactions to obtain the balanced redox equation: \(Cr(OH)_3(s) + 2ClO^-(aq) + 5OH^-(aq) \rightarrow CrO_4^{2-}(aq) + Cl_2(g) + 3H_2O(l)\)

10

Calculate the standard emf (E掳)

Using the standard electrode potential table, we can find the E掳 for each half-reaction: E掳(CrO鈧劼测伝/Cr(OH)鈧�) = +0.13 V E掳(ClO鈦�/Cl鈧�) = +1.64 V Next, calculate the overall E掳 for the reaction: E掳(cell) = E掳(ClO鈦�/Cl鈧�) - E掳(CrO鈧劼测伝/Cr(OH)鈧�) = +1.64 V - (+0.13 V) = +1.51 V

11

Calculate the standard Gibbs free energy change (螖G掳)

Since there are 3 electrons transferred in this reaction, n=3. Now, calculate 螖G掳 using 螖G掳 = -nFE掳: 螖G掳 = -(3 mol e鈦�)(96485 C/mol e鈦�)(1.51 V) = -437 kJ/mol

12

Calculate the equilibrium constant (K)

Now, use the relationship 螖G掳 = -RTlnK to calculate the equilibrium constant: \(-437 kJ/mol = -(8.314 J/mol*K)(298 K)lnK\) Solving for K, we get: K = 2.1 脳 10虏鹿

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Key Concepts

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

Balanced Chemical Equations

A balanced chemical equation is essential in any redox reaction because it ensures the conservation of mass and charge. This means that the number of atoms for each element is the same on both sides of the reaction, and the total charge is also equal. Balancing redox reactions involves separating the oxidation and reduction processes into half-reactions, then adjusting coefficients to ensure equal numbers of exchanged electrons.
This can involve adding substances like water, hydrogen ions (H鈦�), or hydroxide ions (OH鈦�), depending on whether the reaction occurs in acidic or basic environments. For instance, in acidic conditions, as demonstrated with copper(I) ion, you might need to add H鈦� ions. In basic conditions, like with the oxidation of Cr(OH)鈧�, adding OH鈦� is necessary to achieve balance.
Thus, writing a balanced chemical equation is about not just the reactants and products, but also the stoichiometric relationships that maintain equal charge and element count.

Standard Electrode Potential

Standard electrode potential ( E掳 ) is a measure of the tendency of a chemical species to be reduced, and it is measured in volts. Each half-reaction in a redox process has its own E掳 value, often found in a standard electrode potential table. These values are critical because they help determine the spontaneity and direction of a redox reaction.

• If the E掳 of a cell is positive, the reaction is spontaneous under standard conditions.
• Conversely, a negative E掳 indicates a non-spontaneous reaction.

E掳 values are calculated by subtracting the reduction potential of the oxidation half-reaction from the reduction potential of the reduction half-reaction: E掳(cell) = E掳(cathode) - E掳(anode).
This value helps us understand and predict the flow of electrons during the reaction, as with the calculation for the Hg_2^{2+}/I^- reaction which yielded a E掳 value of 0.31 V.

Gibbs Free Energy

Gibbs free energy ( 螖G掳 ) is another crucial concept in redox reactions. It indicates the maximum amount of work a chemical reaction can perform at constant temperature and pressure. For redox reactions, it is directly related to the standard cell potential E掳 and the number of electrons transferred, n . The equation connecting these quantities is: 螖G掳 = -nFE掳 , where F is the Faraday constant ( 96485 C/mol ).

• A negative 螖G掳 signifies that the reaction is spontaneous, capable of doing work.
• A positive 螖G掳 means the reaction is non-spontaneous and requires energy input.

For instance, the 螖G掳 calculated for the iodide oxidation showed -59.4 kJ/mol, indicating spontaneity. Understanding 螖G掳 helps predict whether a reaction will proceed and is essential for determining the reaction's driving force.

Equilibrium Constant

The equilibrium constant ( K ) of a redox reaction at a specific temperature is a measure of the extent to which the reaction proceeds to form products. There is a close relationship between K and 螖G掳 , given by the formula: 螖G掳 = -RTlnK , where R is the gas constant ( 8.314 J/mol鈰匥 ) and T is the temperature in Kelvin. This equation allows us to calculate K once 螖G掳 is known.

• A large K value indicates that the reaction strongly favors the formation of products.
• A small K signifies that reactants are predominant at equilibrium.

For example, the reaction of iodide being oxidized to iodine by mercury(II) showed a K of 3.3 脳 10^{12} , illustrating a reaction that proceeds nearly to completion. Calculating K is vital for understanding the dynamic balance of reversible reactions and predicting the final concentrations of reactants and products in a system.