Problem 50 - Chemistry: The Central Science

Question

For 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 \(\mathrm{Hg}_{2}{\underline{\phantom{xx}}}^{2+}(a q)\). (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

Answer

(a) Balanced equation: \(2 \: I^{-}(aq) + Hg_{2}^{2+}(aq) \rightarrow I_{2}(s) + 2 \: Hg(l)\). Standard emf: \(E°(cell) = 0.315 V\). Gibbs free energy change: \(\Delta G^{\circ} = -60709 \:J/mol\). Equilibrium constant: \(K = 1.12 \times 10^{13}\). (b) Balanced equation: \(2 \: Cu^{+}(aq) + 2 \: NO_{3}^{-}(aq) + 4 \: H^{+}(aq) \rightarrow 2 \: Cu^{2+}(aq) + 2 \: NO_{2}(g) + 2 \: H_{2}O(l)\). Standard emf: \(E°(cell) = 1.109 V\). Gibbs free energy change: \(\Delta G^{\circ} = -214131 \:J/mol\). Equilibrium constant: \(K = 3.13 \times 10^{36}\). (c) Balanced equation: \(2 \: Cr(OH)_{3}(s) + 6 \: ClO^{-}(aq) + 6 \: H_{2}O(L) \rightarrow 2 \: CrO_{4}^{2-}(aq) + 6 \: Cl^{-}(aq) + 12 \: OH^{-}(aq)\). Standard emf: \(E°(cell) = 1.531 V\). Gibbs free energy change: \(\Delta G^{\circ} = -888994 \:J/mol\). Equilibrium constant: \(K = 1.25 \times 10^{57}\).

Step 1: Balancing the reaction

To balance the given reaction, just mention the coefficients that make the number of atoms equal on both sides. \(2 \: I^{-}(aq) + Hg_{2}^{2+}(aq) \rightarrow I_{2}(s) + 2 \: Hg(l)\)

Step 2: Standard emf calculation

To calculate the standard emf, look up the given half-cell reactions and their potentials in the standard reduction potential table and use the formula: \(E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}\). Reduction half-cell reaction: \( Hg_{2}^{2+}(aq) + 2 \: e^{-} \rightarrow 2 \: Hg(l) \) E°(Hg2^2+/Hg) = 0.851 V Oxidation half-cell reaction: \( 2 \: I^{-}(aq) \rightarrow I_{2}(s) + 2 \: e^{-} \) E°(I2/I-) = 0.536 V E°(cell) = 0.851 V - 0.536 V = 0.315 V

Step 3: Gibbs free energy change calculation

To calculate the Gibbs free energy change, use the formula: \(\Delta G^{\circ} = -nFE_{cell}^{\circ}\), where n is the number of moles of electrons transferred, F is the Faraday constant (96485 C/mol), and E°(cell) is the standard emf. \(\Delta G^{\circ} = -(2)(96485\:C/mol)(0.315\: V)\) \(\Delta G^{\circ} = -60709 \:J/mol\)

Step 4: Equilibrium constant calculation

To calculate the equilibrium constant K, use the relationship: \(\Delta G^{\circ} = -RT \ln K\), where R is the gas constant (8.314 J/mol-K), and T is the temperature (298K). \(-60709\:J/mol = -(8.314\:J/mol \cdot K)(298\: K) \ln K\) \(K = 1.12 \times 10^{13}\) (b) In acidic solution, copper(I) ion is oxidized to copper(II) ion by nitrate ion We will now repeat the steps for this reaction.

Step 5: Balancing the reaction

First, determine the oxidation half-cell reaction and reduction half-cell reaction in their acidic conditions. Oxidation half-cell reaction: \( Cu^{+}(aq) \rightarrow Cu^{2+}(aq) + e^{-}\) Reduction half-cell reaction: \( NO_{3}^{-}(aq) + 2\:H^{+}(aq) + e^{-} \rightarrow NO_2(g) + H_2O(l)\) Then, balance the overall reaction, applying coefficients on the half-cell reactions to equalize the number of electrons transferred. \(2 \: Cu^{+}(aq) + 2 \: NO_{3}^{-}(aq) + 4 \: H^{+}(aq) \rightarrow 2 \: Cu^{2+}(aq) + 2 \: NO_{2}(g) + 2 \: H_{2}O(l)\)

Step 6: Standard emf calculation

Look up the given half-cell reactions in the standard reduction potential table. Reduction half-cell reaction: \( NO_{3}^{-}(aq) + 2\:H^{+}(aq) + e^{-} \rightarrow NO_2(g) + H_2O(l)\) E°(NO3-/NO2) = 0.956 V Oxidation half-cell reaction: \( Cu^{+}(aq) \rightarrow Cu^{2+}(aq) + e^{-}\) E°(Cu2+/Cu+) = - 0.153 V E°(cell) = 0.956 V - (- 0.153 V) = 1.109 V

Step 7: Gibbs free energy change calculation

\(\Delta G^{\circ} = -(2)(96485\:C/mol)(1.109\: V)\) \(\Delta G^{\circ} = -214131 \:J/mol\)

Step 8: Equilibrium constant calculation

\(-214131\:J/mol = -(8.314\:J/mol \cdot K)(298\: K) \ln K\) \(K = 3.13 \times 10^{36}\) (c) In basic solution, Cr(OH)3(s) is oxidized to CrO42-(aq) by ClO-(aq) We will follow the steps for this third reaction.

Step 9: Balancing the reaction

First, determine the oxidation half-cell reaction and reduction half-cell reaction in a basic environment. Oxidation half-cell reaction: \( Cr(OH)_{3}(s) + 3\: H_{2}O(l) \rightarrow CrO_{4}^{2-}(aq) + 6 \: OH^{-}(aq) + 3 \: e^{-}\) Reduction half-cell reaction: \( ClO^{-}(aq) + H_{2}O(l) + 2\: e^{-} \rightarrow Cl^{-}(aq) + 2 \: OH^{-}(aq)\) Then, balance the overall reaction, applying coefficients on the half-cell reactions to equalize the number of electrons transferred. \(2 \: Cr(OH)_{3}(s) + 6 \: ClO^{-}(aq) + 6 \: H_{2}O(L) \rightarrow 2 \: CrO_{4}^{2-}(aq) + 6 \: Cl^{-}(aq) + 12 \: OH^{-}(aq)\)

Step 10: Standard emf calculation

Check the given half-cell reactions in the standard reduction potential table. Reduction half-cell reaction: \( ClO^{-}(aq) + H_{2}O(l) + 2\: e^{-} \rightarrow Cl^{-}(aq) + 2 \: OH^{-}(aq)\) E°(ClO-/Cl-) = 1.401 V Oxidation half-cell reaction: \( Cr(OH)_{3}(s) + 3\: H_{2}O(l) \rightarrow CrO_{4}^{2-}(aq) + 6 \: OH^{-}(aq) + 3 \: e^{-}\) E°(CrO42-/Cr(OH)3) = -0.13 V E°(cell) = 1.401 V - (-0.13 V) = 1.531 V

Step 11: Gibbs free energy change calculation

\(\Delta G^{\circ} = -(6)(96485\:C/mol)(1.531\: V)\) \(\Delta G^{\circ} = -888994 \:J/mol\)

Step 12: Equilibrium constant calculation

\(-888994\:J/mol = -(8.314\:J/mol \cdot K)(298\: K) \ln K\) \(K = 1.25 \times 10^{57}\)

Key Concepts

Balanced Chemical Equations

A fundamental part of electrochemistry involves writing balanced chemical equations for redox reactions. Redox reactions, or reduction-oxidation reactions, involve the transfer of electrons between two species. To balance these reactions, you must ensure that the number of electrons lost in oxidation equals the number gained in reduction.
First, split the reaction into two half reactions: oxidation and reduction. Balance the number of atoms other than O and H first, then add H2O to balance O atoms and H+ or OH- to balance H atoms depending on the solution (acidic or basic). Finally, balance the charge by adding electrons and equalize the number of electrons in both half-reactions before combining them back. - **Example 1:** In an acidic solution, copper(I) ion (Cu⁺) is oxidized to copper(II) ion (Cu²⁺) by nitrate ion (NO₃⁻). Here, you balance copper and nitrate separately and then adjust coefficients to equalize electrons. - **Example 2:** In a basic environment, the reaction of Cr(OH)₃ to CrO₄²⁻ is balanced by first accounting for the OH⁻ ions.

Standard Electrode Potential

The standard electrode potential measures the ability of a chemical species to gain or lose electrons under standard conditions. Every half-reaction has a unique standard reduction potential (E°), measured in volts (V). This value can be found in standard tables and is critical for calculating the emf (electromotive force) of a cell.
To compute the overall cell potential, identify and write down the half-cell reactions involved, and then use the formula: \[ E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} \]Here, the cathode is where reduction happens and the anode is where oxidation occurs. A positive E° cell value indicates a spontaneous reaction under standard conditions.- **Example:** When iodide ion (I⁻) is oxidized by mercury ion (Hg₂²⁺), the standard potentials for the involved half-cells determine the total cell voltage. The overall cell potential informs on the energy exchange taking place during the reaction.

Gibbs Free Energy

Gibbs free energy (\( \Delta G^{\circ} \)) helps in understanding the spontaneity of a reaction. It connects the electrical energy from electrochemistry to the fundamental thermodynamic principles.
The equation to find Gibbs free energy change is:\[ \Delta G^{\circ} = -nFE_{cell}^{\circ}\]Where \( n \) is the number of moles of electrons transferred in the reaction, \( F \) is the Faraday constant (96485 C/mol), and \( E_{cell}^{\circ} \) is the standard cell potential. A negative \( \Delta G^{\circ} \) value indicates a spontaneous process, which means the reaction can proceed without needing external energy.- **Example:** For the copper and nitrate reaction, the calculation of \( \Delta G^{\circ} \) tells us that the reaction is highly favorable and releases energy as it occurs.

Equilibrium Constant

The equilibrium constant (K) is a crucial concept that relates the concentrations of reactants and products for a reaction at equilibrium. For electrochemistry, it provides insight into the extent of the reaction under standard conditions.
The formula linking Gibbs free energy and K is:\[ \Delta G^{\circ} = -RT \ln K\]where \( R \) is the gas constant (8.314 J/mol-K) and \( T \) is the temperature in Kelvin (K). By rearranging, you can solve for \( K \):\[ K = e^{-\Delta G^{\circ}/RT}\]A larger \( K \) indicates a reaction that heavily favors the products at equilibrium. For reactions like the oxidation of Cr(OH)₃ by ClO⁻, a very high K value shows that the reaction almost proceeds completely to form products under the given conditions.