Question

Consider the reaction below at \(25^{\circ} \mathrm{C}\) : $$ 3 \mathrm{SO}_{4}{ }^{2-}(a q)+12 \mathrm{H}^{+}(a q)+2 \mathrm{Cr}(s) \longrightarrow 3 \mathrm{SO}_{2}(g)+2 \mathrm{Cr}^{3+}(a q)+6 \mathrm{H}_{2} \mathrm{O} $$ Use Table \(18.1\) to answer the following questions. Support your answers with calculations. (a) Is the reaction spontaneous at standard conditions? (b) Is the reaction spontaneous at a \(\mathrm{pH}\) of \(3.00\) with all other ionic species at \(0.100 \mathrm{M}\) and gases at \(1.00\) atm? (c) Is the reaction spontaneous at a pH of \(8.00\) with all other ionic species at \(0.100 \mathrm{M}\) and gases at \(1.00 \mathrm{~atm}\) ? (d) At what \(\mathrm{pH}\) is the reaction at equilibrium with all other ionic species at \(0.100 \mathrm{M}\) and gases at \(1.00 \mathrm{~atm}\) ?

Short answer

What is the pH at which the reaction is at equilibrium? Answer: The reaction is non-spontaneous at standard conditions, as the standard cell potential is negative (-0.95 V). To determine the spontaneity of the reaction at pH 3.00 and pH 8.00, calculate the cell potential using the Nernst equation and given concentrations. If the cell potential is positive, the reaction is spontaneous; if the cell potential is negative, the reaction is non-spontaneous. The pH at which the reaction reaches equilibrium can be found by setting the cell potential equal to zero and solving for the pH using the Nernst equation.

Step-by-step solution

Calculate standard cell potential

To determine whether the reaction is spontaneous at standard conditions, we must calculate the standard cell potential of the reaction. Using table 18.1, note that the standard reduction potential for the \(\mathrm{SO}_4{ }^{2-}\) couple and the \(\mathrm{Cr}\) couple are as follows: \(\mathrm{SO}_4{ }^{2-}(a q)+4 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-}\longrightarrow \mathrm{SO}_{2}(g)+2 \mathrm{H}_{2}\mathrm{O}\) with \(E^\circ=+0.17 \mathrm{V}\) \(\mathrm{Cr}^{3+}(a q)+3 \mathrm{e}^{-}\longrightarrow \mathrm{Cr}(s)\) with \(E^\circ=-0.74 \mathrm{V}\) To get the reaction given in the exercise, we need to multiply the first reaction by 3 and the second reaction by 2. This means their respective standard reduction potentials will also be multiplied by these factors. Add these potentials to find the overall standard cell potential: \(E^\circ_{cell} = 3(0.17 \mathrm{V}) - 2(0.74 \mathrm{V}) = -0.95 \mathrm{V}\)

Determine spontaneity at standard conditions

If the standard cell potential is negative, the reaction is non-spontaneous at standard conditions. In this case, \(E^\circ_{cell} = -0.95 \mathrm{V}\), which is negative. So, the reaction is non-spontaneous at standard conditions.

Use the Nernst equation to find the cell potential at given pH

To answer parts (b) and (c), we must calculate the cell potential at the given pH values. Use the Nernst equation for this purpose: $$ E_{cell} = E^\circ_{cell} - \frac{0.0592}{n} \log Q $$ where n is the number of moles of electrons transferred in the reaction, and Q is the reaction quotient. For the given reaction, n = 6 moles of electrons. The reaction quotient Q can be expressed as: $$ Q = \frac{[\mathrm{Cr}^{3+}]^2[\mathrm{H}^{+}]^{12}}{[\mathrm{SO}_4{ }^{2-}]^3} $$ Calculate \(E_{cell}\) for pH = 3.00 (part b) and pH = 8.00 (part c) using the given concentrations for other ionic species and partial pressures for gases.

Determine spontaneity at given pH

If \(E_{cell} > 0\), the reaction is spontaneous; if \(E_{cell} < 0\), the reaction is non-spontaneous. Use the calculated \(E_{cell}\) values from Step 3 to determine the spontaneity of the reaction at the given pH values.

Find the pH at which the reaction is at equilibrium

At equilibrium, \(E_{cell} = 0\). Use the Nernst equation to solve for pH when the reaction is at equilibrium with the given concentrations for other ionic species and partial pressures for gases. $$ 0 = E^\circ_{cell} - \frac{0.0592}{n} \log Q $$ Solve for pH to find the equilibrium value.

Key concepts

Standard Cell Potential

Understanding the standard cell potential is crucial for assessing the spontaneity of a chemical reaction under standard conditions, which include solute concentrations at 1 M, a pressure of 1 atm for gases, and a temperature of 25°C or 298 K. This potential, denoted as \(E^\text{\textdegree}_{cell}\), is a measure of the electromotive force of an electrochemical cell when all reagents are at their standard states.

To calculate \(E^\text{\textdegree}_{cell}\), one must recognize that it is the difference between the standard reduction potentials of the cathode and the anode. The standard reduction potential is a measure of how strongly a species gains electrons, with a more positive value indicating a greater tendency to be reduced. In the example given in the exercise, the reaction between sulfate ions, protons, and solid chromium to produce sulfur dioxide, trivalent chromium ions, and water has a \(E^\text{\textdegree}_{cell}\) calculated by combining the reduction potentials of each half-reaction. Since a negative \(E^\text{\textdegree}_{cell}\) suggests the reaction is non-spontaneous at standard conditions, understanding and calculating this value is the key step in predicting the behavior of the reaction without carrying it out.

Nernst Equation

While standard cell potential offers insight at standard conditions, the Nernst equation allows us to predict cell potential under non-standard conditions by accounting for temperature, concentration, and partial pressures of the reactants and products involved in a chemical reaction. The Nernst equation is expressed as:
\[ E_{cell} = E^\text{\textdegree}_{cell} - \frac{0.0592}{n} \times \text{log}Q \]
Here, \(E_{cell}\) represents the actual cell potential, \(E^\text{\textdegree}_{cell}\) is the standard cell potential, \(n\) is the number of moles of electrons exchanged in the redox reaction, and \(Q\) is the reaction quotient reflecting the ratio of the products' concentrations to the reactants' concentrations, raised to their stoichiometric coefficients. The Factor 0.0592 arises from combining constants from the gas constant, Faraday's constant, and the standard temperature (298 K). Applying the Nernst Equation makes it possible to determine the cell potential, and thus the spontaneity of a reaction, at any given pH, allowing us to peer into conditions far from standard.

Reaction Quotient

The reaction quotient, \(Q\), plays a pivotal role in understanding chemical equilibria and predicting the direction of a reaction. It is calculated in a similar way to the equilibrium constant, \(K\), but with the current concentrations of the reactants and products, not necessarily those at equilibrium. The reaction quotient is given by the expression:
\[ Q = \frac{[C]^c[D]^d}{[A]^a[B]^b} \]
where square brackets denote molar concentrations, and the lowercase letters represent stoichiometric coefficients. When evaluating spontaneity at various pH levels, the hydrogen ion concentration, \([H^+]\), changes, directly influencing \(Q\) since it's part of the reaction in our example. If \(Q < K\), the reaction will proceed forward to reach equilibrium, but if \(Q > K\), the reaction will progress in the reverse direction. Therefore, by calculating the reaction quotient, we can use the Nernst Equation to determine whether a reaction will be spontaneous under varied conditions.

Equilibrium pH

The concept of equilibrium pH is essential when considering reactions involving acid-base equilibria, as in the dissolution or formation of ions in water. It is the pH at which a chemical reaction achieves equilibrium, meaning the rate of the forward reaction equals that of the reverse reaction, and the net change of reactants and products is zero. At this specific pH, the reaction quotient \(Q\) equals the equilibrium constant \(K\) and the cell potential \(E_{cell}\) is zero. This is important in electrochemical reactions, as it determines the pH at which there's no driving force for the reaction to occur in either direction. As seen in the exercise, finding the equilibrium pH involves manipulating the Nernst equation to solve for the pH that satisfies this condition. This calculation is critical for understanding the acidity or basicity conditions needed for a reaction to stay at equilibrium and is valuable for process optimization in industrial applications and environmental predictions.