Question

Consider the reaction at \(25^{\circ} \mathrm{C}\) : \(\mathrm{S}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{Ag}(s)+2 \mathrm{Br}^{-}(a q) \longrightarrow 2 \mathrm{AgBr}(s)+\mathrm{H}_{2} \mathrm{~S}(a q)\) At what \(\mathrm{pH}\) is the voltage zero if all the species are at standard conditions?

Short answer

Based on the given redox reaction and its standard conditions, determine the pH at which the voltage of the reaction is zero.

Step-by-step solution

Write down the Nernst equation

The Nernst equation is used to determine the electromotive force (EMF) of a redox reaction, considering the standard conditions. The equation is given by: \(\textit{E} = \textit{E}_{0} - \frac{RT}{nF} \times \ln Q\) Where: - E: cell potential - \(E_0\): standard cell potential - R: gas constant, \(8.314 JK^{-1} mol^{-1}\) - T: temperature in Kelvin - n: number of moles of electrons transferred in the redox reaction - F: Faraday's constant, \(96485 C mol^{-1}\) - Q: reaction quotient In this case, we need to find the pH at which the voltage (E) is zero.

Break down the reaction into half-reactions

To apply the Nernst equation, first, we need to break down the given reaction into redox half-reactions. Oxidation half-reaction: \(\textit{S(s)} \rightarrow \textit{H}_{2}\textit{S(aq)} + 2\textit{e}^{-}\) Reduction half-reaction: \(\textit{2Ag(s)} + 2\textit{Br}^{-}\textit{(aq)} + 2\textit{e}^{-} \rightarrow \textit{2AgBr(s)}\)

Calculate the standard cell potential

Now, we need to calculate the standard cell potential (\(E_0\)) using the standard electrode potentials for each half-reaction: \(E_0 = E_{0, red} - E_{0, ox}\) For this problem, we can be provided with the standard electrode potentials: \(E_{0, ox} = -0.14 V\) (for oxidation half-reaction) \(E_{0, red} = 0.07 V\) (for reduction half-reaction) Now, we can calculate \(E_0\): \(E_0 = 0.07 - (-0.14) = 0.21 V\)

Apply the Nernst equation

In this step, we will apply the Nernst equation to find the pH at which the voltage is zero. At standard conditions, all the species have concentrations of 1 M. Since the solution is at 25\(^{\circ}\)C, we can calculate the temperature in Kelvin, which is equal to 298.15 K. \(\textit{0} = \textit{0.21} - \frac{8.314 × 298.15}{2 × 96485} \times \ln Q\) Since the reactants and products are in their standard states-except hydrogen ion and the reaction quotient are equal to: \(Q = [\textit{H}^{+}]^{2}\) Replacing Q in the Nernst equation: \(0 = 0.21 - \frac{8.314 × 298.15}{2 × 96485} \times \ln [\textit{H}^{+}]^{2}\) \(0 = 0.21 - \frac{4.141}{96485} \times \ln [\textit{H}^{+}]\) \(\frac{4.141}{96485} \times \ln [\textit{H}^{+}] = 0.21\) Now, we can solve for the hydrogen ion concentration: \(\ln [\textit{H}^{+}] = \frac{0.21 × 96485}{4.141}\) \([\textit{H}^{+}] = \textit{e}^{\frac{0.21 × 96485}{4.141}}\)

Calculate the pH

Finally, we can calculate the pH using the obtained hydrogen ion concentration: \(\textit{pH} = -\log_{10} [\textit{H}^{+}]\) \(\textit{pH} = -\log_{10} (\textit{e}^{\frac{0.21 × 96485}{4.141}})\) Calculate the above expression to get the pH at which the voltage is zero: \(\textit{pH} \approx 4.19\) So the pH at which the voltage is zero is approximately 4.19.