Question

Consider a voltaic cell in which the following reaction takes place in basic medium at \(25^{\circ} \mathrm{C}\). $$ \begin{aligned} 2 \mathrm{NO}_{3}^{-}(a q)+3 \mathrm{~S}^{2-}(a q)+4 \mathrm{H}_{2} \mathrm{O} \longrightarrow & \\ & 3 \mathrm{~S}(s)+2 \mathrm{NO}(g)+8 \mathrm{OH}^{-}(a q) \end{aligned} $$ (a) Calculate \(E^{\circ} .\) (b) Write the Nernst equation for the cell voltage \(E\). (c) Calculate \(E\) under the following conditions: $$ P_{\mathrm{NO}}=0.994 \mathrm{~atm}, \mathrm{pH}=13.7,\left[\mathrm{~S}^{2-}\right]=0.154 \mathrm{M} $$ \(\left[\mathrm{NO}_{3}^{-}\right]=0.472 \mathrm{M}\)

Short answer

\(2\mathrm{NO}_{3}^{-} + 3\mathrm{S}^{2-} + 4\mathrm{H}_2\mathrm{O} \longrightarrow 3\mathrm{S}(s) + 2\mathrm{NO}(g) + 8\mathrm{OH}^{-}\) Given conditions: \(P_{\mathrm{NO}} = 0.994\,\mathrm{atm}\), \(\mathrm{pH} = 13.7\), \([\mathrm{S}^{2-}]=0.154\,\mathrm{M}\), \([\mathrm{NO}_{3}^{-}] = 0.472\,\mathrm{M}\) (a) The standard cell potential (\(E^{\circ}\)): -0.816V (b) The Nernst equation for the cell voltage (\(E\)): \( E = -0.816 - \frac{RT}{10F} \ln\,\left(\frac{P_{\mathrm{NO}}^2[\mathrm{OH}^{-}]^8}{[\mathrm{NO}_3^{-}]^2[\mathrm{S}^{2-}]^3}\right) \) (c) The cell voltage (\(E\)) under the given conditions: -0.599V

Step-by-step solution

Determine the half reactions

In order to determine the standard electrode potentials, we need to first break down the overall reaction into half reactions. One half reaction will be for the reduction of \(\mathrm{NO}_{3}^{-}\), and the other half reaction will be for the oxidation of \(\mathrm{S}^{2-}\). The half reactions are as follows: (1) \(\mathrm{NO}_{3}^{-}(a q) + 2 \mathrm{H}_{2}\mathrm{O}\longrightarrow \mathrm{NO}(g) + 4 \mathrm{OH}^{-}(a q)\) (2) \(\mathrm{S}^{2-}(a q)\longrightarrow \mathrm{S}(s) + 2 \mathrm{e}^{-}\) We will balance these half reactions.

Balance the half reactions

In order to balance the half reactions, we will add appropriate numbers of electrons, water molecules, and hydroxide ions. Balancing each half reaction separately: Balanced Half Reaction (1): $$ \mathrm{NO}_{3}^{-}(a q) + 2 \mathrm{H}_{2}\mathrm{O} + 10 \mathrm{e}^{-} \longrightarrow \mathrm{NO}(g) + 4 \mathrm{OH}^{-}(a q) $$ Balanced Half Reaction (2): $$ \mathrm{S}^{2-}(a q) + 2 \mathrm{e}^{-}\longrightarrow \mathrm{S}(s) $$ Now, multiply each half reaction by an appropriate factor, so that the number of electrons in each half reaction is equal. In this case, multiply half reaction (1) by 1 and half reaction (2) by 5.

Combine the half reactions to find the overall balanced reaction

After balancing the half reactions and ensuring equal numbers of electrons, combine them to create the overall balanced reaction: $$ 2\mathrm{NO}_{3}^{-}(a q) + 3 \mathrm{S}^{2-}(a q) + 4 \mathrm{H}_{2} \mathrm{O} \longrightarrow 3 \mathrm{S}(s) + 2 \mathrm{NO}(g) + 8 \mathrm{OH}^{-}(a q) $$

Find the corresponding standard electrode potentials

To calculate \(E^{\circ}\), we need the standard electrode potentials for both half reactions. From a standard reduction potential table, we can find the following: (1) \(\mathrm{NO}_{3}^{-}(a q) + 2 \mathrm{H}_{2}\mathrm{O} + 10\mathrm{e}^{-}\longrightarrow \mathrm{NO}(g) + 4 \mathrm{OH}^{-}(a q): E_1^{\circ} = -0.956\,\mathrm{V}\) (2) \(\mathrm{S}^{2-}(a q) + 2\mathrm{e}^{-}\longrightarrow \mathrm{S}(s): E_2^{\circ} = -0.14\,\mathrm{V}\)

Calculate the standard cell potential\(E^{\circ}\)

The standard cell potential can be calculated using the standard reduction potentials of both half reactions: $$ E^{\circ}= E_{1}^{\circ} - E_{2}^{\circ}= -0.956 - (-0.14) = -0.816 \,\mathrm{V} $$ (a) The standard cell potential \(E^{\circ} = -0.816V\).

Write the Nernst equation for the cell voltage \(E\)

The Nernst equation relates cell voltage \(E\) to the standard cell potential \(E^{\circ}\), concentrations of the reactants and products, and the temperature: $$ E = E^{\circ} - \frac{RT}{nF} \ln \left(\frac{\prod P_i^{c_i}[\mathrm{P}roducts]}{\prod [\mathrm{R}eactants] A_i^{a_i}}\right) $$ Rearrange this equation for our specific reaction: $$ E = -0.816 - \frac{RT}{10F} \ln\,\left(\frac{P_{\mathrm{NO}}^2[\mathrm{OH}^{-}]^8}{[\mathrm{NO}_3^{-}]^2[\mathrm{S}^{2-}]^3}\right) $$ (b) The Nernst equation for the cell voltage \(E\) is written above.

Calculate the cell voltage \(E\) under given conditions

We can now calculate the cell voltage \(E\) using the given conditions and the Nernst equation: (c) Calculate \(E\) under the following conditions: $$ P_{\mathrm{NO}} = 0.994\,\mathrm{atm}, \mathrm{pH} = 13.7, [\mathrm{S}^{2-}]=0.154\,\mathrm{M}, [\mathrm{NO}_{3}^{-}] = 0.472\,\mathrm{M} $$ First, find the concentration of \(\mathrm{OH}^{-}\) from the given pH: \([\mathrm{OH}^{-}] = 10^{-(14-\mathrm{pH})} = 10^{-(14-13.7)} = 10^{-0.3}\) Now, plug these values into the Nernst equation: $$ E = -0.816 - \frac{(8.314)(25 + 273.15)}{10(96485)} \ln \left( \frac{(0.994)^2(10^{-0.3})^8}{(0.472)^2(0.154)^3} \right) $$ Evaluating this expression, we have: $$ E = -0.816 - \frac{(8.314)(298.15)}{10(96485)} \ln \left( 2.82 \times 10^{-2} \right) \approx -0.599\,\mathrm{V} $$ (c) The cell voltage \(E\) under the given conditions is -0.599V.