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Chapter 18: Problem 53

Estimate \(\mathscr{E}^{\circ}\) for the half-reaction $$2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}+2 \mathrm{OH}^{-}$$ given the following values of \(\Delta G_{\mathrm{f}}^{\circ} :\) $$\quad\quad\quad \mathrm{H}_{2} \mathrm{O}(l)=-237 \mathrm{kJ} / \mathrm{mol}$$ $$\mathrm{H}_{2}(g)=0.0$$ $$\quad\quad\quad \mathrm{OH}^{-}(a q)=-157 \mathrm{kJ} / \mathrm{mol}$$ $$\quad \mathrm{e}^{-}=0.0$$ Compare this value of \(\mathscr{E}^{\circ}\) with the value of \(\mathscr{E}^{\circ}\) given in Table 18.1

Short Answer

Expert verified
The estimated standard electrode potential (E°) for the given half-reaction \(2H_2O + 2e^- \rightarrow H_2 + 2OH^-\) is \(0.83\,V\). This value is similar to the E° value found in Table 18.1 (\(0.828\,V\)), indicating that our calculated estimate is accurate and reasonable.

Step by step solution

01

Write the reaction equation with the given values of ∆Gf°

Rewrite the half-reaction and substitute the given Gibbs free energy of formation values: \(2H_2O + 2e^- \rightarrow H_2 + 2OH^-\) \(-2(237) + 2(0) \rightarrow 0 + 2(-157)\)
02

Calculate the total change in Gibbs free energy (∆G°) for the reaction

Determine the total change in Gibbs free energy for the half-reaction using the stoichiometric coefficients. The change in Gibbs free energy for the complete reaction is the sum of ∆Gf° products minus the sum of ∆Gf° reactants: \(\Delta G^{\circ} = [1(0) + 2(-157)] - [2(-237) + 2(0)]\) \(\Delta G^{\circ} = (-314) - (-474)\) \(\Delta G^{\circ} = 160\,kJ/mol\)
03

Convert ∆G° to standard electrode potential (E°)

Given the relationship between Gibbs free energy and standard electrode potential: \(\Delta G^{\circ} = -nFE^{\circ}\) where: \(\Delta G^{\circ}\) = change in Gibbs free energy (in J/mol) n = number of moles of electrons transferred (mol) F = Faraday constant (\(96,485\,C/mol\)) \(E^{\circ}\) = standard electrode potential (V) In our case, the number of moles of electrons transferred (n) is 2 as we can see from the balanced half-reaction. Also, convert the ΔG° value from kJ/mol to J/mol: \(160 \,kJ/mol = 160,000 \,J/mol\) Now we can solve for \(E^{\circ}\): \(-160,000 \,J/mol = -2*\text{(}96,485 \,C/mol\text{)} E^{\circ}\)
04

Calculate E° for the reaction

To determine the E° value, we will divide both sides of the equation by -2 multiplied by the Faraday constant: \(E^{\circ} = \frac{-160,000 \,J/mol}{(-2)(96,485 \,C/mol)}\) \(E^{\circ} = 0.83\,V\) Finally, the estimated standard electrode potential (E°) for the given half-reaction is \(0.83\,V\).
05

Compare to value in Table 18.1

In Table 18.1, the E° value found for the given half-reaction is approximately \(0.828\,V\). This is similar to the E° value calculated in this exercise (\(0.83\,V\)), demonstrating that our calculated estimate is accurate and reasonable.

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Most popular questions from this chapter

Calculate \(\mathscr{E}^{\circ}\) for the following half-reaction: $$\mathrm{AgI}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)+\mathrm{I}^{-}(a q)$$ (Hint: Reference the \(K_{\mathrm{sp}}\) value for AgI and the standard reduction potential for \(\mathrm{Ag}^{+}\).)

What is electrochemistry? What are redox reactions? Explain the difference between a galvanic and an electrolytic cell.

Consider a concentration cell that has both electrodes made of some metal M. Solution A in one compartment of the cell contains 1.0\(M \mathrm{M}^{2+} .\) Solution \(\mathrm{B}\) in the other cell compartment has a volume of 1.00 L. At the beginning of the experiment 0.0100 mole of \(\mathrm{M}\left(\mathrm{NO}_{3}\right)_{2}\) and 0.0100 mole of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) are dissolved in solution \(\mathrm{B}\) (ignore volume changes), where the reaction $$\mathrm{M}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q) \rightleftharpoons \mathrm{MSO}_{4}(s)$$ occurs. For this reaction equilibrium is rapidly established, whereupon the cell potential is found to be 0.44 \(\mathrm{V}\) at \(25^{\circ} \mathrm{C} .\) Assume that the process $$\mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M}$$ has a standard reduction potential of \(-0.31 \mathrm{V}\) and that no other redox process occurs in the cell. Calculate the value of \(K_{\mathrm{sp}}\) for \(\mathrm{MSO}_{4}(s)\) at \(25^{\circ} \mathrm{C} .\)

Consider the following electrochemical cell: a. If silver metal is a product of the reaction, is the cell a galvanic cell or electrolytic cell? Label the cathode and anode, and describe the direction of the electron flow. b. If copper metal is a product of the reaction, is the cell a galvanic cell or electrolytic cell? Label the cathode and anode, and describe the direction of the electron flow. c. If the above cell is a galvanic cell, determine the standard cell potential. d. If the above cell is an electrolytic cell, determine the minimum external potential that must be applied to cause the reaction to occur.

Calculate \(\mathscr{E}^{\circ}\) values for the following cells. Which reactions are spontaneous as written (under standard conditions)? Balance the equations that are not already balanced. Standard reduction potentials are found in Table 18.1. a. \(\mathrm{H}_{2}(g) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{H}^{-}(a q)\) b. $\mathrm{Au}^{3+}(a q)+\mathrm{Ag}(s) \longrightarrow \mathrm{Ag}^{+}(a q)+\mathrm{Au}(s)$
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