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Chapter 18: Problem 19

Calculate \(E^{\circ}\) for the following voltaic cells: (a) \(\mathrm{MnO}_{2}(s)+4 \mathrm{H}^{+}(a q)+2 \mathrm{I}^{-}(a q) \longrightarrow\) \(\mathrm{Mn}^{2+}(a q)+2 \mathrm{H}_{2} \mathrm{O}+\mathrm{I}_{2}(s)\) (b) \(\mathrm{H}_{2}(g)+2 \mathrm{OH}^{-}(a q)+\mathrm{S}(s) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{S}^{2-}(a q)\) (c) an \(\mathrm{Ag}-\mathrm{Ag}^{+}\) half-cell and an \(\mathrm{Au}-\mathrm{AuCl}_{4}^{-}\) half-cell.

Short Answer

Expert verified
Question: Calculate the standard cell potentials for the following voltaic cells: (a) MnO₂(s) + 4H⁺(aq) + 2I⁻(aq) → Mn²⁺(aq) + 2H₂O(l) + I₂(s) (b) S(s) + 2H₂(g) → S²⁻(aq) + 4H⁺(aq) (c) half-cell 1: Ag⁺ → Ag half-cell 2: Au³⁺ + 4Cl⁻ → AuCl₄⁻ Answer: The standard cell potentials for the given voltaic cells are as follows: (a) \(E^{\circ}_a = 0.69\,\text{V}\) (b) \(E^{\circ}_b = -0.14\,\text{V}\) (c) \(E^{\circ}_c = 0.13\,\text{V}\)

Step by step solution

01

Identify the half-reactions for (a) and (b)

First, we need to split the given reactions (a) and (b) into their respective half-reactions. For (a), we can see that MnO2 is reduced, and I⁻ is oxidized. For (b), H₂ is oxidized and S is reduced. (a) Reduction half-reaction: MnO₂(s) + 4H⁺(aq) + 2e⁻ → Mn²⁺(aq) + 2H₂O(l) Oxidation half-reaction: 2I⁻(aq) → I₂(s) + 2e⁻ (b) Reduction half-reaction: S(s) + 2e⁻ → S²⁻(aq) Oxidation half-reaction: H₂(g) → 2H⁺(aq) + 2e⁻ Now that we have the half-reactions for (a) and (b), we will look up their standard reduction potentials.
02

Determine the standard reduction potentials

Using a standard reduction potential table, we can find the following values for the half-reactions: For (a) Reduction: \(E_{red,a}^\circ(MnO_2) = 1.23\,\text{V}\) For (a) Oxidation: \(E_{red,a}^\circ(I^-) = 0.54\,\text{V}\) For (b) Reduction: \(E_{red,b}^\circ(S) = -0.14\,\text{V}\) For (b) Oxidation: \(E_{red,b}^\circ(H_2) = 0\,\text{V}\)
03

Calculate the standard cell potentials

To calculate the standard cell potentials, we can now use the following formula: \(E^{\circ} = E_{red}^\circ(\text{reduction}) - E_{red}^\circ(\text{oxidation})\) For (a): \(E^{\circ}_a = 1.23\,\text{V} - 0.54\,\text{V} = 0.69\,\text{V}\) For (b): \(E^{\circ}_b = -0.14\,\text{V} - 0\,\text{V} = -0.14\,\text{V}\)
04

Determine the standard reduction potentials for (c)

For (c), we have given half-cells, so we can directly look up their standard reduction potentials: \(E_{red,c}^\circ(Ag^+) = 0.80\,\text{V}\) \(E_{red,c}^\circ(AuCl_4^-) = 0.93\,\text{V}\)
05

Calculate the standard cell potential for (c)

Following the same formula as in Step 3, we can calculate the standard cell potential for (c): \(E^{\circ}_c = 0.93\,\text{V} - 0.80\,\text{V} = 0.13\,\text{V}\) Now we have calculated the standard cell potentials for all three voltaic cells: (a) \(E^{\circ}_a = 0.69\,\text{V}\) (b) \(E^{\circ}_b = -0.14\,\text{V}\) (c) \(E^{\circ}_c = 0.13\,\text{V}\)

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Most popular questions from this chapter

Which of the following reactions is (are) spontaneous at standard conditions? (a) \(2 \mathrm{NO}_{3}^{-}(a q)+8 \mathrm{H}^{+}(a q)+6 \mathrm{Cl}^{-}(a q) \longrightarrow\) \(2 \mathrm{NO}(g)+4 \mathrm{H}_{2} \mathrm{O}+3 \mathrm{Cl}_{2}(g)\) (b) \(\mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)+4 \mathrm{Cl}^{-}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{Cl}_{2}(g)\) (c) \(3 \mathrm{Fe}(s)+2 \mathrm{AuCl}_{4}^{-}(a q) \longrightarrow 2 \mathrm{Au}(s)+8 \mathrm{Cl}^{-}(a q)+3 \mathrm{Fe}^{2+}(a q)\)

For the cell $$ \mathrm{Zn}\left|\mathrm{Zn}^{2+}\right| \mathrm{Cu}^{2+} \mid \mathrm{Cu} $$ \(E^{\circ}\) is \(1.10 \mathrm{~V}\). A student prepared the same cell in the lab at standard conditions. Her experimental \(E^{\circ}\) was \(1.0 \mathrm{~V}\). A possible explanation for the difference is that (a) a larger volume of \(\mathrm{Zn}^{2+}\) than \(\mathrm{Cu}^{2+}\) was used. (b) the zinc electrode had twice the mass of the copper electrode. (c) \(\left[\mathrm{Zn}^{2+}\right]\) was smaller than \(1 M\). (d) \(\left[\mathrm{Cu}^{2+}\right]\) was smaller than \(1 M\). (e) the copper electrode had twice the surface area of the zinc electrode.

A hydrogen-oxygen fuel cell operates on the reaction: $$ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) $$ If the cell is designed to produce \(1.5\) amp of current and if the hydrogen is contained in a 1.0-L tank at 200 atm pressure and \(25^{\circ} \mathrm{C}\), how long can the fuel cell operate before the hydrogen runs out? Assume that oxygen gas is in excess.

A solution containing a metal ion \(\left(\mathrm{M}^{3+}(a q)\right)\) is electrolyzed by a current of \(5.0 \mathrm{~A}\). After \(10.0\) minutes, \(1.19 \mathrm{~g}\) of the metal is plated out. (a) How many coulombs are supplied by the battery? (b) What is the metal? (Assume \(100 \%\) efficiency.)

Consider a voltaic cell at \(25^{\circ} \mathrm{C}\) in which the following reaction takes place. $$ 3 \mathrm{H}_{2} \mathrm{O}_{2}(a q)+6 \mathrm{H}^{+}(a q)+2 \mathrm{Au}(s) \longrightarrow 2 \mathrm{Au}^{3+}(a q)+6 \mathrm{H}_{2} \mathrm{O} $$ (a) Calculate \(E\). (b) Write the Nernst equation for the cell. (c) Calculate \(E\) when \(\left[\mathrm{Au}^{3+}\right]=0.250 \mathrm{M},\left[\mathrm{H}^{+}\right]=1.25 \mathrm{M},\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]=\) \(1.50 M\)
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