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Chapter 20: Problem 99

Predict whether the following reactions will be spontaneous in acidic solution under standard conditions: (a) oxidation of \(\mathrm{Cu}\) to \(\mathrm{Cu}^{2+}\) by \(\mathrm{I}_{2}\) (to form \(\mathrm{I}^{-}\) ), \((\mathbf{b})\) reduction of \(\mathrm{Fe}^{2+}\) to \(\mathrm{Fe}\) by \(\mathrm{H}_{2}\) (to form \(\mathrm{H}^{+}\) ), \(\left(\mathbf{c}\right.\) ) reduction of \(\mathrm{I}_{2}\) to \(\mathrm{I}^{-}\) by $\mathrm{H}_{2} \mathrm{O}_{2},(\mathbf{d})\( reduction of \)\mathrm{Ni}^{2+}\( to \)\mathrm{Ni}$ by \(\mathrm{Sn}^{2+}\left(\right.\) to form \(\left.\mathrm{Sn}^{4+}\right)\).

Short Answer

Expert verified
Among the four reactions, only the oxidation of Cu to Cu²⁺ by I₂ to form I⁻ (a) is spontaneous in acidic solution under standard conditions. Eº_cell for this reaction is positive (0.20 V). The other reactions (b, c, and d) have negative Eº_cell values, indicating that they are not spontaneous under the given conditions.

Step by step solution

01

Recall the standard potentials for the reactions under consideration

: To predict whether a given reaction will be spontaneous in acidic solution under standard conditions, we can refer to standard reduction potential tables. (a) Cu to Cu²⁺ and I₂ to I⁻. Eº(Cu²⁺/Cu) = +0.34 V Eº(I₂/I⁻) = +0.54 V (b) Fe²⁺ to Fe and H₂ to H⁺. Eº(Fe²⁺/Fe) = -0.44 V Eº(H₂/H⁺) = 0 V (since hydrogen is the reference) (c) I₂ to I⁻ and H₂O₂ to H₂O. Eº(I₂/I⁻) = +0.54 V Eº(H₂O₂/H₂O) = +1.78 V (d) Ni²⁺ to Ni and Sn²⁺ to Sn⁴⁺. Eº(Ni²⁺/Ni)= -0.26 V Eº(Sn⁴⁺/Sn²⁺) = +0.15 V
02

Calculate the overall cell potential for each reaction

: (a) Eº_cell = Eº(I₂/I⁻) - Eº(Cu²⁺/Cu) = 0.54 V - 0.34 V = 0.20 V Since Eº_cell is positive, the reaction is spontaneous. (b) Eº_cell = Eº(Fe²⁺/Fe) - Eº(H₂/H⁺) = -0.44 V - 0 V = -0.44 V Since Eº_cell is negative, the reaction is not spontaneous. (c) Eº_cell = Eº(I₂/I⁻) - Eº(H₂O₂/H₂O) = 0.54 V - 1.78 V = -1.24 V Since Eº_cell is negative, the reaction is not spontaneous. (d) Eº_cell = Eº(Ni²⁺/Ni) - Eº(Sn⁴⁺/Sn²⁺) = -0.26 V - 0.15 V = -0.41 V Since Eº_cell is negative, the reaction is not spontaneous. In conclusion, among the four reactions, only the first reaction (oxidation of Cu to Cu²⁺ by I₂ to form I⁻) is spontaneous in acidic solution under standard conditions.

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Most popular questions from this chapter

Consider the following table of standard electrode potentials for a series of hypothetical reactions in aqueous solution: $$ \begin{array}{lr} \hline \text { Reduction Half-Reaction } & {E^{\circ}(\mathrm{V})} \\ \hline \mathrm{A}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{A}(s) & 1.33 \\\ \mathrm{~B}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{B}(s) & 0.87 \\\ \mathrm{C}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{C}^{2+}(a q) & -0.12 \\ \mathrm{D}^{3+}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{D}(s) & -1.59 \\\ \hline \end{array} $$ (a) Which substance is the strongest oxidizing agent? Which is weakest? (b) Which substance is the strongest reducing agent? Which is weakest? (c) Which substance(s) can oxidize \(\mathrm{C}^{2+} ?\)

Given the following half-reactions and associated standard reduction potentials: $$ \begin{aligned} \mathrm{AuBr}_{4}^{-}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}(s)+4 \mathrm{Br}^{-}(a q) & \\ E_{\mathrm{red}}^{\circ}=-0.86 \mathrm{~V} \\ \mathrm{Eu}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Eu}^{2+}(a q) & \\\ E_{\mathrm{red}}^{\circ}=-0.43 \mathrm{~V} \\ \mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow \mathrm{I}^{-}(a q)+2 \mathrm{OH}^{-}(a q) & \\ E_{\mathrm{red}}^{\circ}=&+0.49 \mathrm{~V} \end{aligned} $$ (a) Write the equation for the combination of these halfcell reactions that leads to the largest positive emf and calculate the value. (b) Write the equation for the combination of half-cell reactions that leads to the smallest positive emf and calculate that value.

A voltaic cell utilizes the following reaction: $4 \mathrm{Fe}^{2+}(a q)+\mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q) \longrightarrow 4 \mathrm{Fe}^{3+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)$ (a) What is the emf of this cell under standard conditions? (b) What is the emf of this cell when $\left[\mathrm{Fe}^{2+}\right]=1.3 \mathrm{M},\left[\mathrm{Fe}^{3+}\right]=\( \)0.010 \mathrm{M}, P_{\mathrm{O}_{2}}=50.7 \mathrm{kPa},\( and the \)\mathrm{pH}$ of the solution in the cathode half-cell is \(3.50 ?\)

A student designs an ammeter (device that measures electrical current) that is based on the electrolysis of water into hydrogen and oxygen gases. When electrical current of unknown magnitude is run through the device for 90 min, \(32.5 \mathrm{~mL}\) of water-saturated \(\mathrm{H}_{2}(g)\) is collected. The temperature of the system is \(20^{\circ} \mathrm{C},\) and the atmospheric pressure is \(101.3 \mathrm{kPa}\). What is the magnitude of the average current in amperes?

A voltaic cell is constructed with two \(\mathrm{Cu}^{2+}-\mathrm{Cu}\) electrodes. The two half-cells have $\left[\mathrm{Cu}^{2+}\right]=0.100 \mathrm{M}\( and \)\left[\mathrm{Cu}^{2+}\right]=1.00 \times 10^{-4} \mathrm{M}$, respectively. (a) Which electrode is the cathode of the cell? (b) What is the standard emf of the cell? (c) What is the cell emf for the concentrations given? (d) For each electrode, predict whether \(\left[\mathrm{Cu}^{2+}\right]\) will increase, decrease, or stay the same as the cell operates.
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