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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Reduction Potential
Standard reduction potentials ( E° ) are essential in determining the likelihood of a chemical reaction occurring under standard conditions (25°C, 1 atm, and 1 M concentrations). A standard reduction potential is measured in volts and indicates a substance's ability to gain electrons, essentially how eagerly it wants to be reduced.

In electrochemistry, every substance has a specific reduction potential, which can be found in standard reduction potential tables. These tables list various chemical reactions and their respective E° values.

Here is how you can utilize them:
  • A higher E° value means a greater tendency to gain electrons and be reduced.
  • If comparing two half-reactions, the one with the higher E° value will act as the cathode (reduction site) in a galvanic cell.
By using standard reduction potentials, you can calculate the overall cell potential ( E°_cell ) of a redox reaction, which determines if the reaction is spontaneous, as positive overall cell potentials indicate spontaneous reactions.
Spontaneous Reaction
A spontaneous reaction is one that occurs without needing additional energy once started. In electrochemistry, whether a reaction is spontaneous is determined using standard reduction potentials.

When you calculate the overall cell potential ( E°_cell ) of a reaction, look for:
  • A positive E°_cell indicates a spontaneous reaction.
  • A negative E°_cell means the reaction is non-spontaneous, and an additional energy input is needed to start it.
To find E°_cell , subtract the E° of the oxidation reaction from the E° of the reduction reaction. Redox reactions are the heart of electrochemical cells, and understanding whether they are spontaneous is crucial for predicting the behavior of these cells in real-world applications.
Oxidation
Oxidation is a chemical process where a substance loses electrons. In redox reactions, it occurs alongside reduction, which is the gain of electrons by another substance.

Remember:
  • The substance that loses electrons is oxidized and termed the reducing agent or reductant.
  • Every oxidation must be paired with a reduction, as electrons cannot exist free in solutions.
In an oxidation process, the oxidized substance's oxidation state increases.

For example, in the reaction highlighted in the solution, copper ( Cu ) is oxidized to copper ions ( Cu^{2+} ). This is essential to forming a complete redox reaction, where Iodine goes to I⁻ , paired with the reduction half to complete the loop of electron transfer. Understanding these electron transfers helps predict the pathway and feasibility of chemical processes.

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Most popular questions from this chapter

Indicate whether each of the following statements is true or false: (a) If something is oxidized, it is formally losing electrons. (b) For the reaction \(\mathrm{Fe}^{3+}(a q)+\mathrm{Co}^{2+}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+\) \(\mathrm{Co}^{3+}(a q), \mathrm{Fe}^{3+}(a q)\) is the reducing agent and \(\mathrm{Co}^{2+}(a q)\) is the oxidizing agent. (c) If there are no changes in the oxidation state of the reactants or products of a particular reaction, that reaction is not a redox reaction.

If the equilibrium constant for a two-electron redox reaction at \(298 \mathrm{~K}\) is \(2.2 \times 10^{5},\) calculate the corresponding \(\Delta G^{\circ}\) and \(E^{\circ}\).

Complete and balance the following half-reactions. In each case, indicate whether the half-reaction is an oxidation or a reduction. (a) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)\) (acidic solution) (b) \(\mathrm{Mn}^{2+}(a q) \longrightarrow \mathrm{MnO}_{4}^{-}(a q)\) (acidic solution) (c) \(\mathrm{I}_{2}(s) \longrightarrow \mathrm{IO}_{3}^{-}(a q)\) (acidic solution) (d) \(\mathrm{S}(s)(a q) \longrightarrow \mathrm{H}_{2} \mathrm{~S}(g)\) (acidic solution) (e) \(\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}^{-}(a q)\) (basic solution) (f) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{OH}^{-}(a q)\) (basic solution)

The capacity of batteries such as a lithium-ion battery is expressed in units of milliamp-hours (mAh). A typical battery of this type yields a nominal capacity of \(2000 \mathrm{mAh}\). (a) What quantity of interest to the consumer is being expressed by the units of \(\mathrm{mAh}\) ? (b) The starting voltage of a fresh lithium-ion battery is \(3.60 \mathrm{~V}\). The voltage decreases during discharge and is \(3.20 \mathrm{~V}\) when the battery has delivered its rated capacity. If we assume that the voltage declines linearly as current is withdrawn, estimate the total maximum electrical work the battery could perform during discharge.

Cytochrome, a complicated molecule that we will represent as \(\mathrm{CyFe}^{2+}\), reacts with the air we breathe to supply energy required to synthesize adenosine triphosphate (ATP). The body uses ATP as an energy source to drive other reactions (Section 19.7). At \(\mathrm{pH} 7.0\) the following reduction potentials pertain to this oxidation of \(\mathrm{CyFe}^{2+}\) $$ \begin{aligned} \mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)+4 \mathrm{e}^{-} & \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) & & E_{\mathrm{red}}^{\circ}=+0.82 \mathrm{~V} \\\ \mathrm{CyFe}^{3+}(a q)+\mathrm{e}^{-} & \longrightarrow \mathrm{CyFe}^{2+}(a q) & E_{\mathrm{red}}^{\circ} &=+0.22 \mathrm{~V} \end{aligned} $$ (a) What is \(\Delta G\) for the oxidation of \(\mathrm{CyFe}^{2+}\) by air? \((\mathbf{b})\) If the synthesis of \(1.00 \mathrm{~mol}\) of ATP from adenosine diphosphate (ADP) requires a \(\Delta G\) of \(37.7 \mathrm{~kJ},\) how many moles of ATP are synthesized per mole of \(\mathrm{O}_{2} ?\)

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