Question

Determine the standard potential for electrochemical cells in which each equation represents the overall cell reaction. Identify the reactions as spontaneous or nonspontaneous as written. $$ \begin{array}{l}{\text { a. } 2 \mathrm{Al}^{3+}(\mathrm{aq})+3 \mathrm{Cu}(\mathrm{s}) \rightarrow 3 \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{Al}(\mathrm{s})} \\ {\text { b. } \mathrm{Hg}^{2+}(\mathrm{aq})+2 \mathrm{Cu}^{+}(\mathrm{aq}) \rightarrow 2 \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Hg}(\mathrm{l})}\end{array} $$ $$ \mathrm{c} \cdot \mathrm{Cd}(\mathrm{s})+2 \mathrm{NO}_{3}-(\mathrm{aq})+4 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow \mathrm{Cd}^{2+}(\mathrm{aq})+2 \mathrm{NO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) $$

Short answer

For the given reactions: a) The standard cell potential is -1.32 V, and the reaction is nonspontaneous. b) The standard cell potential is +0.70 V, and the reaction is spontaneous. c) The standard cell potential is +0.56 V, and the reaction is spontaneous.

Step-by-step solution

For each cell reaction, we need to identify the half-reactions, which are the reduction and oxidation processes. We will determine the oxidizing (loses electrons) and the reducing agents (gains electrons). For reaction a: \( 2 Al^{3+}(aq) + 3 Cu(s) \rightarrow 3 Cu^{2+}(aq) + 2 Al(s) \) Oxidation half-reaction: \( 3 Cu(s) \rightarrow 3 Cu^{2+}(aq) + 6 e^{-} \) (Cu loses electrons, becoming Cu²⁺) Reduction half-reaction: \( 2 Al^{3+}(aq) + 6 e^{-} \rightarrow 2 Al(s) \) (Al³⁺ gains electrons, becoming Al) For reaction b: \( Hg^{2+}(aq) + 2 Cu^{+}(aq) \rightarrow 2 Cu^{2+}(aq) + Hg(l) \) Oxidation half-reaction: \( 2 Cu^{+}(aq) \rightarrow 2 Cu^{2+}(aq) + 2 e^{-} \) (Cu⁺ loses electrons, becoming Cu²⁺) Reduction half-reaction: \( Hg^{2+}(aq) + 2 e^{-} \rightarrow Hg(l) \) (Hg²⁺ gains electrons, becoming Hg) For reaction c: \( Cd(s) + 2 NO_{3}^{-}(aq) + 4 H^{+}(aq) \rightarrow Cd^{2+}(aq) + 2 NO_{2}(g) + 2 H_{2}O(l) \) Oxidation half-reaction: \( Cd(s) \rightarrow Cd^{2+}(aq) + 2 e^{-} \) (Cd loses electrons, becoming Cd²⁺) Reduction half-reaction: \( 2 NO_{3}^{-}(aq) + 4 H^{+}(aq) + 2 e^{-} \rightarrow 2 NO_{2}(g) + 2 H_{2}O(l) \) (NO₃⁻ gains electrons, becoming NO₂) ##Step 2: Determine the standard reduction potentials##

Use the standard reduction potential table available in textbooks or online to find the standard reduction potentials for the given half-reactions. Remember that the potentials are given for the reduction process, so if our half-reaction is oxidation, we need to change the sign of the potential. For reaction a: Standard reduction potential for Cu²⁺/Cu: +0.34 V Standard reduction potential for Al³⁺/Al: -1.66 V For reaction b: Standard reduction potential for Cu²⁺/Cu⁺: +0.15 V Standard reduction potential for Hg²⁺/Hg: +0.85 V For reaction c: Standard reduction potential for Cd²⁺/Cd: -0.40 V Standard reduction potential for NO₃⁻/NO₂ and H⁺/H₂O: +0.96 V ##Step 3: Calculate the standard cell potentials##

The overall standard cell potential is the sum of the standard reduction potentials of the half-reactions. For reaction a: E°(cell) = E°(reduction half-reaction) + E°(oxidation half-reaction) = -1.66 + 0.34 = -1.32 V For reaction b: E°(cell) = E°(reduction half-reaction) + E°(oxidation half-reaction) = 0.85 + (-0.15) = +0.70 V For reaction c: E°(cell) = E°(reduction half-reaction) + E°(oxidation half-reaction) = 0.96 + (-0.40) = +0.56 V ##Step 4: Determine spontaneity##

If the standard cell potential is positive, it is spontaneous. If it is negative, it is nonspontaneous for the given reaction. For reaction a: E°(cell) = -1.32 V, so the reaction is nonspontaneous. For reaction b: E°(cell) = +0.70 V, so the reaction is spontaneous. For reaction c: E°(cell) = +0.56 V, so the reaction is spontaneous.

Key concepts

Standard Reduction Potentials

Understanding standard reduction potentials is crucial for studying electrochemical cells. These potentials give us insights into how likely a substance is to gain electrons (or be reduced). They are measured in volts (V) and are typically referenced against the standard hydrogen electrode, which has a potential of 0.00 V.

Each substance in an electrochemical series has a potential listed for when it undergoes a reduction reaction. For instance, a higher (more positive) standard reduction potential indicates a greater tendency to gain electrons, making the substance a strong oxidizing agent.

When working with electrochemical cells, these standard potentials help predict the direction of electron flow. If you are given an oxidation half-reaction, remember to take the negative of the reduction potential to use in your calculations.

Half-Reactions

Electrochemical cells involve complex reactions, but you can simplify them using half-reactions. A half-reaction focuses on either oxidation (loss of electrons) or reduction (gain of electrons). By breaking the full cell reaction into these parts, you gain clarity on what happens to each species involved.

Consider reaction (a) from your exercise: the oxidation half-reaction shows copper ( Cu(s) ) losing electrons, while the reduction half-reaction shows aluminum ( Al^{3+}(aq) ) gaining electrons. This separation makes calculations, like determining the cell potential, easier.

Recognizing half-reactions also helps determine the oxidizing and reducing agents. The oxidizing agent gains electrons (gets reduced), while the reducing agent loses electrons (is oxidized). This principle can aid in predicting the spontaneity and the driving force behind a reaction in a cell.

Cell Spontaneity

Determining cell spontaneity is crucial when analyzing electrochemical reactions. It refers to whether a reaction will occur without continuous input of energy. To ascertain this, you can look at the standard cell potential ( E°(cell) ).

Spontaneous reactions have a positive standard cell potential ( E°(cell) > 0 ). This implies that the reaction can produce electrical energy as it proceeds in the given direction. Conversely, if the potential is negative, the reaction is nonspontaneous in that direction, requiring energy input to proceed.

In your example, reaction (b) is spontaneous with a positive 0.70 V, while reaction (a), with its -1.32 V, is nonspontaneous as written. By understanding this concept, you can predict and manipulate conditions to drive reactions in desired ways.

Oxidation and Reduction Processes

Oxidation and reduction processes are what power electrochemical cells. Oxidation refers to the loss of electrons, while reduction is the gain of electrons. During a redox reaction, these processes occur concurrently in different parts of the cell.

For example, in reaction (a), copper is oxidized as it loses electrons, changing from Cu to Cu²⁺. Simultaneously, aluminum ions ( Al^{3+} ) are reduced, gaining electrons to form Al metal. This duality of processes allows the movement of electrons, which is central to generating electrical energy.

Understanding which species is oxidized or reduced helps identify the direction of electron flow and can dictate how to connect components in an electrochemical cell for desired outcomes. Knowing this can be a game-changer in designing efficient electrochemical systems like batteries.