Question

Given the following half-reactions and associated standard reduction potentials: $$ \begin{aligned} \mathrm{AuBr}_{4}^{-}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}(s)+& 4 \mathrm{Br}^{-}(a q) \\ & E_{\text {red }}^{\circ}=-0.858 \mathrm{~V} \\ \mathrm{Eu}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Eu}^{2+}(a q) & \\\ E_{\text {red }}^{\circ}=-0.43 \mathrm{~V} \end{aligned} $$ $$ \begin{aligned} \mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow \mathrm{I}^{-}(\mathrm{aq}) &+2 \mathrm{OH}^{-}(a q) \\ & E_{\mathrm{red}}^{\circ}=+0.49 \mathrm{~V} \\ \mathrm{Sn}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \operatorname{Sn}(s) & \\\ E_{\mathrm{red}}^{\circ}=-0.14 \mathrm{~V} \end{aligned} $$ (a) Write the cell reaction for the combination of these half-cell reactions that leads to the largest positive cell emf, and calculate the value. (b) Write the cell reaction for the combination of half-cell reactions that leads to the smallest positive cell emf, and calculate that value.

Short answer

(a) The largest positive cell emf is obtained by combining the IO⁻ half-reaction as the cathode and AuBr₄⁻ half-reaction as the anode. The overall reaction is \(2\text{AuBr}_{4}^{-}(aq)+6\text{IO}^{-}(aq) + 3\text{H}_{2}\text{O}(l) \rightarrow 2\text{Au}(s) + 8\text{Br}^{-}(aq) + 6\text{I}^{-}(aq) + 6\text{OH}^{-}(aq)\) and the cell emf is \(1.348\text{ V}\). (b) The smallest positive cell emf is obtained by combining the Sn²⁺ half-reaction as the cathode and Eu³⁺ half-reaction as the anode. The overall reaction is \(\text{Sn}^{2+}(aq) + 2\text{Eu}^{3+}(aq) \rightarrow \text{Sn}(s) + 2\text{Eu}^{2+}(aq)\) and the cell emf is \(0.29\text{ V}\).

Step-by-step solution

Identifying the Reduction/Oxidation Half-Reactions

First, we must identify which of the given half-reactions are reduction reactions and which are oxidation reactions. Remember that reduction refers to the gain of electrons, while oxidation refers to the loss of electrons. The reduction half-reactions are: $$\begin{aligned} \text { AuBr }_{4}^{-}(a q)+3 \text { e }^{-} \longrightarrow \text { Au }(s)+ 4 \text { Br }^{-}(a q) \quad E_{\text { red }}^{\circ}=-0.858 \mathrm{~ V} \\ \text { Eu }^{3+}(a q)+\text { e }^{-} \longrightarrow \text { Eu }^{2+}(a q) \quad E_{\text { red }}^{\circ}=-0.43 \mathrm{~ V } \\ \text { IO }^{-}(a q)+\text { H }_{2} \mathrm{ O }(l)+2 \mathrm{ e }^{-} \longrightarrow \text { I }^{-}(\text { aq })+2 \text { OH }^{-}(a q) \quad E_{\mathrm{red}}^{\circ}=+0.49 \mathrm{~ V } \\ \text { Sn }^{2+}(a q)+2 \mathrm{ e }^{-} \longrightarrow \mathrm{ Sn }(s) \quad E_{\mathrm{ red }}^{\circ}=-0.14 \mathrm{~ V } \\ \end{aligned}$$

Calculate the Net Cell Potential for Each Combination

Now that we have identified the reduction/oxidation half-reactions, we must combine these half-reactions in pairs and apply the Nernst equation to calculate the net cell potential for each combination: $$E_{cell} = E_{cathode} - E_{anode}$$ We want the largest positive cell emf, so we will choose the highest reduction potential for the cathode and the lowest reduction potential for the anode.

(a) Largest Positive Cell EMF

The largest positive cell emf can be obtained by combining the half-reaction with the highest reduction potential (IO⁻) as the cathode and the half-reaction with the lowest reduction potential (AuBr₄⁻) as the anode. Cathode: IO⁻(aq) + H₂O(l) + 2e⁻ → I⁻(aq) + 2OH⁻(aq); E_red = +0.49 V Anode: AuBr₄⁻(aq) + 3e⁻ → Au(s) + 4Br⁻(aq); E_red = -0.858 V To balance the number of electrons, we need to multiply the cathode reaction by 3 and the anode reaction by 2. Cathode: 3(IO⁻(aq) + H₂O(l) + 2e⁻ → I⁻(aq) + 2OH⁻(aq)); E_red = +0.49 V Anode: 2(AuBr₄⁻(aq) + 3e⁻ → Au(s) + 4Br⁻(aq)); E_red = -0.858 V Now we can add the half-reactions and calculate the net cell potential. Overall reaction: 2AuBr₄⁻(aq) + 6IO⁻(aq) + 3H₂O(l) → 2Au(s) + 8Br⁻(aq) + 6I⁻(aq) + 6OH⁻(aq) E_cell = E_cathode - E_anode E_cell = (+0.49 V) - (-0.858 V) = 1.348 V Answer: The largest positive cell emf is 1.348 V.

(b) Smallest Positive Cell EMF

The smallest positive cell emf can be obtained by combining the half-reaction with the lowest positive reduction potential (Sn²⁺) as the cathode and the half-reaction with the highest negative reduction potential (Eu³⁺) as the anode. Cathode: Sn²⁺(aq) + 2e⁻ → Sn(s); E_red = -0.14 V Anode: Eu³⁺(aq) + e⁻ → Eu²⁺(aq); E_red = -0.43 V To balance the electron number, we need to multiply the cathode reaction by 1 and the anode reaction by 2. Cathode: 1(Sn²⁺(aq) + 2e⁻ → Sn(s)); E_red = -0.14 V Anode: 2(Eu³⁺(aq) + e⁻ → Eu²⁺(aq)); E_red = -0.43 V Now we can add the half-reactions and calculate the net cell potential. Overall reaction: Sn²⁺(aq) + 2Eu³⁺(aq) → Sn(s) + 2Eu²⁺(aq) E_cell = E_cathode - E_anode E_cell = (-0.14 V) - (-0.43 V) = 0.29 V Answer: The smallest positive cell emf is 0.29 V.

Key concepts

Standard Reduction Potentials

Standard Reduction Potentials are a measure of the tendency of a chemical species to acquire electrons and be reduced. This is often denoted as \( E_{\text{red}}^{\circ} \). These potentials are measured in volts (V) under standard conditions, which include a 1 M concentration for aqueous solutions, 1 atm pressure for gases, and pure substances in their standard state.

Reduction potentials help determine which substances are better at oxidizing or reducing other substances. A more positive \( E_{\text{red}}^{\circ} \) value indicates a greater tendency to gain electrons, while a more negative value suggests a lesser tendency to do so. In practice:

• If \( E_{\text{red}}^{\circ} \) is positive, it means the species is more likely to be reduced.
• If \( E_{\text{red}}^{\circ} \) is negative, the species is less likely to be reduced.

Electrochemical series, a table of these potential values, allows chemists to predict how substances will react in redox reactions. This knowledge is crucial when constructing electrochemical cells.

Cell EMF Calculation

Calculating the electromotive force (EMF) of a cell involves using the standard reduction potentials of the half-reactions. The EMF, also known as the cell potential, is given by:
\[ E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} \]

• \( E_{\text{cathode}} \) is the reduction potential of the cathode.
• \( E_{\text{anode}} \) is the reduction potential of the anode.

The cathode is where reduction occurs, and the anode is where oxidation happens. To find the cell potential, subtract the anode potential from the cathode potential.

Cell EMF tells us how much voltage a cell can produce. A positive EMF indicates a spontaneous reaction, while a negative EMF suggests a non-spontaneous one. It's essential to make sure your equations are balanced so that the electrons lost and gained are equal, ensuring accurate potential calculation.

For example, if a half-reaction at the cathode has a potential of +0.49 V and at the anode -0.858 V, the EMF is:
\(E_{\text{cell}} = (+0.49 \, \text{V}) - (-0.858 \, \text{V}) = 1.348 \, \text{V}\)This positive value shows a feasible and spontaneous reaction.

Redox Reactions

Redox reactions are chemical reactions where oxidation and reduction occur simultaneously. In these processes, one species loses electrons (oxidized) while another gains electrons (reduced). Understanding these reactions involves recognizing the transfer of electrons between species.

Key points include:

• Oxidation involves the loss of electrons, and the species is said to be oxidized.
• Reduction involves the gain of electrons, and the species is reduced.
• The substance that gives away electrons is called the reducing agent, while the one that accepts electrons is the oxidizing agent.

A complete redox reaction can be broken down into two half-reactions: the oxidation half-reaction and the reduction half-reaction. These help in balancing the overall equation by ensuring that the number of electrons lost equals the number of electrons gained.

Balancing redox reactions involves making sure both mass and charge are conserved, which might require multiplying half-reactions by appropriate integers to equalize the electrons transferred in each half. It is this careful accounting of electrons that allows chemists to predict the feasibility and spontaneity of electrochemical processes, shaping applications in batteries, corrosion prevention, and electroplating.