Question

Given the following half-reactions and associated standard reduction potentials: $$ \begin{aligned} \mathrm{AuBr}_{4}^{-}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}(s)+4 \mathrm{Br}^{-}(a q) \\ \mathrm{Eu}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Eu}^{2+}(a q) & E_{\mathrm{red}}^{\circ}=-0.858 \mathrm{~V} \\ & E_{\mathrm{red}}^{\circ}=-0.43 \mathrm{~V} \\ \mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow \mathrm{I}^{-}(\mathrm{aq})+2 \mathrm{OH}^{-}(a q) \\ \mathrm{En}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Sn}(s) & E_{\mathrm{red}}^{\circ}=+0.49 \mathrm{~V} \\ E_{\mathrm{red}}^{\circ}=-0.14 \mathrm{~V} \end{aligned} $$ (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

The smallest positive cell EMF can be obtained by combining the half-reactions of AuBr4- and Eu2+, which gives the following balanced cell reaction: $$3AuBr_4^-(aq) + 9Eu^{2+}(aq) \rightarrow 3Au(s) + 12Br^-(aq) + 9Eu^{3+}(aq)$$ Calculating the corresponding cell EMF results in a value of -9.012 V.

Step-by-step solution

Identify the Half-Reactions with the Highest and Lowest Reduction Potentials

The combinations with the largest positive cell EMF will involve the half-reactions with the highest and lowest reduction potentials. From the given table, we can see that the highest reduction potential belongs to Sn2+(aq) + 2e- -> Sn(s) with E° = +0.49 V and the lowest reduction potential belongs to Eu3+(aq) + e- -> Eu2+(aq) with E° = -0.858 V.

Combining Half-Reactions for Largest Positive Cell EMF

For combining the half-reactions, we need to balance the electrons for both oxidation and reduction processes. As the oxidation half-reaction is the one with the lowest reduction potential, we will reverse the sign to obtain the oxidation potential. $$Eu^{2+}(aq) \rightarrow Eu^{3+}(aq) + e^-$$ The oxidation potential for this reaction will be +0.858 V. If we compare the reactions, we can notice that we already have equal numbers of electrons in both half reactions: $$Sn^{2+}(aq) + 2e^- \rightarrow Sn(s)$$ $$Eu^{2+}(aq) \rightarrow Eu^{3+}(aq) + e^-$$ Now, we combine these half-reactions: $$Sn^{2+}(aq) + Eu^{2+}(aq) \rightarrow Sn(s) + Eu^{3+}(aq)$$

Calculate the Largest Positive Cell EMF

We can now calculate the largest positive cell EMF by substracting the oxidation potential from the reduction potential: $$E_{cell}^\circ = E_{red}^\circ - E_{ox}^\circ = 0.49 - 0.858 = -0.368 V$$ However, we made a mistake by choosing the largest positive cell EMF. It was given we are looking for the smallest positive EMF. Therefore, we need to combine the half-reactions with the highest and second-lowest reduction potentials, which are: $$AuBr_4^-(aq) + 3e^- \rightarrow Au(s) + 4Br^-(aq)$$ $$Eu^{3+}(aq) + e^- \rightarrow Eu^{2+}(aq)$$

Combining Half-Reactions for the Smallest Positive Cell EMF

To combine these half-reactions, we need to balance the electrons in both processes. The oxidation half-reaction is still the one with the lowest reduction potential, and therefore, the sign will be reversed: $$Eu^{2+}(aq) \rightarrow Eu^{3+}(aq) + e^-$$ In order to balance the electrons, we should also multiply the Au half-reaction by 3: $$3[AuBr_4^-(aq) + 3e^- \rightarrow Au(s) + 4Br^-(aq)]$$ Now, the half-reactions look like this: $$3AuBr_4^-(aq) + 9e^- \rightarrow 3Au(s) + 12Br^-(aq)$$ $$Eu^{2+}(aq) \rightarrow Eu^{3+}(aq) + e^-$$ We need to multiply the second half-reaction by 9 to equal the electron quantity: $$9[Eu^{2+}(aq) \rightarrow Eu^{3+}(aq) + e^-]$$ These half-reactions can now be combined: $$3AuBr_4^-(aq) + 9Eu^{2+}(aq) \rightarrow 3Au(s) + 12Br^-(aq) + 9Eu^{3+}(aq)$$

Calculate the Smallest Positive Cell EMF

Lastly, we can calculate the smallest positive cell EMF by substracting the oxidation potential from the reduction potential. Since we multiply the second half-reaction by 9, we also need to multiply the corresponding E° values by 9 (0.858 * 9 = 7.722). $$E_{cell}^\circ = E_{red}^\circ - E_{ox}^\circ = 3*(-0.43) - 7.722 = -1.29 - 7.722 = -9.012 V$$ So, the smallest positive cell EMF is 9.012 V.

Key concepts

Standard Reduction Potentials

Standard reduction potentials are crucial in understanding how different substances can gain electrons during chemical reactions. These potentials are measured in volts ( V ) and tell us how easily a substance can be reduced, or how eager it is to accept electrons.
The standard reduction potential is determined under standard conditions: a concentration of 1 M for solutions, a pressure of 1 atm for gases, and a temperature of 298 K (25 °C).
These potentials are listed in tables for various half-reactions, helping us compare different substances. A higher potential means easier reduction, while a lower potential suggests it's harder for that substance to gain electrons.

• Comparing these values allows us to predict which half-reaction will occur as a reduction.
• It also helps predict which substance will undergo oxidation, by reversing its process.

Half-Reactions

In electrochemistry, a half-reaction represents one part of a redox reaction.
A redox reaction involves the transfer of electrons between substances—one gains electrons (reduction) and the other loses electrons (oxidation).
Every redox reaction can be split into two half-reactions: one that shows oxidation and another that shows reduction. A half-reaction includes all the electrons gained or lost in the process. It can be balanced to ensure that the number of electrons lost equals the number gained.

• Balancing electrons is key when combining half-reactions to form a full redox reaction.
• The sum of the half-reactions gives the overall chemical process.

Cell EMF

Cell Electromotive Force (EMF) is a measure of the energy provided by a cell or battery per coulomb of charge.
In the context of electrochemistry, the cell EMF indicates the voltage or potential difference between two electrodes in a galvanic cell when no current flows.
You calculate the standard cell EMF ( E_{cell}^ { °}) by taking the difference between the reduction potential of the cathode and the reduction potential of the anode.

• The anode is where oxidation occurs, and the cathode is where reduction takes place.
• For a cell to be spontaneous, the cell EMF must be positive.

The use of reversing the sign for oxidation potential is critical in calculations.

Oxidation and Reduction Processes

The oxidation and reduction processes are fundamental to understanding electrochemical reactions.
Oxidation occurs when a substance loses electrons, while reduction happens when a substance gains electrons. An easy way to remember this is with the acronym "OIL RIG" which stands for "Oxidation Is Loss, Reduction Is Gain".
These processes always occur together in redox reactions because the electrons lost by one substance are gained by another. Understanding the oxidation and reduction processes in half-reactions helps in forming the overall redox equation.

• In any given reaction, identify the species being oxidized and reduced to balance the chemical equation properly.
• Balancing involves ensuring the same number of electrons is transferred in both half-reactions.

This ensures the conservation of charge and mass in electrochemical reactions.