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Chapter 18: Problem 54

An electrochemical cell consists of a standard hydrogen electrode and a second half-cell in which a cadmium electrode is immersed in a \(1.00 M\) solution of \(\mathrm{Cd}^{2+}\) ions. a. What is the value of \(E_{\text {cell }} ?\) b. Which electrode is the anode? c. Which is a product of the cell reaction: \(\mathrm{Cd}^{2+}\) ions or \(\mathrm{Cd}\) metal?

Short Answer

Expert verified
a. The cell potential is -0.403 V. b. The anode is the standard hydrogen electrode. c. The product of the cell reaction is Cd虏鈦� ions.

Step by step solution

01

Identify the half-reactions

First, we need to find the half-reactions for both the standard hydrogen electrode and the cadmium electrode. For the standard hydrogen electrode, the half-reaction is: $$ \mathrm{2H}^{+}(aq)+2e^{-}\rightarrow\mathrm{H}_{2}(g) $$ For the cadmium electrode, the half-reaction is: $$ \mathrm{Cd}^{2+}(aq)+2e^{-}\rightarrow\mathrm{Cd}(s) $$
02

Determine standard reduction potentials

Next, we need to find the standard reduction potentials for both half-reactions. For the standard hydrogen electrode, the reduction potential is defined as \(0 V\). For the cadmium half-cell, the standard reduction potential is given as \(-0.403 V\).
03

Calculate the cell potential

Now we can calculate the cell potential using the Nernst equation, which states that \(E_{\text{cell}}=E_{\text{cathode}}-E_{\text{anode}}\). Since we want the cell potential to be positive and spontaneous, we'll have the hydrogen half-cell as the anode and the cadmium half-cell as the cathode. So, $$ E_{\text{cell}}=(-0.403~V)-0~V=-0.403~V $$
04

Identify the anode

Since we determined that the hydrogen half-cell acts as the anode in this case (as it has a lower reduction potential than the cadmium half-cell), we can say that the anode is the standard hydrogen electrode.
05

Determine the product of the cell reaction

From the half-reactions, the overall cell reaction can be written as: $$ \mathrm{2H}^{+}(aq)+\mathrm{Cd}(s)\rightarrow\mathrm{H}_{2}(g)+\mathrm{Cd}^{2+}(aq) $$ Thus, the product of the cell reaction is \(\mathrm{Cd}^{2+}\) ions. a. The value of \(E_{\text {cell}}\) is \(-0.403~V\). b. The anode is the standard hydrogen electrode. c. The product of the cell reaction is \(\mathrm{Cd}^{2+}\) ions.

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Key Concepts

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

Standard Hydrogen Electrode
The standard hydrogen electrode (SHE) is a central reference point in electrochemistry. Imagine it as the 'zero' on a scale from which we measure all other electrode potentials. In practice, it consists of a platinum electrode surrounded by an acid solution, with hydrogen gas bubbling around it at a pressure of one atmosphere.

The key thing about SHE is that its standard reduction potential is set at exactly 0 volts. This doesn't mean that it doesn't participate in reactions; rather, it acts as a baseline. Think of it like sea level on a map - it's the point from which all elevations are measured. By comparing other half-reactions against the SHE, we can determine which are more or less likely to gain electrons (reduction) or lose electrons (oxidation).

Understanding SHE is critical not just for calculating cell potentials but for getting a grasp on the mechanics of redox reactions, setting the stage for more complex concepts in electrochemistry.
Cell Potential Calculation
Cell potential, often symbolized as Ecell, is the measure of the potential difference between two half-cells in an electrochemical cell. It's essential for predicting whether a reaction will be spontaneous. To put it simply, it's like the 'push' that drives the electron flow in a battery.

To calculate Ecell, we look at the standard reduction potentials of the two half-reactions. The equation is straightforward: Ecell = Ecathode - Eanode. The reduction potential of the cathode is subtracted from the anode鈥檚. We ensure positive cell potential for a spontaneous reaction, which is why it's crucial to assign the electrodes correctly. If your Ecell turns out to be negative, it means electrons naturally want to travel in the opposite direction, and the reaction won't occur without external energy.

Always remember, a positive cell potential signifies a reaction that can happen on its own - kind of like a ball rolling downhill.
Nernst Equation
The Nernst equation is an essential formula in electrochemistry that allows us to calculate the actual cell potential at any given conditions, not just the standard ones. It accounts for changes like temperature, pressure, and concentration.

At its core, the Nernst equation fine-tunes our cell potential calculation to reflect real-world conditions. The equation is E = Eo - (RT/nF)ln(Q), where E is the cell potential under non-standard conditions, Eo is the standard cell potential, R is the universal gas constant, T is the temperature in Kelvin, n is the number of moles of electrons transferred, F is Faraday's constant, and Q is the reaction quotient.

Applying the Nernst equation to the exercise, if for example, anything about our setup was non-standard - such as temperature or concentrations different from 1 M - we would need this equation to recalculate the cell potential, ensuring precision in the real world. It's like adjusting your recipe depending on the size of your guest list - the core principles are the same, but the details matter.

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Most popular questions from this chapter

Chlorine dioxide \(\left(\mathrm{ClO}_{2}\right)\) is produced by the following reaction of chlorate \(\left(\mathrm{ClO}_{3}^{-}\right)\) with \(\mathrm{Cl}^{-}\) in acid solution: \(\begin{aligned} 2 \mathrm{ClO}_{3}^{-}(a q)+2 \mathrm{Cl}^{-}(a q)+4 \mathrm{H}^{+}(a q) \rightarrow & \\ 2 \mathrm{ClO}_{2}(g)+\mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(\ell) \end{aligned}\) a. Determine \(E^{\circ}\) for the reaction. b. The reaction produces a mixture of gases in the reaction vessel in which \(P_{\mathrm{C} 10_{2}}=2.0 \mathrm{atm} ; P_{\mathrm{C}_{2}}=1.00 \mathrm{atm}\) Calculate \(\left[\mathrm{ClO}_{3}^{-}\right]\) if, at equilibrium \(\left(T=25^{\circ} \mathrm{C}\right),\left[\mathrm{H}^{+}\right]=\) \(\left[\mathrm{Cl}^{-}\right]=10.0 \mathrm{M}\)

Calculate the \(E_{\text {cell value at } 298} \mathrm{K}\) for the cell based on the reaction $$ mathrm{Cu}(s)+2 \mathrm{Ag}^{+}(a q) \rightarrow \mathrm{Cu}^{2+}(a q)+2 \mathrm{Ag}(s)$$ when \(\left[\mathrm{Ag}^{+}\right]=2.56 \times 10^{-3} \mathrm{Mand}\left[\mathrm{Cu}^{2+}\right]=8.25 \times 10^{-4} \mathrm{M}\)

An element that is a good reducing agent is also ______ a. easily oxidized b. a good oxidizing agent c. easily reduced d. a noble gas

Super Iron Batteries In \(1999,\) scientists in Israel developed a battery based on the following cell reaction with iron(VI), nicknamed "super iron": \(\beth \mathrm{K}_{2} \mathrm{FeO}_{4}(a q)+3 \mathrm{Zn}(s) \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{ZnO}(s)+2 \mathrm{K}_{2} \mathrm{ZnO}_{2}(a q)\) a. Determine the number of electrons transferred in the cell reaction. b. What are the oxidation states of the transition metals in the reaction? c. Draw the cell diagram.

Electrolysis of Seawater Magnesium metal is obtained by the electrolysis of molten \(\mathrm{Mg}^{2+}\) salts from evaporated seawater. a. Would elemental Mg form at the cathode or anode? b. Do you think the principal ingredient in sea salt (NaCl) would need to be separated from the \(\mathrm{Mg}^{2+}\) salts before electrolysis? Explain your answer. c. Would electrolysis of an aqueous solution of \(\mathrm{Mg} \mathrm{Cl}_{2}\) also produce elemental Mg? d. If your answer to part (c) was no, what would be the products of electrolysis?
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