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Chapter 20: Problem 42

A voltaic cell consists of a strip of cadmium metal in a solution of \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\) in one beaker, and in the other beaker a platinum electrode is immersed in a NaCl solution, with \(\mathrm{Cl}_{2}\) gas bubbled around the electrode. A salt bridge connects the two beakers. (a) Which electrode serves as the anode, and which as the cathode? (b) Does the Cd electrode gain or lose mass as the cell reaction proceeds? (c) Write the equation for the overall cell reaction. (d) What is the emf generated by the cell under standard conditions?

Short Answer

Expert verified
Anode: Cd electrode. Cd loses mass. Overall reaction: Cd + Cl鈧� 鈫� Cd虏鈦� + 2Cl鈦�. EMF = 1.76 V.

Step by step solution

01

Identify the Anode and Cathode

In a voltaic cell, the anode is where oxidation occurs, and the cathode is where reduction takes place. In this cell, cadmium (Cd) is oxidized, losing electrons to form Cd ions ( Cd^{2+} ). At the platinum electrode, chlorine gas (Cl鈧�) is reduced to chloride ions (Cl鈦�). Thus, the cadmium electrode is the anode, and the platinum electrode is the cathode.
02

Determine Mass Change at the Cadmium Electrode

Since oxidation occurs at the cadmium electrode, Cd is converted into Cd虏鈦� ions, releasing electrons. As a result, the cadmium electrode loses mass as the cell reaction proceeds.
03

Write the Overall Cell Reaction

Combining the oxidation and reduction half-reactions gives the overall cell reaction. Oxidation at the anode: Cd ightarrow Cd^{2+} + 2e^{-} . Reduction at the cathode: Cl_{2} + 2e^{-} ightarrow 2Cl^{-} . The overall cell reaction is: Cd + Cl_{2} ightarrow Cd^{2+} + 2Cl^{-} .
04

Calculate the EMF of the Cell

Under standard conditions, the standard reduction potentials for Cd and Cl鈧�/Cl鈦� are -0.40 ext{ V for } Cd^{2+}/Cd and +1.36 ext{ V for } Cl_{2}/Cl^{-} , respectively. The EMF of the cell is calculated as E_{cell} = E_{cathode} - E_{anode} = 1.36 - (-0.40) = 1.76 ext{ V} .

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Key Concepts

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

Electrode Reactions
In a voltaic cell, reactions at the electrodes are fundamental in driving the electrochemical process. The cell consists of two electrodes: the anode and the cathode. Each electrode hosts a specific type of reaction - oxidation or reduction.
To determine which electrode plays which role, we need to look at the nature of reactions happening there:
  • Anode: This is the site of oxidation. In the given voltaic cell, the anode is the cadmium electrode where the cadmium metal loses electrons to form cadmium ions (\( \text{Cd} \rightarrow \text{Cd}^{2+} + 2e^- \)). As a result, electrons are released into the external circuit.
  • Cathode: Reduction occurs here. For our voltaic cell, at the cathode, chlorine gas gains electrons, transforming into chloride ions (\( \text{Cl}_{2} + 2e^{-} \rightarrow 2\text{Cl}^{-} \)). Electrons entering at this electrode are used up by this reduction process.
Understanding these electrode reactions is crucial as they dictate the flow of electrons outside the cell and help us comprehend which material undergoes a physical change inside the cell.
Oxidation and Reduction
Oxidation and reduction are core concepts in understanding voltaic cells and their functionalities. These processes occur simultaneously in every electrochemical cell and are inseparable.
Breaking these terms down:
  • Oxidation: This process involves the loss of electrons by a molecule, atom, or ion. Within the context of our example, the cadmium metallic strip undergoes oxidation, losing electrons to form positively charged cadmium ions.
  • Reduction: Conversely, reduction represents the gain of electrons by another species. In this case, the chlorine gas is reduced as it gains electrons and forms chloride ions.
Together, these two reactions create a complete electrochemical reaction, driving the flow of current within the cell. Understanding which species undergoes oxidation and which undergoes reduction helps in constructing overall balanced chemical equations for electrochemical reactions.
Standard Reduction Potential
The standard reduction potential is a measure used to predict the direction of electron flow in a voltaic cell. This potential is determined under standard conditions and is specific for different half-reactions.
The potential influences several aspects:
  • Magnitude: A more positive reduction potential signifies a stronger tendency for a species to gain electrons and undergo reduction. For example, in our voltaic set-up, the chlorine reduction has a potential of \(+1.36 \text{ V}\), indicating a strong affinity for electrons.
  • Direction of Electron Flow: The species with a higher standard reduction potential is more likely to act as the cathode, whereas the one with a lower potential, acting as the anode.
In our voltaic cell example, cadmium's standard reduction potential of \(-0.40 \text{ V}\) is less than that of chlorine, hence it undergoes oxidation while chlorine gets reduced.
Cell EMF Calculation
The electromotive force (EMF) of a cell is the measure of the cell's potential to produce electricity. To compute the EMF of a voltaic cell, we subtract the standard reduction potential of the anode from that of the cathode:
\[ E_{cell} = E_{cathode} - E_{anode} \]
  • In our example, **E(cathode)** for the chlorine reduction is given as \(+1.36 \text{ V}\).
  • For the cadmium reaction, **E(anode)** is \(-0.40 \text{ V}\).
  • Thus,\[E_{cell} = 1.36 - (-0.40) = 1.76 \text{ V}\]This positive EMF value indicates a spontaneous reaction, which is characteristic of a functioning voltaic cell.
Calculating the cell's EMF is crucial in determining the cell's ability to perform work and provide electric current through an external circuit.

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Most popular questions from this chapter

Aqueous solutions of ammonia \(\left(\mathrm{NH}_{3}\right)\) and bleach (active ingredient \(\mathrm{NaOCl}\) ) are sold as cleaning fluids, but bottles of both of them warn: "Never mix ammonia and bleach, as toxic gases may be produced." One of the toxic gases that can be produced is chloroamine, \(\mathrm{NH}_{2} \mathrm{Cl}\). (a) What is the oxidation number of chlorine in bleach? (b) What is the oxidation number of chlorine in chloramine? (c) Is Cl oxidized, reduced, or neither, upon the conversion of bleach to chloramine? (d) Another toxic gas that can be produced is nitrogen trichloride, \(\mathrm{NCl}_{3}\). What is the oxidation number of \(\mathrm{N}\) in nitrogen trichloride? (e) Is N oxidized, reduced, or neither, upon the conversion of ammonia to nitrogen trichloride?

Indicate whether each statement is true or false: (a) The cathode is the electrode at which oxidation takes place. (b) A galvanic cell is another name for a voltaic cell. (c) Electrons flow spontaneously from anode to cathode in a voltaic cell.

In the Bronsted-Lowry concept of acids and bases, acidbase reactions are viewed as proton-transfer reactions. The stronger the acid, the weaker is its conjugate base. If we were to think of redox reactions in a similar way, what particle would be analogous to the proton? Would strong oxidizing agents be analogous to strong acids or strong bases? [Sections 20.1 and 20.2\(]\)

Gold exists in two common positive oxidation states, +1 and +3 . The standard reduction potentials for these oxidation states are $$ \begin{array}{l} \mathrm{Au}^{+}(a q)+\mathrm{e}^{-} \quad \longrightarrow \mathrm{Au}(s) \quad E_{\mathrm{red}}^{\circ}=+1.69 \mathrm{~V} \\ \mathrm{Au}^{3+}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}(s) \quad E_{\mathrm{red}}^{\circ}=+1.50 \mathrm{~V} \end{array} $$ (a) Can you use these data to explain why gold does not tarnish in the air? (b) Suggest several substances that should be strong enough oxidizing agents to oxidize gold metal. (c) Miners obtain gold by soaking gold-containing ores in an aqueous solution of sodium cyanide. A very soluble complex ion of gold forms in the aqueous solution because of the redox reaction $$ \begin{aligned} 4 \mathrm{Au}(s)+8 \mathrm{NaCN}(a q) &+2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g) \\ & \longrightarrow 4 \mathrm{Na}\left[\mathrm{Au}(\mathrm{CN})_{2}\right](a q)+4 \mathrm{NaOH}(a q) \end{aligned} $$ What is being oxidized, and what is being reduced in this reaction? (d) Gold miners then react the basic aqueous product solution from part (c) with \(\mathrm{Zn}\) dust to get gold metal. Write a balanced redox reaction for this process. What is being oxidized, and what is being reduced?

The purification process of silicon involves the reaction of silicon tetrachloride vapor \(\left(\mathrm{SiCl}_{4}(g)\right)\) with hydrogen to \(1250^{\circ} \mathrm{C}\) to form solid silicon and hydrogen chloride. (a) Write a balanced equation for this reaction. (b) What is being oxidized, and what is being reduced? (c) Which substance is the reductant, and which is the oxidant?
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