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Chapter 20: Problem 29

A voltaic cell similar to that shown in Figure 20.5 is constructed. One electrode half-cell consists of a magnesium strip placed in a solution of \(\mathrm{MgCl}_{2}\), and the other has a nickel strip placed in a solution of \(\mathrm{NiCl}_{2}\). The overall cell reaction is $$ \mathrm{Mg}(s)+\mathrm{Ni}^{2+}(a q) \longrightarrow \mathrm{Ni}(s)+\mathrm{Mg}^{2+}(a q) $$ (a) What is being oxidized, and what is being reduced? (b) Write the half- reactions that occur in the two half-cells. (c) Which electrode is the anode, and which is the cathode?(d) Indicate the signs of the electrodes. (e) Do electrons flow from the magnesium electrode to the nickel electrode or from the nickel to the magnesium? (f) In which directions do the cations and anions migrate through the solution?

Short Answer

Expert verified
a) Magnesium is being oxidized and Nickel ion is being reduced. b) Oxidation half-reaction: \(\mathrm{Mg}(s) \longrightarrow \mathrm{Mg^{2+}}(aq) + 2e^-\), Reduction half-reaction: \(\mathrm{Ni^{2+}}(aq) + 2e^- \longrightarrow \mathrm{Ni}(s)\) c) The Mg electrode is the anode, and the Ni electrode is the cathode. d) Anode (Mg electrode) is negative, and Cathode (Ni electrode) is positive. e) Electrons flow from the magnesium electrode to the nickel electrode. f) Cations (\(\mathrm{Mg^{2+}}\)) migrate towards the Ni electrode (cathode), and anions (\(\mathrm{Cl^-}\)) migrate towards the Mg electrode (anode).

Step by step solution

01

Identify oxidation and reduction processes

We can use the acronym "OIL RIG" (Oxidation Is Loss, Reduction Is Gain) to remember that oxidation is the loss of electrons, while reduction is the gain of electrons. The overall cell reaction is given: \[\mathrm{Mg}(s)+\mathrm{Ni^{2+}}(aq) \longrightarrow \mathrm{Ni}(s)+\mathrm{Mg^{2+}}(aq)\] From the cell reaction, we can see that \(\mathrm{Mg}\) loses two electrons and becomes \(\mathrm{Mg^{2+}}\), while \(\mathrm{Ni^{2+}}\) gains two electrons and turns into \(\mathrm{Ni}\). Therefore: a) Magnesium is being oxidized and Nickel ion is being reduced.
02

Write half-reactions

We will write the half-reactions for both the oxidation and reduction processes: b) Oxidation half-reaction: \[\mathrm{Mg}(s) \longrightarrow \mathrm{Mg^{2+}}(aq) + 2e^-\] Reduction half-reaction: \[ \mathrm{Ni^{2+}}(aq) + 2e^- \longrightarrow \mathrm{Ni}(s)\]
03

Identify Anode and Cathode

Since oxidation happens at the anode and reduction occurs at the cathode: c) The Mg electrode is the anode, and the Ni electrode is the cathode.
04

Determine signs of electrodes

Anode is where oxidation happens, and is the site where electrons are released, thus it's considered as the negative electrode. Cathode is where reduction occurs, and it's the site where electrons are gained, thus it is the positive electrode. d) Anode (Mg electrode) is negative, and Cathode (Ni electrode) is positive.
05

Determine direction of electron flow

Electrons flow from the negative electrode (anode) to the positive electrode (cathode). e) Electrons flow from the magnesium electrode to the nickel electrode.
06

Determine cation and anion migration

In a voltaic cell, cations (\(\mathrm{Mg^{2+}}\) in this case) migrate towards the cathode, and anions (\(\mathrm{Cl^-}\) in this case) migrate towards the anode to maintain charge neutrality. f) Cations (\(\mathrm{Mg^{2+}}\)) migrate towards the Ni electrode (cathode), and anions (\(\mathrm{Cl^-}\)) migrate towards the Mg electrode (anode).

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Most popular questions from this chapter

(a) Aluminum metal is used as a sacrificial anode to protect offshore pipelines in salt water from corrosion. Why is the aluminum referred to as a "sacrificial anode"? (b) Looking in Appendix E, suggest what metal the pipelines could be made from in order for aluminum to be successful as a sacrificial anode.

(a) Suppose that an alkaline battery was manufactured using cadmium metal rather than zinc. What effect would this have on the cell emf? (b) What environmental advantage is provided by the use of nickel-metal hydride batteries over nickel-cadmium batteries?

If the equilibrium constant for a one-electron redox reaction at $298 \mathrm{~K}\( is \)2.2 \times 10^{-5},\( calculate the corresponding \)\Delta G^{\circ}\( and \)E^{\circ}$.

Given the following reduction half-reactions: $$ \begin{aligned} \mathrm{Fe}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+}(a q) & E_{\mathrm{red}}^{\circ}=+0.77 \mathrm{~V} \\ \mathrm{~S}_{2} \mathrm{O}_{6}^{2-}(a q)+4 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{SO}_{3}(a q) & E_{\mathrm{red}}^{\circ}=+0.60 \mathrm{~V} \\ \mathrm{~N}_{2} \mathrm{O}(g)+2 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & E_{\mathrm{red}}^{\circ}=-1.77 \mathrm{~V} \\ \mathrm{VO}_{2}^{+}(a q)+2 \mathrm{H}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{VO}^{2+}+\mathrm{H}_{2} \mathrm{O}(l) & E_{\mathrm{red}}^{\circ}=+1.00 \mathrm{~V} \end{aligned} $$ (a) Write balanced chemical equations for the oxidation of $\mathrm{Fe}^{2+}(a q)\( by \)\mathrm{S}_{2} \mathrm{O}_{6}^{2-}(a q),\( by \)\mathrm{N}_{2} \mathrm{O}(a q),\( and by \)\mathrm{VO}_{2}^{+}(a q)\( (b) Calculate \)\Delta G^{\circ}\( for each reaction at \)298 \mathrm{~K}$. (c) Calculate the equilibrium constant \(K\) for each reaction at \(298 \mathrm{~K}\).

Indicate whether the following balanced equations involve oxidation-reduction. If they do, identify the elements that undergo changes in oxidation number. (a) $2 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow 2 \mathrm{HNO}_{3}(a q)$ (b) $\mathrm{FeS}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{FeCl}_{2}(a q)+\mathrm{H}_{2} \mathrm{~S}(g)$ (c) $\mathrm{Fe}(s)+2 \mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+ 2 \mathrm{NO}_{2}(g)+\mathrm{FeO}(s)$
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