A voltaic cell similar to that shown in Figure 20.5 is constructed. One half- cell consists of an aluminum strip placed in a solution of \(\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\) , and the other has a nickel strip placed in a solution of \(\mathrm{NiSO}_{4}\) . The overall cell reaction is $$ 2 \mathrm{Al}(s)+3 \mathrm{Ni}^{2+}(a q) \longrightarrow 2 \mathrm{Al}^{3+}(a q)+3 \mathrm{Ni}(s) $$ (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 aluminum electrode to the nickel electrode or from the nickel to the aluminum? (f) In which directions do the cations and anions migrate through the solution? Assume the Al is not coated with its oxide.

Short Answer

Expert verified
Aluminum is being oxidized, and nickel is being reduced. The oxidation half-reaction is \(2 \mathrm{Al}(s) \longrightarrow 2 \mathrm{Al}^{3+}(a q) +6 e^-\), and the reduction half-reaction is \(3 \mathrm{Ni}^{2+}(a q) + 6 e^- \longrightarrow 3 \mathrm{Ni}(s)\). The aluminum electrode is the anode (negative), and the nickel electrode is the cathode (positive). Electrons flow from the aluminum electrode to the nickel electrode. \(\mathrm{Ni}^{2+}\) and \(\mathrm{Al}^{3+}\) ions migrate towards the nickel electrode (cathode), while \(\mathrm{NO}_{3}^-\) and \(\mathrm{SO}_{4}^{2-}\) ions migrate towards the aluminum electrode (anode).

Step by step solution

01

(a) Identifying Oxidation and Reduction

In the overall cell reaction: $$ 2 \mathrm{Al}(s)+3 \mathrm{Ni}^{2+}(a q) \longrightarrow 2 \mathrm{Al}^{3+}(a q)+3 \mathrm{Ni}(s) $$ 1. Observe that the oxidation state of Al changes from 0 to +3, which indicates that the aluminum is losing electrons (oxidation). 2. The oxidation state of Ni changes from +2 to 0, which indicates that nickel is gaining electrons (reduction). Thus, aluminum is being oxidized, and nickel is being reduced.
02

(b) Writing Half-reactions

We can separate the overall cell reaction into two half-reactions: one for oxidation and one for reduction. 1. Oxidation half-reaction (Aluminum): $$ 2 \mathrm{Al}(s) \longrightarrow 2 \mathrm{Al}^{3+}(a q) +6 e^- $$ 2. Reduction half-reaction (Nickel): $$ 3 \mathrm{Ni}^{2+}(a q) + 6 e^- \longrightarrow 3 \mathrm{Ni}(s) $$
03

(c) Determining the Anode and Cathode

In a voltaic cell: 1. Anode: The electrode where oxidation occurs (loss of electrons). In this case, the aluminum electrode is the anode. 2. Cathode: The electrode where reduction occurs (gain of electrons). In this case, the nickel electrode is the cathode.
04

(d) Identifying Electrode Signs

In a voltaic cell: 1. Anode is negative because it releases electrons (oxidation). 2. Cathode is positive because it attracts electrons (reduction). Thus, the aluminum electrode (anode) is negative, and the nickel electrode (cathode) is positive.
05

(e) Determining the Direction of Electron Flow

Electrons flow from the anode to the cathode in a voltaic cell. In this case, electrons flow from the aluminum electrode (anode) to the nickel electrode (cathode).
06

(f) Determining the Direction of Ion Migration

In a voltaic cell solution: 1. Cations (positive ions) migrate towards the cathode. In this case, \(\mathrm{Ni}^{2+}\) and \(\mathrm{Al}^{3+}\) ions will migrate towards the nickel electrode (cathode). 2. Anions (negative ions) migrate towards the anode. In this case, \(\mathrm{NO}_{3}^-\) and \(\mathrm{SO}_{4}^{2-}\) ions will migrate towards the aluminum electrode (anode).

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Most popular questions from this chapter

In some applications nickel-cadmium batteries have been replaced by nickel- zinc batteries. The overall cell reaction for this relatively new battery is: $$ \begin{aligned} 2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{NiO}(\mathrm{OH})(s)+\mathrm{Zn}(s) & \\ & \longrightarrow 2 \mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{Zn}(\mathrm{OH})_{2}(s) \end{aligned} $$ (a)What is the cathode half-reaction? (b)What is the anode half-reaction? (c) A single nickel-cadmium cell has a voltage of 1.30 \(\mathrm{V}\) . Based on the difference in the standard reduction potentials of \(\mathrm{Cd}^{2+}\) and \(\mathrm{Zn}^{2+},\) what voltage would you estimate a nickel-zinc battery will produce? (d) Would you expect the specific energy density of a nickel-zinc battery to be higher or lower than that of a nickel-cadmium battery?

A voltaic cell is constructed with two silver-silver chloride electrodes, each of which is based on the following half-reaction: $$ \operatorname{AgCl}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)+\mathrm{Cl}^{-}(a q) $$ The two half-cells have \(\left[\mathrm{Cl}^{-}\right]=0.0150 \mathrm{M}\) and \(\left[\mathrm{Cl}^{-}\right]=\) \(2.55 \mathrm{M},\) respectively. (a) Which electrode is the cathode of the cell? (b) What is the standard emf of the cell? (c) What is the cell emf for the concentrations given? (d) For each electrode, predict whether \(\left[\mathrm{Cl}^{-}\right]\) will increase, decrease, or stay the same as the cell operates.

In the Bronsted-Lowry concept of acids and bases, acid-base 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\(]\)

(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?

Indicate whether each statement is true or false: (a) The anode is the electrode at which oxidation takes place. (b) A voltaic cell always has a positive emf. (c) A salt bridge or permeable barrier is necessary to allow a voltaic cell to operate.

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