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Chapter 18: Problem 32

Consider the following electrochemical cell: a. If silver metal is a product of the reaction, is the cell a galvanic cell or electrolytic cell? Label the cathode and anode, and describe the direction of the electron flow. b. If copper metal is a product of the reaction, is the cell a galvanic cell or electrolytic cell? Label the cathode and anode, and describe the direction of the electron flow. c. If the above cell is a galvanic cell, determine the standard cell potential. d. If the above cell is an electrolytic cell, determine the minimum external potential that must be applied to cause the reaction to occur.

Short Answer

Expert verified
a. If silver metal is a product, then the cell is a galvanic cell. The cathode is the silver electrode where Ag+ is reduced to Ag(s), and the anode is the copper electrode where Cu(s) is oxidized to Cu2+. Electrons flow from the anode (copper) to the cathode (silver). b. If copper metal is a product, then the cell is an electrolytic cell. The cathode is the copper electrode where Cu2+ is reduced to Cu(s), and the anode is the silver electrode where Ag(s) is oxidized to Ag+. Electrons flow from the anode (silver) to the cathode (copper) due to the external potential applied. c. For the galvanic cell, the standard cell potential is: \[E_{cell} = 0.80 - 0.34 = +0.46V\] d. For the electrolytic cell, the minimum external potential required is: \[|E_{cell}| = |{-0.46}| = 0.46V\]

Step by step solution

01

Scenario A: Silver metal is a product.

In an electrochemical cell, the silver would be reduced if it is a product, so the half-reaction would be: \[Ag^{+} + e^{-} \rightarrow Ag(s)\]
02

Scenario B: Copper metal is a product.

If copper metal is a product, copper ions would be reduced in the reaction, so the half-reaction would be: \[Cu^{2+} + 2e^{-} \rightarrow Cu(s)\] #Step 2: Label the cathode and anode and identify electron flow direction#
03

Scenario A: Silver metal is a product (Galvanic cell).

Since silver ions are being reduced in this half-reaction, this would be the cathode. Consequently, the copper electrode will be the anode, with its half-reaction: \[Cu(s) \rightarrow Cu^{2+} + 2e^-\] The electrons flow from the anode (copper) to the cathode (silver).
04

Scenario B: Copper metal is a product (Electrolytic cell).

In this case, since the copper ions are being reduced in the half-reaction, this would be the cathode. Consequently, the silver electrode will be the anode, with its half-reaction: \[Ag(s) \rightarrow Ag^{+} + e^{-}\] The electrons flow from the anode (silver) to the cathode (copper) due to the external potential applied. #Step 3: Determine the standard cell potential or the minimum external potential#
05

Scenario A: Galvanic cell potentials.

For a galvanic cell (scenario A), we determine the standard cell potential by taking the difference in standard electrode potentials of the two half cells. The standard electrode potential for \(Ag^{+}/Ag\) is \(+0.80V\) and for \(Cu^{2+}/Cu\) is \(+0.34V\). So the standard cell potential will be: \[E_{cell} = E_{cathode} - E_{anode} = 0.80 - 0.34 = +0.46V\]
06

Scenario B: Minimum external potential for electrolytic cell.

For an electrolytic cell (scenario B), the minimum external potential required to cause the reaction to occur will be equal to the absolute value of the standard cell potential of the electrolytic cell. We can calculate this by finding the difference in the standard electrode potentials of the two half-cells for scenario B. Here, the silver electrode will act as anode and copper electrode will act as cathode. So the standard cell potential will be: \[E_{cell} = E_{cathode} - E_{anode} = 0.34 - 0.80 = -0.46V\] The minimum external potential required for the electrolytic cell will be the absolute value of the standard cell potential: \[|E_{cell}| = |{-0.46}| = 0.46V\]

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Most popular questions from this chapter

An electrochemical cell is set up using the following unbalanced reaction: $$\mathrm{M}^{a+}(a q)+\mathrm{N}(s) \longrightarrow \mathrm{N}^{2+}(a q)+\mathrm{M}(s)$$ The standard reduction potentials are: $$\mathrm{M}^{a+}+a \mathrm{e}^{-} \longrightarrow \mathrm{M} \quad \mathscr{E}^{\circ}=0.400 \mathrm{V}$$ $$\mathrm{N}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{N} \quad \mathscr{E}^{\circ}=0.240 \mathrm{V}$$ The cell contains 0.10\(M \mathrm{N}^{2+}\) and produces a voltage of 0.180 \(\mathrm{V}\) . If the concentration of \(\mathrm{M}^{a+}\) is such that the value of the reaction quotient \(Q\) is \(9.32 \times 10^{-3},\) calculate \(\left[\mathrm{M}^{a+}\right] .\) Calculate \(w_{\text { max }}\) for this electrochemical cell.

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Consider the cell described below: $$\text { Al }\left|\mathrm{Al}^{3+}(1.00 M)\right|\left|\mathrm{Pb}^{2+}(1.00 M)\right| \mathrm{Pb}$$ Calculate the cell potential after the reaction has operated long enough for the [Al \(^{3+} ]\) to have changed by 0.60 \(\mathrm{mol} / \mathrm{L}\) . (Assume \(T=25^{\circ} \mathrm{C} .\)

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Consider a concentration cell that has both electrodes made of some metal M. Solution A in one compartment of the cell contains 1.0\(M \mathrm{M}^{2+} .\) Solution \(\mathrm{B}\) in the other cell compartment has a volume of 1.00 L. At the beginning of the experiment 0.0100 mole of \(\mathrm{M}\left(\mathrm{NO}_{3}\right)_{2}\) and 0.0100 mole of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) are dissolved in solution \(\mathrm{B}\) (ignore volume changes), where the reaction $$\mathrm{M}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q) \rightleftharpoons \mathrm{MSO}_{4}(s)$$ occurs. For this reaction equilibrium is rapidly established, whereupon the cell potential is found to be 0.44 \(\mathrm{V}\) at \(25^{\circ} \mathrm{C} .\) Assume that the process $$\mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M}$$ has a standard reduction potential of \(-0.31 \mathrm{V}\) and that no other redox process occurs in the cell. Calculate the value of \(K_{\mathrm{sp}}\) for \(\mathrm{MSO}_{4}(s)\) at \(25^{\circ} \mathrm{C} .\)
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