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Chapter 18: Problem 7

Sketch a cell that forms iron metal from iron(II) while changing chromium metal to chromium(III). Calculate the voltage, show the electron flow, label the anode and cathode, and balance the overall cell equation.

Short Answer

Expert verified
In the galvanic cell, the anode half-reaction is : \(3Fe^{2+}(aq) \rightarrow 3Fe(s) + 6e^{-}\) and the cathode half-reaction is: \(2Cr(s) + 6e^{-} \rightarrow 2Cr^{3+}(aq)\). The cell potential (voltage) is -1.18 V. The balanced overall cell equation is: \(3Fe^{2+}(aq) + 2Cr(s) \rightarrow 3Fe(s) + 2Cr^{3+}(aq)\). The electron flow is from anode (iron half-cell) to cathode (chromium half-cell).

Step by step solution

01

Identify Half-Reactions and Electrodes

Anode (where oxidation occurs): Iron(II) (Fe^2+) is oxidized to form iron metal (Fe) $$Fe^{2+}(aq) \rightarrow Fe(s) + 2e^{-}$$ Cathode (where reduction occurs): Chromium metal (Cr) changes to chromium(III) (Cr^3+) $$Cr(s) + 3e^{-} \rightarrow Cr^{3+}(aq)$$
02

Calculate the Cell Potential

We can find the cell potential by using the standard reduction potentials of the two half-reactions. Standard reduction potential for Fe^2+ to Fe: -0.44 V Standard reduction potential for Cr^3+ to Cr: -0.74 V $$E_{cell} = E_{cathode} - E_{anode}$$ When we consider the oxidation and reduction process in the cell, reverse the anode reaction first: $$Fe(s) - 2e^{-} \rightarrow Fe^{2+}(aq)$$ Now, the standard reduction potential for the anode becomes +0.44 V (reversal). $$E_{cell} = (-0.74) - (+0.44) = -1.18 V$$
03

Balance the Overall Cell Equation

In order to balance the overall cell equation, we need to balance both charge and atoms. We will balance the electrons first. For the anode half: $$Fe^{2+}(aq) \rightarrow Fe(s) + 2e^{-}$$ For the cathode half: $$Cr(s) + 3e^{-} \rightarrow Cr^{3+}(aq)$$ Multiply the first half-reaction by 3 and second half-reaction by 2 to get the same number of electrons exchanged in both half-reactions. Anode half-reaction (x3): $$3Fe^{2+}(aq) \rightarrow 3Fe(s) + 6e^{-}$$ Cathode half-reaction (x2): $$2Cr(s) + 6e^{-} \rightarrow 2Cr^{3+}(aq)$$ Now, add these balanced half-reactions to get the overall cell equation: $$3Fe^{2+}(aq) + 2Cr(s) \rightarrow 3Fe(s) + 2Cr^{3+}(aq)$$
04

Sketch the Cell and Label Components

In the sketch: - Label the anode side: "Anode: Fe^2+ to Fe". - Label the cathode side: "Cathode: Cr to Cr^3+". - Label the direction of electron flow from anode to cathode (electrons are produced at the anode and consumed at the cathode). - Include a salt bridge which allows the movement of ions to keep the cell electrically neutral (e.g., Na^+ and SO_4^2- ions). - Label the cell potential as "E = -1.18 V". The sketched cell should clearly show the electron flow, anode, cathode, and each half-reaction with their respective labels.

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Most popular questions from this chapter

A solution at \(25^{\circ} \mathrm{C}\) contains \(1.0 M \mathrm{Cd}^{2+}, 1.0 M \mathrm{Ag}^{+}, 1.0 \mathrm{M}\) \(\mathrm{Au}^{3+},\) and 1.0 \(\mathrm{M} \mathrm{Ni}^{2+}\) in the cathode compartment of an electrolytic cell. Predict the order in which the metals will plate out as the voltage is gradually increased.

The overall reaction in the lead storage battery is \(\mathrm{Pb}(s)+\mathrm{PbO}_{2}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{HSO}_{4}^{-}(a q) \longrightarrow\) \(2 \mathrm{PbSO}_{4}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)\) Calculate \(\mathscr{E}\) at \(25^{\circ} \mathrm{C}\) for this battery when \(\left[\mathrm{H}_{2} \mathrm{SO}_{4}\right]=4.5 \mathrm{M}\), that is, \(\left[\mathrm{H}^{+}\right]=\left[\mathrm{HSO}_{4}^{-}\right]=4.5 \mathrm{M} . \mathrm{At} 25^{\circ} \mathrm{C}, \mathscr{E}^{\circ}=2.04 \mathrm{V}\) for the lead storage battery.

Specify which of the following equations represent oxidation鈥搑eduction reactions, and indicate the oxidizing agent, the reducing agent, the species being oxidized, and the species being reduced. a. \(\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightarrow \mathrm{CO}(g)+3 \mathrm{H}_{2}(g)\) b. \(2 \mathrm{AgNO}_{3}(a q)+\mathrm{Cu}(s) \rightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{Ag}(s)\) c. \(\mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \rightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g)\) d. \(2 \mathrm{H}^{+}(a q)+2 \mathrm{CrO}_{4}^{2-}(a q) \rightarrow \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\)

Consider the galvanic cell based on the following halfreactions: $$\begin{array}{ll}{\mathrm{Au}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}} & {\mathscr{E}^{\circ}=1.50 \mathrm{V}} \\ {\mathrm{T} 1^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Tl}} & {\mathscr{E}^{\circ}=-0.34 \mathrm{V}}\end{array}$$ a. Determine the overall cell reaction and calculate \(\mathscr{E}_{\text { cell }}\) b. Calculate \(\Delta G^{\circ}\) and \(K\) for the cell reaction at \(25^{\circ} \mathrm{C}\) c. Calculate \(\mathscr{E}_{\text { cell }}\) at \(25^{\circ} \mathrm{C}\) when \(\left[\mathrm{Au}^{3+}\right]=1.0 \times 10^{-2} \mathrm{M}\) and \(\left[\mathrm{T} 1^{+}\right]=1.0 \times 10^{-4} \mathrm{M}\)

Give the balanced cell equation and determine \(\mathscr{E}^{\circ}\) for the galvanic cells based on the following half-reactions. Standard reduction potentials are found in Table 18.1. a. \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}\) \(\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}\) b. \(2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2}\) \(\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Al}\)
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