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Chapter 18: Problem 5

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
The overall balanced cell equation for the formation of iron metal (Fe) from iron(II) ions (Fe^2+) and the conversion of chromium metal (Cr) to chromium(III) ions (Cr^3+) is: 3Fe^2+ (aq) + Cr(s) → 3Fe(s) + Cr^3+ (aq) The standard cell potential (E° cell) is calculated as: E° cell = E°(Fe^2+/Fe) - E°(Cr^3+/Cr) = (-0.44) - (-0.74) = 0.30 V The galvanic cell consists of a chromium metal anode (where oxidation occurs) on the left side, and an iron metal cathode (where reduction occurs) on the right side. Electrons flow from the anode to the cathode, and ions flow through a salt bridge to maintain charge neutrality.

Step by step solution

01

Identify half-reactions for iron and chromium ions

The half-reactions for iron and chromium ions are as follows: Iron(II) ions to form iron metal: Fe^2+ (aq) → Fe(s) Chromium metal to form chromium(III) ions: Cr(s) → Cr^3+ (aq)
02

Balance the half-reactions

In order to balance the half-reactions, we must ensure that both the atoms and charges are balanced. For iron half-reaction: Fe^2+ (aq) → Fe(s) It is already balanced. For chromium half-reaction: Initial: Cr(s) → Cr^3+ (aq) Balance the charge by adding 3 electrons to the right side: Cr(s) → Cr^3+ (aq) + 3e^-
03

Determine the overall reaction

To get the overall reaction, we need to make the electron transfer equal in both half-reactions. Since we have 3 electrons in the chromium half-reaction, we must multiply the iron half-reaction by 3 to balance the electrons: 3(Fe^2+ (aq) → Fe(s)) Resulting in: 3Fe^2+ (aq) → 3Fe(s) + 6e^- Now we can add the balanced half-reactions to get the overall reaction: 3Fe^2+ (aq) + Cr(s) → 3Fe(s) + Cr^3+ (aq)
04

Calculate the voltage (E° cell)

To calculate the E° cell or the voltage, we can use the formula: E° cell = E° cathode - E° anode We know that Fe^2+ (aq) → Fe(s) is the reduction half-reaction (formation of iron metal) and Cr(s) → Cr^3+ (aq) is the oxidation half-reaction (changing chromium metal to chromium(III)). Look up their standard reduction potentials in a table: Fe^2+ (aq) + 2e^- → Fe(s), E°= -0.44 V Cr^3+ (aq) + 3e^- → Cr(s), E°= -0.74 V Substitute into the formula: E° cell = E°(Fe^2+/Fe) - E°(Cr^3+/Cr) = (-0.44) - (-0.74) = 0.30 V
05

Sketch the galvanic cell with electron flow, labeling anode and cathode

To sketch the galvanic cell, follow these guidelines: - Label the anode (oxidation) and cathode (reduction) sides of the cell - Show the electron flow from the anode to the cathode - Include the salt bridge and anions and cations Here is the sketch description of the galvanic cell: Left side (Anode): Chromium metal electrode where oxidation of chromium metal occurs: Cr(s) → Cr^3+ (aq) + 3e^- Right side (Cathode): Iron metal electrode where reduction of iron(II) ions occurs: 3Fe^2+ (aq) + 6e^- → 3Fe(s) Salt bridge: Connect the two half-cells, allowing ion flow between them to maintain charge neutrality Electron flow: Electrons flow from chromium metal (anode) to iron metal (cathode) Anions and cations: Flow towards appropriate sides (anode or cathode) via the salt bridge; Cr^3+ ions produced at the anode side and Fe^2+ ions reduced at the cathode side.

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Most popular questions from this chapter

In the electrolysis of a sodium chloride solution, what volume of \(\mathrm{H}_{2}(g)\) is produced in the same time it takes to produce \(257 \mathrm{~L}\) \(\mathrm{Cl}_{2}(\mathrm{~g})\), with both volumes measured at \(50 .{ }^{\circ} \mathrm{C}\) and \(2.50 \mathrm{~atm} ?\)

A galvanic cell is based on the following half-reactions: $$\begin{array}{ll}\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(s) & \mathscr{E}^{\circ}=-0.440 \mathrm{~V} \\ 2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}(g) & \mathscr{E}^{\circ}=0.000 \mathrm{~V} \end{array}$$ where the iron compartment contains an iron electrode and \(\left[\mathrm{Fe}^{2+}\right]=1.00 \times 10^{-3} M\) and the hydrogen compartment contains a platinum electrode, \(P_{\mathrm{H}_{2}}=1.00 \mathrm{~atm}\), and a weak acid, \(\mathrm{HA}\), at an initial concentration of \(1.00 M .\) If the observed cell potential is \(0.333 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\), calculate the \(K_{\mathrm{a}}\) value for the weak acid HA.

Consider the galvanic cell based on the following half-reactions: $$\begin{array}{ll}\mathrm{Au}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au} & \mathscr{E}^{\circ}=1.50 \mathrm{~V} \\ \mathrm{Tl}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Tl} & \mathscr{E}^{\circ}=-0.34 \mathrm{~V} \end{array}$$ a. Determine the overall cell reaction and calculate \(\mathscr{E}_{\mathrm{ccll}}^{\circ}\) 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 \(\left[\mathrm{Au}^{3+}\right]=1.0 \times 10^{-2} M\) and \(\left[\mathrm{Tl}^{+}\right]=1.0 \times 10^{-4} \mathrm{M}\)

In making a specific galvanic cell, explain how one decides on the electrodes and the solutions to use in the cell.

When copper reacts with nitric acid, a mixture of \(\mathrm{NO}(\mathrm{g})\) and \(\mathrm{NO}_{2}(g)\) is evolved. The volume ratio of the two product gases depends on the concentration of the nitric acid according to the equilibrium \(2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q)+\mathrm{NO}(g) \rightleftharpoons 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) Consider the following standard reduction potentials at \(25^{\circ} \mathrm{C}\) : \(3 \mathrm{e}^{-}+4 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) $$\begin{array}{r}\mathscr{E}^{\circ}=0.957 \mathrm{~V}\end{array}$$ \(\mathrm{e}^{-}+2 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) $${8}^{\circ}=0.775 \mathrm{~V}$$ a. Calculate the equilibrium constant for the above reaction. b. What concentration of nitric acid will produce a NO and \(\mathrm{NO}_{2}\) mixture with only \(0.20 \% \mathrm{NO}_{2}\) (by moles) at \(25^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) ? Assume that no other gases are present and that the change in acid concentration can be neglected.
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