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Chapter 18: Problem 20

Calculate \(E^{\circ}\) for the following voltaic cells: (a) \(\mathrm{Pb}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Pb}^{2+}(a q)+2 \mathrm{Ag}(s)\) (b) \(\mathrm{O}_{2}(\mathrm{~g})+4 \mathrm{Fe}^{2+}(a q)+4 \mathrm{H}^{+}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}+4 \mathrm{Fe}^{3+}(a q)\) (c) a Cd-Cd \(^{2+}\) half-cell and a \(\mathrm{Zn}-\mathrm{Zn}^{2+}\) half-cell

Short Answer

Expert verified
Question: Calculate the standard cell potential, \(E^{\circ}\), for the given redox reactions. Answer: (a) 0.93 V (b) 0.46 V (c) 0.36 V

Step by step solution

01

Identify the reduction and oxidation half-reactions

For each voltaic cell, identify the reduction and oxidation half-reactions. (a) Reduction: \(2 \mathrm{Ag}^{+}(aq) + 2 e^{-} \longrightarrow 2 \mathrm{Ag}(s)\) Oxidation: \(\mathrm{Pb}(s) \longrightarrow \mathrm{Pb}^{2+}(aq) + 2 e^{-}\) (b) Reduction: $\mathrm{O}_{2}(g) + 4 e^{-} + 4 \mathrm{H}^{+}(aq) \longrightarrow 2 \mathrm{H}_{200}` Oxidation: \(4 \mathrm{Fe}^{2+}(aq) \longrightarrow 4 \mathrm{Fe}^{3+}(aq) + 4 e^{-}\) (c) For the Cd-Cd\(^{2+}\) half-cell and the Zn-Zn\(^{2+}\) half-cell: Reduction: \(\mathrm{Cd}^{2+}(aq) + 2 e^{-} \longrightarrow \mathrm{Cd}(s)\) Oxidation: \(\mathrm{Zn}(s) \longrightarrow \mathrm{Zn}^{2+}(aq) + 2 e^{-}\)
02

Use the standard reduction potentials to calculate the standard cell potential

Using the standard reduction potentials for each half-cell, find the \(E^{\circ}\) for each reaction. (a) \(E^{\circ}_{cell} = E^{\circ}_{reduction} - E^{\circ}_{oxidation} = (0.80 V) - (-0.13 V) = 0.93 V\) (b) \(E^{\circ}_{cell} = E^{\circ}_{reduction} - E^{\circ}_{oxidation} = (1.23 V) - (0.77 V) = 0.46 V\) (c) \(E^{\circ}_{cell} = E^{\circ}_{reduction} - E^{\circ}_{oxidation} = (-0.40 V) - (-0.76 V) = 0.36 V\) So, the standard cell potentials for each cell are: (a) 0.93 V (b) 0.46 V (c) 0.36 V

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Most popular questions from this chapter

Which of the following species will be oxidized by \(1 \mathrm{MHCl}\) ? (a) \(\mathrm{Au}\) (b) \(\mathrm{Mg}\) (c) Cu (d) \(\mathrm{F}^{-}\)

Draw a diagram for a salt bridge cell for each of the following reactions. Label the anode and cathode, and indicate the direction of current flow throughout the circuit. (a) \(\mathrm{Zn}(s)+\mathrm{Cd}^{2+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{Cd}(s)\) (b) \(2 \mathrm{AuCl}_{4}^{-}(a q)+3 \mathrm{Cu}(s) \longrightarrow 2 \mathrm{Au}(s)+8 \mathrm{Cl}^{-}(a q)+3 \mathrm{Cu}^{2+}(a q)\) (c) \(\mathrm{Fe}(s)+\mathrm{Cu}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{Cu}(s)+\mathrm{Fe}(\mathrm{OH})_{2}(s)\)

A solution containing a metal ion \(\left(M^{2+}(a q)\right)\) is electrolyzed by a current of \(7.8\) A. After \(15.5\) minutes, \(2.39 \mathrm{~g}\) of the metal is plated out. (a) How many coulombs are supplied by the battery? (b) What is the metal? (Assume \(100 \%\) efficiency.)

Consider three metals, \(\mathrm{X}, \mathrm{Y}\), and \(\mathrm{Z}\), and their salts, \(\mathrm{XA}, \mathrm{YA}\), and \(\mathrm{ZA}\). Three experiments take place with the following results: \- \(\mathrm{X}+\mathrm{hot} \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}_{2}\) bubbles \(\mathrm{X}+\mathrm{YA} \longrightarrow\) no reaction I \(\mathrm{X}+\mathrm{ZA} \longrightarrow \mathrm{X}\) discolored \(+\mathrm{Z}\) Rank metals \(\mathrm{X}, \mathrm{Y}\), and \(\mathrm{Z}\), in order of decreasing strength as reducing agents.

Calculate \(E^{\circ}\) for the following voltaic cells: (a) \(\mathrm{MnO}_{2}(s)+4 \mathrm{H}^{+}(a q)+2 \mathrm{I}^{-}(a q) \longrightarrow\) \(\mathrm{Mn}^{2+}(a q)+2 \mathrm{H}_{2} \mathrm{O}+\mathrm{I}_{2}(s)\) (b) \(\mathrm{H}_{2}(g)+2 \mathrm{OH}^{-}(a q)+\mathrm{S}(s) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{S}^{2-}(a q)\) (c) an \(\mathrm{Ag}-\mathrm{Ag}^{+}\) half-cell and an \(\mathrm{Au}-\mathrm{AuCl}_{4}^{-}\) half-cell.
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