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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

In biological systems, acetate ion is converted to ethyl alcohol in a two- step process: \(\mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+3 \mathrm{H}^{+}(a q)+2 e^{-} \longrightarrow \mathrm{CH}_{3} \mathrm{CHO}(a q)+\mathrm{H}_{2} \mathrm{O}\) \(E^{o \prime}=-0.581 \mathrm{~V}\) \(\mathrm{CH}_{3} \mathrm{CHO}(a q)+2 \mathrm{H}^{+}(a q)+2 e^{-} \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q) \quad E^{o \prime}=-0.197 \mathrm{~V}\) \(\left(E^{\circ \prime}\right.\) is the standard reduction voltage at \(25^{\circ} \mathrm{C}\) and \(\mathrm{pH}\) of \(7.00 .\) ) (a) Calculate \(\Delta G^{\circ \prime}\) for each step and for the overall conversion. (b) Calculate \(E^{\circ \prime}\) for the overall conversion.

Use the following half-equations to write three spontaneous reactions. Justify your answers by calculating \(E^{\circ}\) for the cells. (1) \(\mathrm{MnO}_{4}^{-}(a q)+8 \mathrm{H}^{+}(a q)+5 e^{-} \longrightarrow \mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}\) \(E^{\circ}=+1.512 \mathrm{~V}\) (2) \(\mathrm{O}_{2}(\mathrm{~g})+4 \mathrm{H}^{+}(\mathrm{aq})+4 e^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O} \quad E^{\circ}=+1.229 \mathrm{~V}\) (3) \(\mathrm{Co}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Co}(s) \quad E^{\circ}=-0.282 \mathrm{~V}\)

Which species in each pair is the stronger oxidizing agent? (a) \(\mathrm{NO}_{3}^{-}\) or \(\mathrm{I}_{2}\) (b) \(\mathrm{Fe}(\mathrm{OH})_{3}\) or \(\mathrm{S}\) (c) \(\mathrm{Mn}^{2+}\) or \(\mathrm{MnO}_{2}\) (d) \(\mathrm{ClO}_{3}^{-}\) in acidic solution or \(\mathrm{ClO}_{3}^{-}\) in basic solution

An electrolytic cell produces aluminum from \(\mathrm{Al}_{2} \mathrm{O}_{3}\) at the rate of ten kilograms a day. Assuming a yield of \(100 \%\), (a) how many moles of electrons must pass through the cell in one day? (b) how many amperes are passing through the cell? (c) how many moles of oxygen \(\left(\mathrm{O}_{2}\right)\) are being produced simultaneously?

For the following half-reactions, answer the questions below: $$ \begin{array}{cc} \mathrm{Ce}^{4+}(a q)+e^{-} \longrightarrow \mathrm{Ce}^{3+}(a q) & E^{\circ}=+1.61 \mathrm{~V} \\ \mathrm{Ag}^{+}(a q)+e^{-} \longrightarrow \mathrm{Ag}(s) & E^{\circ}=+0.80 \mathrm{~V} \\ \mathrm{Hg}_{2}^{2+}(a q)+2 e^{-} \longrightarrow 2 \mathrm{Hg}(l) & E^{\circ}=+0.80 \mathrm{~V} \\ \mathrm{Sn}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Sn}(s) & E^{\circ}=-0.14 \mathrm{~V} \\ \mathrm{Ni}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Ni}(s) & E^{\circ}=-0.24 \mathrm{~V} \\ \mathrm{Al}^{3+}(a q)+3 e^{-} \longrightarrow \mathrm{Al}(s) & E^{o}=-1.68 \mathrm{~V} \end{array} $$ (a) Which is the weakest oxidizing agent? (b) Which is the strongest oxidizing agent? (c) Which is the strongest reducing agent? (d) Which is the weakest reducing agent? (e) Will \(\mathrm{Sn}(s)\) reduce \(\mathrm{Ag}^{+}(\mathrm{aq})\) to \(\mathrm{Ag}(s) ?\) (f) Will \(\mathrm{Hg}(l)\) reduce \(\mathrm{Sn}^{2+}(a q)\) to \(\mathrm{Sn}(s) ?\) (g) Which ion(s) can be reduced by \(\operatorname{Sn}(s)\) ? (h) Which metal(s) can be oxidized by \(\mathrm{Ag}^{+}(a q)\) ?
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