Chapter 18: Problem 19
What is electrochemistry? What are redox reactions? Explain the difference between a galvanic and an electrolytic cell.
Chapter 18: Problem 19
What is electrochemistry? What are redox reactions? Explain the difference between a galvanic and an electrolytic cell.
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Get started for freeSketch the galvanic cells based on the following overall reactions. Show the direction of electron flow, the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced equation. Assume that all concentrations are \(1.0 M\) and that all partial pressures are \(1.0 \mathrm{~atm}\).
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: $$ \begin{array}{ll} \mathrm{M}^{a+}+a \mathrm{e}^{-} \longrightarrow \mathrm{M} & \mathscr{E}^{\circ}=0.400 \mathrm{~V} \\ \mathrm{~N}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{N} & \mathscr{E}^{\circ}=0.240 \mathrm{~V} \end{array} $$ The cell contains \(0.10 \mathrm{M} \mathrm{N}^{2+}\) 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 \(w_{\max }\) for this electrochemical cell.
Hydrazine is somewhat toxic. Use the half-reactions shown below to explain why household bleach (a highly alkaline solution of sodium hypochlorite) should not be mixed with household ammonia or glass cleansers that contain ammonia. \(\mathrm{ClO}^{-}+\mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{OH}^{-}+\mathrm{Cl}^{-} \quad \mathscr{E}^{\circ}=0.90 \mathrm{~V}\) \(\mathrm{N}_{2} \mathrm{H}_{4}+2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{NH}_{3}+2 \mathrm{OH}^{-} \quad \mathscr{E}^{\circ}=-0.10 \mathrm{~V}\)
An experimental fuel cell has been designed that uses carbon monoxide as fuel. The overall reaction is $$ 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g) $$ The two half-cell reactions are $$ \begin{array}{l} \mathrm{CO}+\mathrm{O}^{2-} \longrightarrow \mathrm{CO}_{2}+2 \mathrm{e}^{-} \\\ \mathrm{O}_{2}+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{O}^{2-} \end{array} $$ The two half-reactions are carried out in separate compartments connected with a solid mixture of \(\mathrm{CeO}_{2}\) and \(\mathrm{Gd}_{2} \mathrm{O}_{3} . \mathrm{Ox}-\) ide ions can move through this solid at high temperatures (about \(\left.800^{\circ} \mathrm{C}\right) . \Delta G\) for the overall reaction at \(800^{\circ} \mathrm{C}\) under certain concentration conditions is \(-380 \mathrm{~kJ}\). Calculate the cell potential for this fuel cell at the same temperature and concentration conditions.
Consider the galvanic cell based on the following halfreactions: $$ \begin{array}{ll} \mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn} & \mathscr{E}^{\circ}=-0.76 \mathrm{~V} \\ \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe} & \mathscr{E}^{\circ}=-0.44 \mathrm{~V} \end{array} $$ a. Determine the overall cell reaction and calculate \(\mathscr{E}_{\text {cell }}^{\circ}\) 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 \(\left[\mathrm{Zn}^{2+}\right]=0.10 M\) and \(\left[\mathrm{Fe}^{2+}\right]=1.0 \times 10^{-5} M\)
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