Chapter 18: Problem 13
If the cell potential is proportional to work and the standard reduction potential for the hydrogen ion is zero, does this mean that the reduction of the hydrogen ion requires no work?
Chapter 18: Problem 13
If the cell potential is proportional to work and the standard reduction potential for the hydrogen ion is zero, does this mean that the reduction of the hydrogen ion requires no work?
All the tools & learning materials you need for study success - in one app.
Get started for freeGold is produced electrochemically from an aqueous solution of \(\mathrm{Au}(\mathrm{CN})_{2}^{-}\) containing an excess of \(\mathrm{CN}^{-}\). Gold metal and oxygen gas are produced at the electrodes. What amount (moles) of \(\mathrm{O}_{2}\) will be produced during the production of \(1.00 \mathrm{~mole}\) of gold?
Consider the cell described below: $$ \mathrm{Al}\left|\mathrm{Al}^{3+}(1.00 M)\right|\left|\mathrm{Pb}^{2+}(1.00 M)\right| \mathrm{Pb} $$ Calculate the cell potential after the reaction has operated long enough for the \(\left[\mathrm{Al}^{3+}\right]\) to have changed by \(0.60 \mathrm{~mol} / \mathrm{L} .\) (Assume \(T=25^{\circ} \mathrm{C}\).)
Sketch the galvanic cells based on the following half-reactions. Show the direction of electron flow, show the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced equation, and determine \(\mathscr{E}^{\circ}\) for the galvanic cells. Assume that all concentrations are \(1.0 M\) and that all partial pressures are \(1.0 \mathrm{~atm}\). a. \(\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.78 \mathrm{~V}\) \(\mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O}_{2}\) \(\mathscr{C}^{\circ}=0.68 \mathrm{~V}\) \(\begin{array}{ll}\text { b. } \mathrm{Mn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn} & \mathscr{E}^{\circ}=-1.18 \mathrm{~V} \\ \mathrm{Fe}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} & \mathscr{E}^{\circ}=-0.036 \mathrm{~V}\end{array}\)
A solution containing \(\mathrm{Pt}^{4+}\) is electrolyzed with a current of \(4.00 \mathrm{~A}\). How long will it take to plate out \(99 \%\) of the platinum in \(0.50 \mathrm{~L}\) of a \(0.010-M\) solution of \(\mathrm{Pt}^{4+}\) ?
Consider a concentration cell that has both electrodes made of some metal M. Solution A in one compartment of the cell contains \(1.0 M \mathrm{M}^{2+}\). Solution B in the other cell compartment has a volume of \(1.00 \mathrm{~L}\). At the beginning of the experiment \(0.0100\) mole of \(\mathrm{M}\left(\mathrm{NO}_{3}\right)_{2}\) and \(0.0100\) mole of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) are dissolved in solution \(\mathrm{B}\) (ignore volume changes), where the reaction $$ \mathrm{M}^{2+}(a q)+\mathrm{SO}_{4}{ }^{2-}(a q) \rightleftharpoons \mathrm{MSO}_{4}(s) $$ occurs. For this reaction equilibrium is rapidly established, whereupon the cell potential is found to be \(0.44 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\). Assume that the process $$ \mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M} $$ has a standard reduction potential of \(-0.31 \mathrm{~V}\) and that no other redox process occurs in the cell. Calculate the value of \(K_{\text {sp }}\) for \(\operatorname{MSO}_{4}(s)\) at \(25^{\circ} \mathrm{C}\).
What do you think about this solution?
We value your feedback to improve our textbook solutions.