Chapter 17: Q39E (page 961)
Why do batteries go dead, but fuel cells do not?
Fuel cells do not go dead because they use chemical energy to produce electrical energy.
Over 30 million students worldwide already upgrade their learning with Vaia!
All the tools & learning materials you need for study success - in one app.
Get started for free
An inventor proposes using a SHE (standard hydrogen electrode) in a new battery for smartphones that also removes toxic carbon monoxide from the air:
Anode:\({\bf{CO(g) + }}{{\bf{H}}_{\bf{2}}}{\bf{O(l)}} \to {\bf{C}}{{\bf{O}}_{\bf{2}}}{\bf{(g) + 2}}{{\bf{H}}^{\bf{ + }}}{\bf{(aq) + 2}}{{\bf{e}}^{\bf{ - }}}\;\;\;{\bf{E}}_{{\bf{anode }}}^{\bf{^\circ }}{\bf{ = - 0}}{\bf{.53\;V}}\)
Cathode:\({\bf{2}}{{\bf{H}}^{\bf{ + }}}{\bf{(aq) + 2}}{{\bf{e}}^{\bf{ - }}} \to {{\bf{H}}_{\bf{2}}}{\bf{(g)}}\;\;\;{\bf{E}}_{{\bf{cathode }}}^{\bf{^\circ }}{\bf{ = 0\;V}}\)
___________________________________________________________
Overall:\({\bf{CO(g) + }}{{\bf{H}}_{\bf{2}}}{\bf{O(l)}} \to {\bf{C}}{{\bf{O}}_{\bf{2}}}{\bf{(g) + }}{{\bf{H}}_{\bf{2}}}{\bf{(g)}}\;\;\;{\bf{E}}_{{\bf{cell }}}^{\bf{^\circ }}{\bf{ = + 0}}{\bf{.53\;V}}\)
Would this make a good battery for smartphones? Why or why not?
A galvanic cell consists of a Mg electrode in \({\bf{1M}}\)\({\bf{Mg}}{\left( {{\bf{N}}{{\bf{O}}_{\bf{3}}}} \right)_{\bf{2}}}\)solution and a Ag electrode in 1M AgNO solution. Calculate the standard cell potential at \({25^\circ }{\rm{C}}\).
Consider the following metals: Ag, Au, \(Mg, Ni,\)\(and\)\(Zn\). Which of these metals could be used as a sacrificial anode in the cathodic protection of an underground steel storage tank? Steel is mostly iron, so use \( - 0.447\;{\rm{V}}\) as the standard reduction potential for steel.
For the \(\Delta {G\circ }\) values given here, determine the standard cell potential for the cell.
(a) \(12\;{\rm{kJ}}/{\rm{mol}},{\rm{n}} = 3\)
(b) \( - 45\;{\rm{kJ}}/{\rm{mol}},{\rm{n}} = 1\)
For each of the following balanced half-reactions, determine whether an oxidation or reduction is occurring.
(a) \({\bf{F}}{{\bf{e}}^{{\bf{3 + }}}}{\bf{ + 3}}{{\bf{e}}^{\bf{ - }}} \to {\bf{Fe}}\)
(b) \({\bf{Cr}} \to {\bf{C}}{{\bf{r}}^{{\bf{3 + }}}}{\bf{ + 3}}{{\bf{e}}^{\bf{ - }}}\)
(c) \({\bf{MnO}}_{\bf{4}}^{{\bf{2 - }}} \to {\bf{MnO}}_{\bf{4}}^{\bf{ - }}{\bf{ + }}{{\bf{e}}^{\bf{ - }}}\)
(d) \({\bf{L}}{{\bf{i}}^{\bf{ + }}}{\bf{ + }}{{\bf{e}}^{\bf{ - }}} \to {\bf{Li}}\)
What do you think about this solution?
We value your feedback to improve our textbook solutions.
