Why would a sacrificial anode made of lithium metal be a bad choice despite its \({\bf{E}}_{{\bf{Li}}}^{\bf{^\circ }}{\bf{ + Li = - 3}}{\bf{.04\;V}}\), which appears to be able to protect all the other metals listed in the standard reduction potential table?

Short Answer

Expert verified

Lithium is a very reactive metal.

Step by step solution

01

Sacrificial anode made of lithium metal

  • Cathodic protection is the method of protecting metal by utilising a sacrificial anode and making the metal the cathode.
  • As a result, it would be protected against rusting.
  • Metals with a lower standard reduction potential are more prone to corrosion.
02

Step 2:Finding the cause of a lithium metal sacrificial anode is a bad idea:

  • Even though Lithium has a very low reduction potential and is therefore more likely to corrode than most metals, it would not be a good choice for a sacrificial anode.
  • The reason behind this is that lithium is a very reactive metal and it would enter a most likely very exothermic chemical reaction before it would even have the chance to corrode.
  • This means that the lithium would react swiftly with other compounds, even if they did not oxidise the metal it was seeking to protect.
  • With such reactivity, the sacrificial anode would be rapidly depleted and would need to be changed on a regular basis.

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Most popular questions from this chapter

Aluminium\(\left( {{\bf{E}}_{{\bf{A}}{{\bf{l}}^{{\bf{3 + }}}}{\bf{/Al}}}^{\bf{^\circ }}{\bf{ = - 2}}{\bf{.07\;V}}} \right)\) is more easily oxidized than iron \(\left( {{\bf{E}}_{{\bf{F}}{{\bf{e}}^{\bf{3}}}}^{\bf{^\circ }}{\bf{/F}}{{\bf{e}}^{\bf{ - }}}{\bf{ = - 0}}{\bf{.477\;V}}} \right){\bf{,}}\) and yet when both are exposed to the environment, untreated aluminium has very good corrosion resistance while the corrosion resistance of untreated iron is poor. Explain this observation.

Identify the species that undergoes oxidation, the species that undergoes reduction, the oxidizing agent, and thereducing agent in each of the reactions of the previous problem.

\(\begin{array}{l}{\bf{(a) H_2O_2 + S}}{{\bf{n}}^{{\bf{2 + }}}} \to {\bf{H_2O + S}}{{\bf{n}}^{{\bf{4 + }}}}\\{\bf{(b) PbO_2 + Hg}} \to {\bf{Hg}}{{\bf{2}}^{{\bf{2 + }}}}{\bf{ + P}}{{\bf{b}}^{{\bf{2 + }}}}\\{\bf{(c) Al + Cr_2O}}{{\bf{7}}^{{\bf{2 - }}}} \to {\bf{A}}{{\bf{l}}^{{\bf{3 + }}}}{\bf{ + C}}{{\bf{r}}^{{\bf{3 + }}}}\end{array}\)

Given the following cell notations, determine the species oxidized, species reduced, and the oxidizing agent and reducing agent, without writing the balanced reactions

a. \({\rm{Mg}}(s)\left| {{\rm{M}}{{\rm{g}}^{2 + }}(aq) || {\rm{C}}{{\rm{u}}^{2 + }}(aq)} \right|{\rm{Cu}}(s)\)

b.\({\rm{Ni}}(s)\left| {{\rm{N}}{{\rm{i}}^{2 + }}(aq) || {\rm{A}}{{\rm{g}}^ + }(aq)} \right|{\rm{Ag}}(s)\)

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) \({\bf{L}}{{\bf{i}}^{\bf{ + }}}{\bf{ + }}{{\bf{e}}^{\bf{ - }}} \to {\bf{Li}}\)

For the standard cell potentials given here, determine the \(\Delta {G\circ }\) for the cell in \({\rm{kJ}}\).

(a) \(0.000\;{\rm{V}},{\rm{n}} = 2\)

(b) \( + 0.434\;{\rm{V}},{\rm{n}} = 2\)

(c) \( - 2.439\;{\rm{V}},{\rm{n}} = 1\)

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