Use the following half-reactions to write three spontaneous reactions, calculate E°cell for each reaction, and rank the strengths of the oxidizing and reducing agents:Au⁺(aq) + e⁻ → Au(s) E=1.69V

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E°cell for each reaction, and rank the strengths of the oxidizing and reducing agents:

Complete the given reaction

[tex]$\mathrm{N}_{2} \mathrm{O}(g)+2 \mathrm{H}^{+}(a q)+2 \mathrm{Au}(s) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{Au}^{+}(a q)$[/tex]

Write the oxidation reaction

[tex]$\mathrm{Au}(s) \longrightarrow \mathrm{Au}^{+}(a q)+e^{-}$[/tex]

[tex]$E_{a c d}=-1.69 \mathrm{~V}$[/tex]

Write the reduction reaction

[tex]$3 \mathrm{~N}_{2} \mathrm{O}(g)+2 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \quad E_{r e d}=+1.77 \mathrm{~V}$[/tex]

Calculate the standard cell voltage

[tex]$E_{c ell}=E_{r e d}+E_{o x d}$[/tex]

Substitute the value

[tex]$\begin{aligned}&E_{c e l l}=1.77-1.69 \\&E_{c e l l}=0.08 \mathrm{~V}\end{aligned}$[/tex]

Hence the standard cell potential is [tex]$0.08 \mathrm{~V}$[/tex]

Oxidizing agent in this reaction is

[tex]$\mathrm{N}_{2}(g), \mathrm{N}_{2} \mathrm{O}(g)$[/tex]

Reducing agent in this reaction is

[tex]$\mathrm{Au}^{+}(a q), \mathrm{Au}(s)$[/tex]

The potential difference between the two electrodes of an electrochemical cell, which develops when electrons are sent via the external circuit of a cell that has not reached equilibrium, is known as the cell potential.

A chemical process known as an oxidation-reduction (redox) reaction includes the exchange of electrons between two substances. Any chemical reaction in which the oxidation number of a molecule, atom, or ion changes due to the gain or loss of an electron is referred to as an oxidation-reduction reaction.

Learn more about oxidation reaction

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