Redox and electrochemistry — Question 1

b)

i) 4H⁺(aq) + 2NO₃

⁻(aq) + 2e⁻→2NO₂(g) + 2H₂O(l)

• provides 2H⁺(aq) + e⁻→H₂O(l) [1 mark]

• provides NO₃

⁻(aq) + e⁻→NO₂(g) [1 mark]

Acceptable response is 2H⁺(aq) + NO₃

⁻(aq) +

e⁻→NO₂(g) + H₂O(l)

$\mathrm{i}\mathrm{i})\,\mathrm{E}^{\mathrm{o}}_{\mathrm{red}}\,=\,0.46\,+\,0.34\,=\,+0.80\,\mathrm{V}$

$•\,\mathrm{d}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{e}\mathrm{s}\,\mathrm{E}^{\mathrm{o}}_{\mathrm{red}}\,=$

$+0.80\,\mathrm{V}\,[1\,\mathrm{m}\mathrm{a}\mathrm{r}\mathrm{k}]$

c)

i) For hydrochloric acid: $\mathrm{E}^{\mathrm{o}}_{\mathrm{cell}}=\mathrm{E}^{\mathrm{o}}_{\mathrm{red}}-\mathrm{E}^{\mathrm{o}}_{\mathrm{ox}}= +0.00 - (+0.34) = -0.34$ Reaction is non-spontaneous, therefore HCl cannot dissolve Cu. For nitric acid: E^o_cell = +0.46 (positive), therefore the reaction is spontaneous. HNO₃ can dissolve Cu.

• determines E^o_cell for HCl equals –0.34 and E^o_cell for HNO₃ equals +0.46 V [1 mark]

• determines reaction between Cu and HCl is not spontaneous and therefore Cu will not dissolve [1 mark]

• indicates reaction between Cu and HNO₃ is spontaneous and therefore Cu will dissolve [1 mark]

Acceptable response is Cl⁻ is more negative, therefore stronger reducing and will be oxidised in preference to Cu.

ii) Reduction:

⁻(aq) + 2e⁻→2NO₂(g) + 2H₂O(l), E^o = +0.80V or

4H⁺(aq) + 2NO₃ H⁺(aq) + 2e⁻→ H₂(g), E^o = 0.00 V The E^o value for NO₃

⁻ is more positive than H⁺(aq), therefore NO₃

⁻ is a

stronger oxidising agent.

Therefore NO₃

⁻ reduced in preference to H⁺ and NO₂(g) formed

• identifies that, in HNO₃, H⁺(aq) and ^– compete to be NO₃ reduced [1 mark]

• indicates that NO₃

⁻ is

stronger oxiding agent [1 mark]

• determines NO₃

⁻ is

preferentially reduced therefore NO₂(g) formed [1 mark]

Acceptable response is NO₃

⁻ is more positive

than Cu, therefore stronger oxidising agent and can be reduced to oxidise Cu to Cu²⁺(aq).