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andselisk ♦
andselisk ♦
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It would really help if you copied the entire half cell reaction. The atoms must balance in a valid half cell reaction! So there are:

$\ce{NO3−(aq) + 2 H+ +  e− <=> NO2(g) +  H2O}\quad\quad\quad\quad\mathrm{E}_0 = +0.80$

$\ce{Hg_2^{2+} + 2 e− <=> 2Hg(l)}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathrm{E}_0 = +0.80$

$\ce{Hg^{2+} + 2 e− <=> Hg(l)}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathrm{E}_0 = +0.85$

$\ce{2 Hg^{2+} + 2 e− <=> Hg2^{2+} }\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathrm{E}_0 = +0.91$

$\ce{NO3−(aq) + 4 H+ + 3 e− <=> NO(g) + 2 H2O(l)}\quad\quad\mathrm{E}_0 = +0.958$\begin{align} \ce{NO3− (aq) + 2 H+ + e− &<=> NO2 (g) + H2O} &\quad &E_0 = \pu{+0.80 V} \\ \ce{Hg2^2+ + 2 e− &<=> 2Hg (l)} &\quad &E_0 = \pu{+0.80 V} \\ \ce{Hg^2+ + 2 e− &<=> Hg (l)} &\quad &E_0 = \pu{+0.85 V} \\ \ce{2 Hg^2+ + 2 e− &<=> Hg2^2+} &\quad &E_0 = \pu{+0.91 V} \\ \ce{NO3−(aq) + 4 H+ + 3 e− &<=> NO (g) + 2 H2O (l)} &\quad &E_0 = \pu{+0.958 V} \\ \end{align}

The reduction of nitrate is interesting. In dissolving copper with nitric acid if the solution is highly acidic you get $\ce{NO}$, if the solution is mildly acid you get $\ce{NO2}$, and if the acidity is somewhere between you get both $\ce{NO}$ and $\ce{NO2}$.

Now if you also consider the Nernst equation, it is obvious that the concentrations of the species and the concentration of acid are important for calculating the half cell potentials. So it is impossible to say yes or no without additional information.

It would really help if you copied the entire half cell reaction. The atoms must balance in a valid half cell reaction! So there are:

$\ce{NO3−(aq) + 2 H+ +  e− <=> NO2(g) +  H2O}\quad\quad\quad\quad\mathrm{E}_0 = +0.80$

$\ce{Hg_2^{2+} + 2 e− <=> 2Hg(l)}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathrm{E}_0 = +0.80$

$\ce{Hg^{2+} + 2 e− <=> Hg(l)}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathrm{E}_0 = +0.85$

$\ce{2 Hg^{2+} + 2 e− <=> Hg2^{2+} }\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathrm{E}_0 = +0.91$

$\ce{NO3−(aq) + 4 H+ + 3 e− <=> NO(g) + 2 H2O(l)}\quad\quad\mathrm{E}_0 = +0.958$

The reduction of nitrate is interesting. In dissolving copper with nitric acid if the solution is highly acidic you get $\ce{NO}$, if the solution is mildly acid you get $\ce{NO2}$, and if the acidity is somewhere between you get both $\ce{NO}$ and $\ce{NO2}$.

Now if you also consider the Nernst equation, it is obvious that the concentrations of the species and the concentration of acid are important for calculating the half cell potentials. So it is impossible to say yes or no without additional information.

It would really help if you copied the entire half cell reaction. The atoms must balance in a valid half cell reaction! So there are:

\begin{align} \ce{NO3− (aq) + 2 H+ + e− &<=> NO2 (g) + H2O} &\quad &E_0 = \pu{+0.80 V} \\ \ce{Hg2^2+ + 2 e− &<=> 2Hg (l)} &\quad &E_0 = \pu{+0.80 V} \\ \ce{Hg^2+ + 2 e− &<=> Hg (l)} &\quad &E_0 = \pu{+0.85 V} \\ \ce{2 Hg^2+ + 2 e− &<=> Hg2^2+} &\quad &E_0 = \pu{+0.91 V} \\ \ce{NO3−(aq) + 4 H+ + 3 e− &<=> NO (g) + 2 H2O (l)} &\quad &E_0 = \pu{+0.958 V} \\ \end{align}

The reduction of nitrate is interesting. In dissolving copper with nitric acid if the solution is highly acidic you get $\ce{NO}$, if the solution is mildly acid you get $\ce{NO2}$, and if the acidity is somewhere between you get both $\ce{NO}$ and $\ce{NO2}$.

Now if you also consider the Nernst equation, it is obvious that the concentrations of the species and the concentration of acid are important for calculating the half cell potentials. So it is impossible to say yes or no without additional information.

fixed typo of Nernst
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MaxW
MaxW
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It would really help if you copied the entire half cell reaction. The atoms must balance in a valid half cell reaction! So there are:

$\ce{NO3−(aq) + 2 H+ +  e− <=> NO2(g) +  H2O}\quad\quad\quad\quad\mathrm{E}_0 = +0.80$

$\ce{Hg_2^{2+} + 2 e− <=> 2Hg(l)}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathrm{E}_0 = +0.80$

$\ce{Hg^{2+} + 2 e− <=> Hg(l)}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathrm{E}_0 = +0.85$

$\ce{2 Hg^{2+} + 2 e− <=> Hg2^{2+} }\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathrm{E}_0 = +0.91$

$\ce{NO3−(aq) + 4 H+ + 3 e− <=> NO(g) + 2 H2O(l)}\quad\quad\mathrm{E}_0 = +0.958$

The reduction of nitrate is interesting. In dissolving copper with nitric acid if the solution is highly acidic you get $\ce{NO}$, if the solution is mildly acid you get $\ce{NO2}$, and if the acidity is somewhere between you get both $\ce{NO}$ and $\ce{NO2}$.

Now if you also consider the NerstNernst equation, it is obvious that the concentrations of the species and the concentration of acid are important for calculating the half cell potentials. So it is impossible to say yes or no without additional information.

It would really help if you copied the entire half cell reaction. The atoms must balance in a valid half cell reaction! So there are:

$\ce{NO3−(aq) + 2 H+ +  e− <=> NO2(g) +  H2O}\quad\quad\quad\quad\mathrm{E}_0 = +0.80$

$\ce{Hg_2^{2+} + 2 e− <=> 2Hg(l)}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathrm{E}_0 = +0.80$

$\ce{Hg^{2+} + 2 e− <=> Hg(l)}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathrm{E}_0 = +0.85$

$\ce{2 Hg^{2+} + 2 e− <=> Hg2^{2+} }\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathrm{E}_0 = +0.91$

$\ce{NO3−(aq) + 4 H+ + 3 e− <=> NO(g) + 2 H2O(l)}\quad\quad\mathrm{E}_0 = +0.958$

The reduction of nitrate is interesting. In dissolving copper with nitric acid if the solution is highly acidic you get $\ce{NO}$, if the solution is mildly acid you get $\ce{NO2}$, and if the acidity is somewhere between you get both $\ce{NO}$ and $\ce{NO2}$.

Now if you also consider the Nerst equation, it is obvious that the concentrations of the species and the concentration of acid are important for calculating the half cell potentials. So it is impossible to say yes or no without additional information.

It would really help if you copied the entire half cell reaction. The atoms must balance in a valid half cell reaction! So there are:

$\ce{NO3−(aq) + 2 H+ +  e− <=> NO2(g) +  H2O}\quad\quad\quad\quad\mathrm{E}_0 = +0.80$

$\ce{Hg_2^{2+} + 2 e− <=> 2Hg(l)}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathrm{E}_0 = +0.80$

$\ce{Hg^{2+} + 2 e− <=> Hg(l)}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathrm{E}_0 = +0.85$

$\ce{2 Hg^{2+} + 2 e− <=> Hg2^{2+} }\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathrm{E}_0 = +0.91$

$\ce{NO3−(aq) + 4 H+ + 3 e− <=> NO(g) + 2 H2O(l)}\quad\quad\mathrm{E}_0 = +0.958$

The reduction of nitrate is interesting. In dissolving copper with nitric acid if the solution is highly acidic you get $\ce{NO}$, if the solution is mildly acid you get $\ce{NO2}$, and if the acidity is somewhere between you get both $\ce{NO}$ and $\ce{NO2}$.

Now if you also consider the Nernst equation, it is obvious that the concentrations of the species and the concentration of acid are important for calculating the half cell potentials. So it is impossible to say yes or no without additional information.

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MaxW
MaxW
  • 22.4k
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  • 38
  • 81

It would really help if you copied the entire half cell reaction. The atoms must balance in a valid half cell reaction! So there are:

$\ce{NO3−(aq) + 2 H+ +  e− <=> NO2(g) +  H2O}\quad\quad\quad\quad\mathrm{E}_0 = +0.80$

$\ce{Hg_2^{2+} + 2 e− <=> 2Hg(l)}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathrm{E}_0 = +0.80$

$\ce{Hg^{2+} + 2 e− <=> Hg(l)}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathrm{E}_0 = +0.85$

$\ce{2 Hg^{2+} + 2 e− <=> Hg2^{2+} }\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathrm{E}_0 = +0.91$

$\ce{NO3−(aq) + 4 H+ + 3 e− <=> NO(g) + 2 H2O(l)}\quad\quad\mathrm{E}_0 = +0.958$

The reduction of nitrate is interesting. In dissolving copper with nitric acid if the solution is highly acidic you get $\ce{NO}$, if the solution is mildly acid you get $\ce{NO2}$, and if the acidity is somesomewhere between you get both $\ce{NO}$ and    $\ce{NO2}$.

Now if you also consider the Nerst equation, it is obvious that the concentrations of the species and the concentration of acid are important for calculating the half cell potentials. So it is impossible to say yes or no without additional information.

It would really help if you copied the entire half cell reaction. The atoms must balance in a valid half cell reaction! So there are:

$\ce{NO3−(aq) + 2 H+ +  e− <=> NO2(g) +  H2O}\quad\quad\quad\quad\mathrm{E}_0 = +0.80$

$\ce{Hg_2^{2+} + 2 e− <=> 2Hg(l)}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathrm{E}_0 = +0.80$

$\ce{Hg^{2+} + 2 e− <=> Hg(l)}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathrm{E}_0 = +0.85$

$\ce{2 Hg^{2+} + 2 e− <=> Hg2^{2+} }\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathrm{E}_0 = +0.91$

$\ce{NO3−(aq) + 4 H+ + 3 e− <=> NO(g) + 2 H2O(l)}\quad\quad\mathrm{E}_0 = +0.958$

The reduction of nitrate is interesting. In dissolving copper with nitric acid if the solution is highly acidic you get $\ce{NO}$, if the solution is mildly acid you get $\ce{NO2}$, and if the acidity is some between you get both $\ce{NO}$ and  $\ce{NO2}$.

Now if you also consider the Nerst equation, it is obvious that the concentrations of the species and the concentration of acid are important for calculating the half cell potentials. So it is impossible to say yes or no without additional information.

It would really help if you copied the entire half cell reaction. The atoms must balance in a valid half cell reaction! So there are:

$\ce{NO3−(aq) + 2 H+ +  e− <=> NO2(g) +  H2O}\quad\quad\quad\quad\mathrm{E}_0 = +0.80$

$\ce{Hg_2^{2+} + 2 e− <=> 2Hg(l)}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathrm{E}_0 = +0.80$

$\ce{Hg^{2+} + 2 e− <=> Hg(l)}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathrm{E}_0 = +0.85$

$\ce{2 Hg^{2+} + 2 e− <=> Hg2^{2+} }\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathrm{E}_0 = +0.91$

$\ce{NO3−(aq) + 4 H+ + 3 e− <=> NO(g) + 2 H2O(l)}\quad\quad\mathrm{E}_0 = +0.958$

The reduction of nitrate is interesting. In dissolving copper with nitric acid if the solution is highly acidic you get $\ce{NO}$, if the solution is mildly acid you get $\ce{NO2}$, and if the acidity is somewhere between you get both $\ce{NO}$ and  $\ce{NO2}$.

Now if you also consider the Nerst equation, it is obvious that the concentrations of the species and the concentration of acid are important for calculating the half cell potentials. So it is impossible to say yes or no without additional information.

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MaxW
MaxW
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