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For the reaction of silver and hydrochloric acid to form silver chloride: would the following be the correct way to couple the below two half reactions?

$\ce{2Ag(s) + 2Cl^- ->2AgCl(s) + 2e^-} ~~E^0 =-0.222 ~V$

$\ce{2e^- + 2H_3^+O -> H_2 + H_2O}{ ~~E^0 =0.000~ V}$

Add the above:

$\ce{2H_3^+O +2Ag(s) + 2Cl^- -> 2AgCl(s) + H_2 + H_2O}{ ~~E^0 =-0.222~ V}$

Implying that the reaction is non-spontaneous as we have a "negative potential"?

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    $\begingroup$ Just for future reference, standard electrode potentials will not always determine correctly whether a reaction will happen or not, because most reactions are not performed in standard conditions. The proper course of action is to perform some correction by taking into account non-standard conditions, often with the Nernst equation, and then analyze the reaction free energy change. $\endgroup$
    – Nicolau Saker Neto
    Commented Jun 15, 2014 at 11:41

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$2~Ag + 2~H^+ ~(+2~Cl^{-}) \rightarrow 2~Ag^{+} + H_2 ~ ( + 2~Cl^{-})$

As two half-reactions, it is seen that the silver is oxidized:

$2~Ag \rightarrow 2~Ag^{+} +2~e^- ~~ E^{\circ}= 0.80\,V$

And the hydrogen is reduced:

$2~H^+ + 2~e^- \rightarrow H_2 ~~ E^{\circ}= 0.0\,V$

To calculate the reduction potential you have to solve the following equation:

$\Delta E = E_{acceptor} - E_{donor} = E_{2H^{+}/H_2}- E_{Ag/Ag^{+}}$

$\Delta E = 0\,V - (0.80\,V) = -0.80\,V$

The reactions doesn't run voluntarily if $\Delta E < 0$

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