Calculating Redox Potential; Coupled Reactions
$2~Ag + 2~H^+ ~(+2~Cl^{-}) \rightarrow 2~Ag^{+} + H_2 ~ ( + 2~Cl^{-})$

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 $\Delta E &lt; 0$