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"?

  • 1
    $\begingroup$ 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 Jun 15 '14 at 11:41

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

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 $\Delta E < 0$

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.