Does silver oxidise in a pH 1 solution? Consider submerging a plate of pure silver in a solution with $\mathrm{pH} = 1$. Does the redox reaction with hydrogen happen spontaneously?
  
  The composition of the solution is not given.


My attempt:


I considered the standard potential for $\mathrm{pH} = 0$. Since the silver must be oxidised, the reaction happens at the anode.


$$
\begin{align}
\ce{Ag+ + e- &amp;-&gt; Ag} &amp;\qquad E^\circ &amp;= \pu{0.80 V} \\
\ce{2 H+ + e- &amp;-&gt; H2} &amp;\qquad E^\circ &amp;= \pu{0.00 V}
\end{align}
$$


At this point, it is obvious that the reaction cannot be spontaneous since the potential for this cell is less than zero. In fact:


$$\Delta E = E_\mathrm{red} - E_\mathrm{ox} = \pu{-0.80 V}$$


From Nernst equation, the potential for the hydrogen at $\mathrm{pH}=1$ must be negative. Thus the reaction is all the more non spontaneous ($E = \pu{-0.059 V}$).


However, this seems to be wrong. I must be applying the $\Delta E$ wrong, because the solution for this problem states clearly that it is given by:

$$\Delta E = E_\mathrm{red} - E_\mathrm{ox} = \pu{-0.059 V} - (\pu{-0.80 V}) = \pu{0.741 V}$$

This is driving me crazy. Where does the minus sign come from?
The answer given in the solution was wrong. It has been corrected, and it eventually proved my point.