Why is this reaction spontaneous in standard conditions?
It has been a while since I took electrochemistry so I'm not sure about the formal procedure that you're using. 

You wrote the correct overall reaction.

$$\ce{2H3O+ + Fe -&gt; H2 + Fe^{2+} +2H2O}\tag{1}$$

and you almost wrote the correct half cells reactions (should be $\ce{Fe^{2+}}$ in equation 3):

$$reduction:  \ \ \ce{2H3O+ + 2e− -&gt;H2}\tag{2}$$

$$oxidation: \ \ \ce{Fe -&gt; 2e^{-} +Fe^{2+}}\tag{3} $$ 
Then you wrote the scheme of the cell: $$\ce{Fe|Fe^{2+}||H_{3}O^{+}|H_{2}}\tag{4}$$ And an equation for the overall cell potential. 
$$E^{0}_{cell}=E^{0}_{right}-E^{0}_{left}\tag{5}$$ 

Your problem is between equations 4 and 5. To subtract $E^{0}_{left}$ the potential must be written as the reduction potential for the half cell. So either you're supposed to be:


Adding the values for the right and left reactions in equation 5


-- or-- 


Writing both half cells as reduction potentials in equation 4. 




The gist here is that the tabular values are always written as reduction potentials. So for iron:

$$\ce{2e^{-} +Fe^{2+} -&gt; Fe }\quad\quad E^{0}_{red} = -0.44\tag{6} $$ 

I was taught to reverse the sign for an oxidation reaction. 

$$\ce{Fe -&gt; 2e^{-} +Fe^{2+}}\quad\quad E^{0}_{ox} = 0.44\tag{7} $$ 

Then add the reduction and oxidation potentials to get the overall cell potential. 

$$E^{0}_{cell}=E^{0}_{red} + E^{0}_{ox}\tag{8}$$