# Why is this reaction spontaneous in standard conditions?

To study if this reaction is spontaneous I proceed in this way: $$\ce{2H_{3}O^{+} +Fe ->H_{2} +Fe^{2+} + 2H_{2}O}$$ First I write the two semireaction: $$reduction: \ \ \ce{2H_{3}O^{+} +2e^{-} -> H_{2}}$$ $$oxidation: \ \ \ce{Fe -> 2e^{-} +Fe^{+}}$$ Then I write the scheme of the cell: $$\ce{Fe|Fe^{2+}||H_{3}O^{+}|H_{2}}$$ I know that: $$E^{0}_{cell}=E^{0}_{right}-E^{0}_{left}$$ $$E^{0}_{cell}=E^{0}_{H_{3}O^{+}|H_{2}}-E^{0}_{Fe|Fe^{2+}}=-E^{0}_{Fe|Fe^{2+}}=-0.44$$ But also: $$\Delta G^{0}=-nFE^{0}_{cell}$$ So $$\Delta G^{0}>0$$ And the reaction is not spontaneous, while my textbook says it is, so I must be wrong. Someone could find my error?

• Note that it is preferred not to use MarhJax/Latex formatting of titles on Chemistry SE site to avoid search problems. – Poutnik Dec 29 '19 at 16:55

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.

$$\ce{2H3O+ + Fe -> H2 + Fe^{2+} +2H2O}\tag{1}$$

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

$$reduction: \ \ \ce{2H3O+ + 2e− ->H2}\tag{2}$$

$$oxidation: \ \ \ce{Fe -> 2e^{-} +Fe^{2+}}\tag{3}$$ 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:

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

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

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

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

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

The main mistake is rather algebraic and forgetting the electrochemical conventions. When we write this equation, we follow the convention that all electrode potential values will be inserted right from the table with their signs intact, whether it is an oxidation or reduction. Don't change the sign of the electrode potential value. The sign of the electrode potential is invariant.

$$E^{0}_{cell}=E^{0}_{right}-E^{0}_{left}$$

So from the table, the Fe(II)/Fe(0) couple is -0.44 V

So, $$E^{0}_{cell}=0.00-(-0.44)= +0.44V$$

when $$E^{0}_{cell}$$ is positive, delta G value is negative.

• I do believe that this is the way that is used in current teaching. The "improvement" is to get the sign change, $-(-0.44)$ on paper rather than doing it mentally. – MaxW Dec 29 '19 at 18:22
• In old school teaching, the equation was E0cell=E0right+E0left, and the student was asked to change the sign. This was the Gibb's or American convention. Rest of the Europe followed Ostwald's approach, which is taught today. – M. Farooq Dec 29 '19 at 18:44
• LOL - My first chemistry class was in 1968 when I was in the 10th grade. "Old school" enough? – MaxW Dec 30 '19 at 1:01
• That is exactly the time I was talking about. Glad that you saw the golden era of science in the US. – M. Farooq Dec 30 '19 at 1:32

Your mistake is that $$E^\circ_\ce{Fe/Fe^{2+}} = \pu{-0.44 V}.$$ So that $$-E^\circ_\ce{Fe/Fe^{2+}} = \pu{+0.44 V}.$$

• no... $E°_{Fe^{2+}/Fe}$ = - $0.44$ V. The table values are always written as reductions. – MaxW Dec 29 '19 at 17:09