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I was confused by this MCQ:

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The answer is A. I thought that the spontaneous reaction was usually between the most negative and least negative. Could someone please explain why this is the answer.

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3 Answers 3

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The important thing is to remember that for a formal redox half-reaction, oxidizing H2 to H+, there is relation of the standard reaction Gibbs energy and the standard redox potential

$$\Delta_r G^\circ = -nFE^\circ .$$

The lower is the potential, the stronger(thermodynamically, not necessarily kinetically) is the respective reductant and weaker the oxidant. And vice versa.

So $\ce{Mn(s)}$ is the strongest reductant, $\ce{Pb(s)}$ the weakest one.

So $\ce{Mn^2+(aq)}$ is the weakest oxidant, $\ce{Pb^2+(aq)}$ the strongest one.

From above must be clear that the only spontaneous reaction is :

$$\ce{Mn(s) + Fe^2+(aq) -> Mn^2+(aq) + Fe(s)}$$

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  • $\begingroup$ Thank you! However, it says that the answer is A, That's why I'm confused. Is it because this reaction is.not in the options? $\endgroup$ Commented Jan 26, 2023 at 14:18
  • $\begingroup$ Sorry, I have confused metals, but the result is the same. $\endgroup$
    – Poutnik
    Commented Jan 26, 2023 at 14:47
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You can also state that, when comparing two redox couples, the metal belonging to the redox couple with the most negative values will react. Here, comparing the three metals $\ce{Mn, Fe, Pb}$, the metallic $\ce{Mn}$ will always react and produce $\ce{Mn^{2+}}$, because of its highest negative value ($E° = - 1.18$ V) and $\ce{Pb}$ will never react: its redox potential is not negative enough.

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One of the easiest ways is to draw the reactions in the direction they occur. For example, considering:

$$\ce{Mn + Fe++ -> Mn++ + Fe}$$

$$\ce{Mn++ + 2e- -> Mn - 1.18 V Does not want 2 electrons}$$

Can be viewed as:

$$\ce{Mn -> Mn++ + 2e- + 1.18 V glad to give them away}$$


$$\ce{Fe++ + 2e- -> Fe. - 0.45 V happy without electrons, but not as much as Mn}$$

Then just add: 1.18 + (-0.45) = + 0.73 V positive, it will occur spontaneously. Electrons flow toward Fe++.

Do this with all the others, only A is positive. The sign (positive or negative) of the sum of the 2 half reactions shows whether or not it is spontaneous. This is how batteries work.

Most importantly, direction of electron flow is relative to the to metals involved, with most positive to most negative (as written!) producing the highest voltage.

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  • $\begingroup$ Sign changing of electrode potentials is something which is disliked by electrochemists. Please search previous discussions on this topic. The sign of the electrode potential is invariant. $\endgroup$
    – ACR
    Commented Jan 26, 2023 at 16:06
  • $\begingroup$ @AChem I have heard this, but it is a very valuable tool for this specific purpose. + favorable ... - unfavorable. Writing out the half reactions in the direction they run and simply summing makes it easy. Try it! $\endgroup$ Commented Jan 26, 2023 at 16:09
  • $\begingroup$ I often compare it to marking left/right river bank naively (on my left/right) or cartografically (on my left/right when looking downstream.) Similarly, marking cathodes/anodes where reduction/oxidation occurs(Phys chem.) vs where reduction/oxidation occur when in active galvanic mode (engineers at design of rechargable cells.) // For potentials, it is unlucky they did not established different formal electrochemical quantities. Perhaps using dual-sign redox potential versus single sign electrode potential. Having flipped signs may confuse many beginners, when confusing both. $\endgroup$
    – Poutnik
    Commented Jan 26, 2023 at 16:46
  • $\begingroup$ @Poutnik points very well taken. Downhill and uphill? But I am more into mechanism than math, which is why half reactions are done first. Rote and rhyme failing, Al metal gives, atomic chlorine takes. $\endgroup$ Commented Jan 26, 2023 at 18:03
  • $\begingroup$ Potentials (not limited to electrochemistry) should have IMHO a single sign, as property of a state, not direction of a process. The flipping nature of redox potentials is abusing of general terminology, unfortunately already burned into foundations of electrochemistry. $\endgroup$
    – Poutnik
    Commented Jan 26, 2023 at 18:23

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