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Oxidation and reductionNational 2016

What is the standard reduction potential for $\ce{Cr^2+(aq)}$ to $\ce{Cr(s)}$ ?

Half-Reaction | $E^\circ$ , V
$\ce{Cr^3+(aq) + 3e- -> Cr(s)}$ | –0.73
$\ce{Cr^3+(aq) + e- -> Cr^2+(aq)}$ | –0.50

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Explanation

draft

To combine two half-reactions into a third, you must add their Gibbs free energies ( $\Delta G = -nFE^{\circ}$ ), not simply add or average the potentials.

Step 1 — Write ΔG for each given half-reaction:

$\ce{Cr^3+ + 3e- -> Cr}$ , $E^{\circ}=-0.73\text{ V}$ , so $\Delta G_1 = -3F(-0.73)=2.19F$

$\ce{Cr^3+ + e- -> Cr^2+}$ , $E^{\circ}=-0.50\text{ V}$ , so $\Delta G_2 = -1F(-0.50)=0.50F$

Step 2 — Get the target reaction by subtraction:

$\ce{Cr^2+ + 2e- -> Cr}$ = (reaction 1) − (reaction 2), so $\Delta G_3 = \Delta G_1 - \Delta G_2 = 2.19F - 0.50F = 1.69F$

Step 3 — Convert back to $E^{\circ}$ using $n=2$ :

$E^{\circ}_3 = -\dfrac{\Delta G_3}{nF} = -\dfrac{1.69F}{2F} = -0.845\text{ V} \approx -0.85\text{ V}$

So the standard reduction potential for $\ce{Cr^2+(aq) + 2e- -> Cr(s)}$ is −0.85 V.

Comes from subtracting the two $E^{\circ}$ values directly ( $-0.73-(-0.50)=-0.23$ ) instead of converting to $\Delta G$ first — potentials aren't additive/subtractive like free energies.

Correct: converts each $E^{\circ}$ to $\Delta G=-nFE^{\circ}$ , combines by subtraction (since the target reaction is reaction 1 minus reaction 2), then divides by the correct $n=2$ to get $E^{\circ}=-0.85$ V.

Adds the two potentials directly ( $-0.73+(-0.50)=-1.23$ ), mixing up sign/operation and ignoring that $E^{\circ}$ values can't simply be summed across reactions with different electron counts.

Correctly computes $\Delta G_3=1.69F$ but forgets to divide by $n=2$ when converting back to $E^{\circ}$ , giving $-1.69$ V instead of $-0.85$ V.

Concept · Combining half-reactions via Gibbs free energy (Hess's law for electrochemical potentials)