Exercises — Equilibrium constant from E° - ln K = nFE° - RT

Sign first. $E^{\circ }$ is positive $\Rightarrow$ the reaction is "downhill" from standard conditions $\Rightarrow$ products favoured $\Rightarrow$ $K>1$. (Look at the map figure: positive $E^{\circ }$ sits on the right, above the $K=1$ line.)

Number. ${\log }_{10}K=\frac{n{E}^{\circ }}{0.0592}=\frac{2\times 0.34}{0.0592}=\frac{0.68}{0.0592}=11.49$ So $K\approx 10^{11.5}\approx 3.1\times 10^{11}$. A large positive answer — consistent with "products favoured".

$\ln K=\frac{(1)(96485)(0.10)}{(8.314)(298.15)}=\frac{9648.5}{2478.8}=3.892$ Then $K=e^{3.892}\approx 49$. Notice $\ln K=38.92n{E}^{\circ }=38.92(1)(0.10)=3.892$ — the shortcut and the raw formula agree.

Step 1 — count electrons. Half-reactions: $\text{Zn}\to {\text{Zn}}^{2+}+2e^{-}$ and ${\text{Cu}}^{2+}+2e^{-}\to \text{Cu}$. Two electrons cross for one formula unit, so ==$n=2$==.

Step 2 — apply the shortcut. ${\log }_{10}K=\frac{n{E}^{\circ }}{0.0592}=\frac{2\times 1.10}{0.0592}=37.16$

Step 3 — read $K$. $K=10^{37.16}\approx 1.4\times 10^{37}$.

Sanity check: huge positive $E^{\circ }$ $\Rightarrow$ huge $K$ $\Rightarrow$ reaction runs essentially to completion (zinc dissolves, copper plates out). Matches the far-right of the map figure.

Rearrange the bridge: $E^{\circ }=\frac{RT}{nF}\ln K$.

$\ln K=\ln (1.0\times 10^8)=8\ln 10=18.42$ $E^{\circ }=\frac{(8.314)(298.15)(18.42)}{(2)(96485)}=\frac{45663}{192970}=0.237\text{V}$

Check with the shortcut: $E^{\circ }=\frac{0.0592{\log }_{10}K}{n}=\frac{0.0592\times 8}{2}=0.237\text{V}$. ✓

Step 1 — shortcut, keep the minus sign. ${\log }_{10}K=\frac{n{E}^{\circ }}{0.0592}=\frac{(1)(-0.03)}{0.0592}=-0.507$

Step 2 — undo with $10^{()}$. $K=10^{-0.507}\approx 0.31$

Interpretation: $K<1$ $\Rightarrow$ reactants (${\text{Fe}}^{3+}$, Ag) are favoured. At equilibrium $\frac{[{\text{Fe}}^{2+}][{\text{Ag}}^{+}]}{[{\text{Fe}}^{3+}]}\approx 0.31$: mostly unreacted ${\text{Fe}}^{3+}$ remains. On the map, $E^{\circ }$ just left of zero pulls $K$ just below 1 — a shallow slide, not a cliff.

$E^{\circ }$ is an intensive property — it does not change when you scale the equation (it's volts, energy per coulomb, not total energy). But $n$ halves.

$\log K_2=\frac{n_2{E}^{\circ }}{0.0592}=\frac{(1)(1.10)}{0.0592}=18.58=\frac{1}{2}\log K_1$

So $\log K_2=\frac{1}{2}\log K_1$, meaning $K_2=K_1^{1/2}=\sqrt{K_1}$. Numerically $K_2=10^{18.58}\approx 3.8\times 10^{18}$, and indeed $(3.8\times 10^{18}{)}^2\approx 1.4\times 10^{37}=K_1$. ✓ Lesson: scaling a reaction raises $K$ to that power; $E^{\circ }$ stays fixed. This is the deep reason $n$ must ride along.

You may not feed a non-standard $E_{\text{cell}}$ into the bridge — that formula demands $E^{\circ }$. First recover $E^{\circ }$ with the Nernst Equation: $E_{\text{cell}}=E^{\circ }-\frac{0.0592}{n}{\log }_{10}Q$

(a) Solve for $E^{\circ }$: $E^{\circ }=E_{\text{cell}}+\frac{0.0592}{n}{\log }_{10}Q=0.150+\frac{0.0592}{2}{\log }_{10}(10^{-4})$ $E^{\circ }=0.150+(0.0296)(-4)=0.150-0.1184=0.0316\text{V}$

(b) Now the bridge is legal: ${\log }_{10}K=\frac{n{E}^{\circ }}{0.0592}=\frac{2\times 0.0316}{0.0592}=1.068\Rightarrow K\approx 10^{1.068}\approx 11.7$

Interpretation: a modest $K$ (~12), products slightly favoured. The measured $E_{\text{cell}}$ was inflated above $E^{\circ }$ because $Q<1$ (few products yet) pushed the cell harder — Le Chatelier in electrical clothing, see Le Chatelier's Principle.

Step 1 — $n$. ${\text{MnO}}_4^{-}$ (Mn $+7\to +2$) gains $5e^{-}$; five ${\text{Fe}}^{2+}$ each lose $1e^{-}$ = five electrons out. Balanced $\Rightarrow$ ==$n=5$==.

Step 2 — $E_{\text{cell}}^{\circ }$ = cathode − anode. Permanganate is reduced (cathode), iron is oxidised (anode): $E_{\text{cell}}^{\circ }=1.51-0.77=0.74\text{V}$

Step 3 — $K$. ${\log }_{10}K=\frac{n{E}^{\circ }}{0.0592}=\frac{5\times 0.74}{0.0592}=62.5\Rightarrow K\approx 10^{62.5}$

Interpretation: an astronomical $K$ — permanganate titrations of ${\text{Fe}}^{2+}$ go essentially to completion, which is exactly why they make sharp, reliable titrations.

Treat $K_{sp}$ as the $K$ in the bridge with $n=1$: $E^{\circ }=\frac{0.0592}{n}{\log }_{10}{K}_{sp}=0.0592{\log }_{10}(1.8\times 10^{-10})$ ${\log }_{10}(1.8\times 10^{-10})={\log }_{10}1.8-10=0.2553-10=-9.745$ $E^{\circ }=0.0592\times (-9.745)=-0.577\text{V}$

Interpretation: the strongly negative $E^{\circ }$ mirrors the tiny $K_{sp}$ — dissolving AgCl is heavily uphill, so almost none dissolves (AgCl is famously insoluble). The bridge turns a solubility number into a voltage and back. This is the same machinery behind Thermodynamics of Electrochemical Cells.

The $0.0592$ constant is $2.303RT/F$ evaluated at 298 K only — it is wrong at any other temperature, so we must use the raw formula.

At $298.15\text{K}$: $\ln K_{298}=\frac{(1)(96485)(0.25)}{(8.314)(298.15)}=\frac{24121}{2478.8}=9.73\Rightarrow K_{298}=e^{9.73}\approx 1.68\times 10^4$

At $350\text{K}$: $\ln K_{350}=\frac{(1)(96485)(0.25)}{(8.314)(350)}=\frac{24121}{2909.9}=8.29\Rightarrow K_{350}=e^{8.29}\approx 3.99\times 10^3$

So $K$ decreases as $T$ rises (for fixed positive $E^{\circ }$). Reason: raising $T$ shrinks $\frac{nF{E}^{\circ }}{RT}$ because $T$ sits in the denominator — the same thermal energy $RT$ that scrambles the reaction toward $K=1$. The exponential-in-$1/T$ shape here is a cousin of the Arrhenius Equation; whenever an energy competes with $RT$, temperature dilutes the extremes.

Sign of $E^{\circ }$ predicts $K$ vs 1? ::: Positive $E^{\circ }\Rightarrow K>1$; negative $\Rightarrow K<1$; zero $\Rightarrow K=1$. Does $E^{\circ }$ change when you scale the equation? ::: No — it is intensive. Only $n$ (and thus $K$) changes. When is the $0.0592$ shortcut valid? ::: Only at $T=298.15\text{K}$; otherwise use $\ln K=nF{E}^{\circ }/RT$. How do you handle a non-standard measured $E_{\text{cell}}$? ::: Use Nernst to recover $E^{\circ }$ first, then apply the bridge. How does $K$ respond to higher $T$ (fixed positive $E^{\circ }$)? ::: It decreases, because $RT$ in the denominator shrinks $\ln K$.