Exercises — Nernst equation E = E° − (RT - nF) ln Q

WHAT we do: list every species and ask "does it appear in $Q$?" WHY: the rule is — pure solids and pure liquids have activity $=1$ and vanish; only aqueous ions and gases stay.

• $\text{Zn}(s)$ → solid → out
• $\text{Cu}(s)$ → solid → out
• ${\text{Cu}}^{2+}(aq)$ → reactant ion → denominator
• ${\text{Zn}}^{2+}(aq)$ → product ion → numerator

$\boxed{Q=\frac{[{\text{Zn}}^{2+}]}{[{\text{Cu}}^{2+}]}}$

The slope term is $\frac{0.0592}{n}$. Increasing $Q$ by a factor of 10 changes ${\log }_{10}Q$ by exactly $+1$. $\Delta E=-\frac{0.0592}{n}\times 1=-\frac{0.0592}{2}=-0.0296\text{V}\approx -29.6\text{mV}$

If $Q<1$ then ${\log }_{10}Q<0$ (log of a number below 1 is negative). So $-\frac{0.0592}{n}\log Q$ is a positive correction. $E=E^{\circ }+(\text{positive number})\Rightarrow E>E^{\circ }$ Meaning: lots of reactants left ($Q$ small) → more downhill to run → voltage above standard.

Step 1 — $Q$: $Q=\frac{[{\text{Zn}}^{2+}]}{[{\text{Cu}}^{2+}]}=\frac{1.0}{0.010}=100$. Step 2 — $n$: each Zn gives 2 electrons, each Cu²⁺ takes 2 → $n=2$. Step 3 — Nernst: $E=1.10-\frac{0.0592}{2}{\log }_{10}(100)=1.10-\frac{0.0592}{2}(2)=1.10-0.0592$ $\boxed{E=1.041\text{V}}$ Below $E^{\circ }$ because product (${\text{Zn}}^{2+}$) is piled up relative to reactant.

Net process is just dilution: ${\text{Ag}}^{+}(1.0)\to {\text{Ag}}^{+}(0.0010)$. $Q=\frac{[{\text{Ag}}^{+}{]}_{\text{dilute (product side)}}}{[{\text{Ag}}^{+}{]}_{\text{concentrated (reactant side)}}}=\frac{0.0010}{1.0}=10^{-3}$ $E=0-\frac{0.0592}{1}{\log }_{10}(10^{-3})=-0.0592(-3)=\boxed{0.1776\text{V}}$ Positive because Nature wants to spread the ions out (high → low concentration).

$\ln (100)=2.3026$. $E=1.10-\frac{0.0257}{2}\ln (100)=1.10-\frac{0.0257}{2}(2.3026)=1.10-0.02959$ $\boxed{E\approx 1.041\text{V}}$ Identical (to rounding), because $0.0592=2.303\times 0.0257$ — the two forms are the same equation dressed in different logs.

As reactants convert to products, $[\text{products}]\uparrow$ and $[\text{reactants}]\downarrow$, so $Q\uparrow$. Since $E=E^{\circ }-\frac{0.0592}{n}\log Q$, a rising $Q$ steadily subtracts more voltage. End state = equilibrium: here $Q=K$ and the driving force is gone. $E_{\text{final}}=0\text{V}$ Look at figure s01: the voltage curve slides down and touches the axis exactly at $Q=K$. A "dead" battery is a cell sitting at $Q=K$.

Set $E=0$; that value of $Q$ is by definition $K$. $0=0.34-\frac{0.0592}{2}{\log }_{10}K\Rightarrow {\log }_{10}K=\frac{2(0.34)}{0.0592}=\frac{0.68}{0.0592}=11.486$ $\boxed{K=10^{11.486}\approx 3.06\times 10^{11}}$ Huge $K$ → reaction runs essentially to completion, matching the sizeable positive $E^{\circ }$.

Nature dilutes the concentrated side and enriches the dilute side.

• Concentrated side (1.0 M): ${\text{Ag}}^{+}$ is plated out onto the electrode → ${\text{Ag}}^{+}+e^{-}\to \text{Ag}(s)$ → this is the cathode, and it gains mass.
• Dilute side (0.0010 M): solid silver dissolves to raise its ion concentration → $\text{Ag}(s)\to {\text{Ag}}^{+}+e^{-}$ → this is the anode, and it loses mass.

Consistent with L2.2 giving $E>0$: the cell spontaneously moves toward equal concentrations.

Step 1 — $Q$: the gas ${\text{H}}_2$ is a product; ${\text{H}}^{+}$ is a reactant. $Q=\frac{P_{{\text{H}}_2}}{[{\text{H}}^{+}{]}^2}=\frac{1}{[{\text{H}}^{+}{]}^2}$ Step 2 — Nernst: $E=0-\frac{0.0592}{2}{\log }_{10}\left(\frac{1}{[{\text{H}}^{+}{]}^2}\right)=-\frac{0.0592}{2}(-2{\log }_{10}[{\text{H}}^{+}])=0.0592{\log }_{10}[{\text{H}}^{+}]$ Step 3 — use $\text{pH}=-{\log }_{10}[{\text{H}}^{+}]$: $\boxed{E=-0.0592\text{pH}}$ So at pH 7, $E=-0.414$ V. This links straight to pH and Half-Cell Potentials.

From the equilibrium link ${\log }_{10}K=\frac{n{E}^{\circ }}{0.0592}$: $E^{\circ }=\frac{0.0592{\log }_{10}K}{n}=\frac{0.0592\times 15.0}{2}=\frac{0.888}{2}=\boxed{0.444\text{V}}$ A neat sanity loop: positive $E^{\circ }$ ⇔ $K>1$ ⇔ spontaneous, exactly the story in Gibbs Free Energy and Spontaneity and Reaction Quotient Q vs Equilibrium Constant K.

The $0.0592$ shortcut is only for 298 K, so we must rebuild the slope: $\frac{RT}{nF}\ln Q=\frac{(8.314)(323)}{(2)(96485)}\ln (100)$ $=\frac{2685.4}{192970}(4.6052)=(0.013916)(4.6052)=0.06409\text{V}$ $E=1.10-0.06409=\boxed{1.036\text{V}}$ Slightly lower than the $25^{\circ }$C answer (1.041 V) because a higher $T$ enlarges the $\frac{RT}{nF}$ correction, so the same $Q$ costs a bit more voltage.

Step 1 — $n$: Ni loses 2 e⁻; two Ag⁺ each take 1 e⁻ → $n=2$. Step 2 — $Q$: solids out. $Q=\frac{[{\text{Ni}}^{2+}]}{[{\text{Ag}}^{+}{]}^2}=\frac{1.0}{[{\text{Ag}}^{+}{]}^2}$ Step 3 — solve Nernst for $\log Q$: $0.98=1.04-\frac{0.0592}{2}{\log }_{10}Q\Rightarrow {\log }_{10}Q=\frac{(1.04-0.98)(2)}{0.0592}=\frac{0.12}{0.0592}=2.027$ $Q=10^{2.027}=106.4$ Step 4 — back out the concentration: $106.4=\frac{1.0}{[{\text{Ag}}^{+}{]}^2}\Rightarrow [{\text{Ag}}^{+}{]}^2=9.40\times 10^{-3}\Rightarrow [{\text{Ag}}^{+}]=\boxed{0.0970\text{M}}$ The voltage dipped below $E^{\circ }$, so $Q>1$, so $[{\text{Ag}}^{+}]$ (a reactant, and squared in the denominator) had to be small — the answer is self-consistent.

Any half-reaction of the form ${\text{H}}^{+}+e^{-}\to \frac{1}{2}{\text{H}}_2$ (i.e. one electron per one proton, $n$ electrons for $n$ protons) gives $Q\propto \frac{1}{[{\text{H}}^{+}{]}^n}\Rightarrow E=E^{\circ }-\frac{0.0592}{n}\log \frac{1}{[{\text{H}}^{+}{]}^n}=E^{\circ }-\frac{0.0592}{n}(n\text{pH})=E^{\circ }-0.0592\text{pH}.$ The $n$ cancels exactly when the number of protons equals the number of electrons, leaving the universal Nernstian slope of $-0.0592$ V/pH regardless of $n$. If protons and electrons were not in a 1:1 ratio (say 2 H⁺ per 1 e⁻), the exponent and the $\frac{1}{n}$ would no longer cancel and the slope would deviate from $0.0592$, making the meter non-linear and harder to calibrate. So the winning design keeps protons = electrons. This is the theory behind pH and Half-Cell Potentials.

$E^{\circ }$ is a fixed material property — the standard height of the hill — and it genuinely never moves. What moves is $Q$. Fresh battery: reactants abundant, products scarce, $Q$ small, so $-\frac{0.0592}{n}\log Q$ is a positive top-up and $E>E^{\circ }\ge 0$. As it discharges, products accumulate, $Q$ climbs toward $K$, the correction term grows more negative, and $E$ slides down (this is the discharge curve of Battery Discharge Curves). At the moment $Q=K$ the correction exactly cancels $E^{\circ }$, giving $E=0$ — a dead cell. Same $E^{\circ }$ throughout; the entire life story is written by $Q$ marching from small to $K$, exactly as figure s01 shows.

At $25^{\circ }$C, per 10× rise in $Q$ with $n=1$, voltage changes by ::: $-0.0592$ V The two log constants and their partners are ::: $0.0257$ with $\ln$, and $0.0592$ with ${\log }_{10}$ At equilibrium the cell voltage $E$ equals ::: $0$ V (because $Q=K$) ${\log }_{10}K$ in terms of $E^{\circ }$ (25°C) is ::: $\frac{n{E}^{\circ }}{0.0592}$