Exercises — Concentration cells

Answer: (b). A concentration cell needs (i) the same metal on both sides AND (ii) different concentrations.

• (a) has different metals (Zn and Cu) → ordinary galvanic cell, not a concentration cell.
• (b) same metal (Cu) + different concentrations (0.1 M vs 2 M) → ✅ concentration cell.
• (c) same metal (Ag) but equal concentrations → $C_{\text{high}}=C_{\text{low}}$, so $\log (1)=0$ and $E_{\text{cell}}=0$. It's the right materials but produces no voltage, so it isn't a working cell.

• Anode = the dilute side = Beaker P (0.02 M). Oxidation here: $\text{Cu}(s)\to {\text{Cu}}^{2+}+2e^{-}$, which adds ions to the weak solution.
• Cathode = the concentrated side = Beaker Q (0.5 M). Reduction here: ${\text{Cu}}^{2+}+2e^{-}\to \text{Cu}(s)$, which removes ions from the strong solution. Nature is trying to equalise the two concentrations — look again at the figure above.

$n=2$ (Cu²⁺ + 2e⁻ → Cu). $E_{\text{cell}}=\frac{0.0592}{2}\log \frac{1.0}{0.010}=0.0296\times \log (100)$ Since $\log (100)=\log (10^2)=2$: $E_{\text{cell}}=0.0296\times 2=0.0592\text{ V}\approx 59\text{ mV}.$

$n=1$ (Ag⁺ + e⁻ → Ag). $E_{\text{cell}}=\frac{0.0592}{1}\log \frac{0.50}{0.0050}=0.0592\times \log (100)=0.0592\times 2=0.1184\text{ V}\approx 0.118\text{ V}.$ Notice: same concentration ratio (100) as L2.1 but twice the voltage — because $n=1$ instead of $n=2$. Fewer electrons per ion means each concentration mismatch pushes harder.

$E_{\text{cell}}=\frac{0.0592}{2}\log (10)=0.0296\times 1=0.0296\text{ V}\approx 29.6\text{ mV}.$ Rule of thumb: for a divalent ion ($n=2$), every factor-of-10 in concentration ratio buys you about 30 mV. For a monovalent ion ($n=1$), about 59 mV per decade.

$n=1$; the anode is the dilute side, so $C_{\text{low}}=0.010$ M and $C_{\text{high}}$ is unknown. $0.118=0.0592\log \frac{C_{\text{high}}}{0.010}$ Divide both sides by 0.0592 to isolate the log: $\log \frac{C_{\text{high}}}{0.010}=\frac{0.118}{0.0592}=1.993$ Undo the log with antilog (10 to the power): $\frac{C_{\text{high}}}{0.010}=10^{1.993}\approx 98.4$ $C_{\text{high}}=0.010\times 98.4\approx 0.98\text{ M}(\approx 1.0\text{ M}).$

• Dilute side (B, 0.020 M) = anode (oxidation, makes ions).
• Concentrated side (A, 0.80 M) = cathode (reduction, removes ions).
• Electrons flow from B → A through the external wire (anode → cathode, always). $E_{\text{cell}}=\frac{0.0592}{2}\log \frac{0.80}{0.020}=0.0296\times \log (40)=0.0296\times 1.602\approx 0.0474\text{ V}.$ ($\log 40=1.602$, giving about 47 mV.)

When $C_{\text{high}}=C_{\text{low}}$, the ratio is $1$ and $\log (1)=0$: $E_{\text{cell}}=\frac{0.0592}{n}\log (1)=0\text{ V}.$ The cell has reached equilibrium — no more push. Physically, the two solutions are now identical, mixing is complete, entropy is maximised, and $\Delta G=0$. There's nothing left to equalise, so no current. This is exactly Le Chatelier's principle and Entropy and free energy speaking through electrochemistry. See the decay curve below.

The reaction is ${\text{H}}^{+}+e^{-}\to \frac{1}{2}{\text{H}}_2$, so $n=1$. Known side is $C_{\text{high}}=1.0$ M; unknown is $C_{\text{low}}$. $0.177=0.0592\log \frac{1.0}{C_{\text{low}}}$ $\log \frac{1.0}{C_{\text{low}}}=\frac{0.177}{0.0592}=2.990\approx 3$ So $\frac{1.0}{C_{\text{low}}}=10^3$, giving $C_{\text{low}}=10^{-3}$ M. $\text{pH}=-\log [{\text{H}}^{+}]=-\log (10^{-3})=\mathrm{3.0.}$ This is precisely how a pH meter works — it's a hidden concentration cell reading H⁺ mismatch as a voltage.

$0.0888=\frac{0.0592}{2}\log \frac{C_{\text{high}}}{0.0010}=0.0296\log \frac{C_{\text{high}}}{0.0010}$ $\log \frac{C_{\text{high}}}{0.0010}=\frac{0.0888}{0.0296}=3.00$ $\frac{C_{\text{high}}}{0.0010}=10^3=1000\Rightarrow C_{\text{high}}=1000\times 0.0010=1.0\text{ M}.$ So a 1.0 M vs 0.0010 M copper cell delivers 89 mV — matching Example 1 of the parent note, reached from the other direction.

(a) Initial ratio $=1.00/0.0100=100$: $E_{\text{cell}}=0.0592\log (100)=0.0592\times 2=0.1184\text{ V}.$ (b) New ratio $=0.991/0.0190=52.2$: $E_{\text{cell}}=0.0592\log (52.2)=0.0592\times 1.718\approx 0.1017\text{ V}.$ The voltage dropped from 118 mV to 102 mV — exactly as L3.3 predicted: as the two concentrations creep together, the log ratio shrinks toward 0 and the cell fades toward its dead state. The decay is fast at first (log is steep near equal-ish ratios) and would crawl to 0 V at equal concentrations.

$\ln (100)=4.605$. $E_{\text{cell}}=\frac{0.0257}{2}\times 4.605=0.01285\times 4.605\approx 0.0592\text{ V}.$ Identical (to 3 sig figs) to the 0.0592 V from L2.1. The two forms are the same equation because $0.0592\log x=\frac{0.0592}{2.303}\ln x=0.0257\ln x$. The factor $2.303=\ln 10$ converts between them — this is the origin of the mysterious "0.0592" (it's $\frac{RT}{F}\ln 10$ at 298 K). See Nernst equation.

$C_{\text{high}}=140$ mM (inside), $C_{\text{low}}=5$ mM (outside). Ratio $=140/5=28$. $E_{\text{cell}}=0.0592\log (28)=0.0592\times 1.447\approx 0.0857\text{ V}\approx 86\text{ mV}.$ This is essentially the Nernst potential for potassium across a cell membrane — the resting membrane potential of neurons (about $-70$ to $-90$ mV) is built from exactly this concentration-cell physics. Full story in Membrane potentials. Living cells are concentration cells you can't switch off.