Nernst equation and concentration effects

A concentration cell is built from the same electrode material dipped into the same electrolyte, with the two half-cells differing only in ion concentration (strictly, activity). Because both half-reactions are the identical redox couple, $E^\circ_{cathode}$ and $E^\circ_{anode}$ are numbers looked up from the same row of a standard-potential table, so they cancel when you subtract:

where: $R$ is the gas constant, $T$ is temperature in kelvin, $n$ is the number of electrons transferred per formula unit, $F$ is the Faraday constant, and $Q$ is the reaction quotient built from the two ion concentrations. With $E^\circ_{cell}$ gone, only $Q$ carries information, and $Q$ is just the ratio of the two concentrations. The direction of spontaneous flow is fixed by thermodynamics, not by which side you happen to label first. Dilution is an oxidation-favouring change (it adds ions to a solution, which is what oxidation at that electrode does), so the dilute side is always the anode; deposition removes ions from solution, so the concentrated side is always the cathode. Electrons flow through the external circuit from the dilute (anode) side to the concentrated (cathode) side, and the whole system relaxes toward equal concentration on both sides — consistent with $\Delta G < 0$ for the net process. Putting $n$ electrons through Faraday's constant and converting the natural log to $\log_{10}$ at $T = 298\ \text{K}$ gives the working formula every concentration-cell problem in this unit reduces to:

where: $c_{concentrated}$ and $c_{dilute}$ are the ion concentrations at the two electrodes, and $n$ is the charge on the metal ion (so $n=1$ for $\text{Ag}^+$, $n=2$ for $\text{Cu}^{2+}$). Since $c_{concentrated} > c_{dilute}$ by construction, the log is always positive and $E_{cell}$ is always positive — a concentration cell is spontaneous by design, once you have correctly assigned anode and cathode. There are two families worth distinguishing at CY11001 level. Electrode concentration cells are the type above — one electrode material, one electrolyte species, two concentrations. Cells with and without transference differ in whether a liquid junction (direct contact or salt bridge between the two solutions) is present; a liquid junction introduces a small extra "liquid-junction potential" that is a real source of error in measured EMFs, though its derivation is beyond this unit — you only need to recognise it as a limitation of the idealised formula above.

Problem. A concentration cell is built from two silver electrodes: one dipped in $0.10\ \text{M}\ \text{AgNO}_3$ and the other in $0.0010\ \text{M}\ \text{AgNO}_3$, connected by a salt bridge. Find the cell EMF at $298\ \text{K}$, and then find the EMF if a second cell of the same type is set up with the concentrated side at $0.50\ \text{M}$ and the dilute side unknown, given that its measured EMF is $0.0296\ \text{V}$ (copper electrodes, $n=2$).

Step 1 — Silver cell, identify electrodes. The dilute side ($0.0010\ \text{M}$) is the anode (oxidation adds $\text{Ag}^+$, raising its concentration); the concentrated side ($0.10\ \text{M}$) is the cathode (reduction removes $\text{Ag}^+$, lowering its concentration). Cell notation: $\text{Ag}(s)\mid\text{Ag}^+(0.0010\ \text{M})\parallel\text{Ag}^+(0.10\ \text{M})\mid\text{Ag}(s)$, with $n=1$.

Step 2 — Apply the formula. $E_{cell} = \dfrac{0.0592}{1}\log_{10}\!\left(\dfrac{0.10}{0.0010}\right) = 0.0592\log_{10}(100) = 0.0592 \times 2 = 0.1184\ \text{V} \approx 0.118\ \text{V}$.

Step 3 — Copper cell, reverse problem. Here $n=2$ and $c_{concentrated} = 0.50\ \text{M}$ is known, but $c_{dilute}$ is unknown; the measured EMF is $0.0296\ \text{V}$. Set up $0.0296 = \dfrac{0.0592}{2}\log_{10}\!\left(\dfrac{0.50}{c_{dil}}\right)$. Dividing both sides by $0.0592/2 = 0.0296$ gives $\log_{10}\!\left(\dfrac{0.50}{c_{dil}}\right) = 1$.

Step 4 — Invert and solve. $\dfrac{0.50}{c_{dil}} = 10^1 = 10$, so $c_{dil} = \dfrac{0.50}{10} = 0.050\ \text{M}$.

Both cells confirm the pattern: EMF from known concentrations gives $E_{cell} = 0.118\ \text{V}$ for the silver cell, and a measured EMF of $0.0296\ \text{V}$ on the copper cell backs out $c_{dil} = 0.050\ \text{M}$.

Competitive stretch

If the silver cell above were instead reported with an EMF of $0.0592\ \text{V}$ for the same two concentrations but you were told only $n$ is unknown, solving $0.0592 = (0.0592/n)\log_{10}(100) = (0.0592/n)(2)$ gives $n = 2$ — a quick consistency check useful when a problem disguises the metal or the ion's charge and asks you to identify $n$ from the numbers alone.

• Assuming $E^\circ_{cell} \neq 0$ for a concentration cell — both half-cells are the same redox couple, so $E^\circ_{cell} = 0$ always, and the entire EMF comes from the log(ratio) term.
• Reversing anode and cathode — the dilute solution is always the anode (oxidation, ion concentration rises) and the concentrated solution is always the cathode (reduction, ion concentration falls); getting this backwards flips the sign of the answer.
• Forgetting that $n$ is the ion's charge, not the number of atoms — $n=1$ for $\text{Ag}^+$, $n=2$ for $\text{Cu}^{2+}$ — and using the wrong $n$ in the $0.0592/n$ formula.
• Treating the idealised formula as exact in a real cell — a liquid junction (salt bridge or direct contact) can add a small liquid-junction potential not captured by $E_{cell} = (0.0592/n)\log_{10}(c_{cathode}/c_{anode})$.

Ask the tutor: if you want to practise setting up a concentration cell from a word problem where the anode/cathode isn't stated outright, or want to see this same algebra reappear with a solubility-product equilibrium substituted in for the concentration ratio, ask the AI tutor to walk through one or two more variants with you.