Chapter 2 · Electrochemistry

2.2 The Nernst Equation, Gibbs Energy and Equilibrium Constant

Change the concentrations and the voltage changes: the Nernst equation $E_{cell}=E^{\circ}_{cell}-\tfrac{0.059}{n}\log Q$ tells you by how much. At equilibrium the EMF is zero, which links $E^{\circ}_{cell}$ to the equilibrium constant and to $\Delta_{r}G^{\circ}=-nFE^{\circ}_{cell}$.

A new torch battery gives a bright beam; after a few evenings the light turns orange and then dies, although the battery still contains plenty of chemicals. What has changed is the concentrations. As a galvanic cell works, its reactants are used up and its products build up, so the reaction's driving force shrinks. When the reaction reaches equilibrium there is no driving force left at all, the EMF is zero, and the battery is "flat". The standard potentials of Section 2.1 describe only the very special case in which every concentration is $1\ \mathrm{mol\,L^{-1}}$.

The Nernst equation (Walther Nernst, 1889) gives the EMF at any concentrations. It turns a voltmeter into a concentration meter (this is how a pH meter works), it tells you how far a battery is from going flat, and, read at equilibrium, it gives the equilibrium constant of a redox reaction from a single voltage. Combined with the electrical work a cell can do, it also links the EMF to the Gibbs energy you met in Class 11 thermodynamics: $\Delta_{r}G=-nFE_{cell}$.

The Nernst Equation

Consider an electrode $\mathrm{M^{n+}(aq)+n\,e^{-}\rightarrow M(s)}$. If the concentration of $\mathrm{M^{n+}}$ is raised above $1\ \mathrm{mol\,L^{-1}}$, the ions have an even greater tendency to take electrons and be reduced, so the electrode potential becomes more positive. If the solution is diluted, the potential falls. The Nernst equation makes this exact.

📐 Law 2.4 — The Nernst equation for an electrode

For a reduction half-reaction $\ \text{oxidised form}+n\,e^{-}\rightarrow\text{reduced form}$, at temperature $T$,

$$E=E^{\circ}-\frac{2.303\,RT}{nF}\log\frac{[\text{reduced form}]}{[\text{oxidised form}]},$$

where each concentration is raised to the power of its coefficient in the half-reaction, $R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$ and $F=96\,487\ \mathrm{C\,mol^{-1}}$ is the Faraday constant (the charge on one mole of electrons). In the ratio, a pure solid or liquid counts as $1$, and a gas enters as its partial pressure in bar.

At $298\ \mathrm{K}$: $\ \dfrac{2.303\,RT}{F}=\dfrac{2.303\times8.314\times298}{96\,487}$ $=0.0591\approx0.059\ \mathrm{V}$, so

$$E=E^{\circ}-\frac{0.059\ \mathrm{V}}{n}\log\frac{[\text{reduced form}]}{[\text{oxidised form}]}.$$

For a metal ion electrode this is $E_{\mathrm{M^{n+}/M}}=E^{\circ}_{\mathrm{M^{n+}/M}}-\dfrac{0.059}{n}\log\dfrac{1}{[\mathrm{M^{n+}}]}$. For the hydrogen electrode with $p_{\mathrm{H_{2}}}=1\ \mathrm{bar}$: $E=-\dfrac{0.059}{2}\log\dfrac{1}{[\mathrm{H^{+}}]^{2}}=-0.059\,\mathrm{pH}$.

✏️ Example 1 — Electrode potentials away from 1 M
At $298\ \mathrm{K}$ find the electrode potential of (a) a copper rod in $0.010\ \mathrm{M}$ $\mathrm{Cu^{2+}(aq)}$; (b) a zinc rod in $0.10\ \mathrm{M}$ $\mathrm{Zn^{2+}(aq)}$; (c) a hydrogen electrode ($p_{\mathrm{H_{2}}}=1\ \mathrm{bar}$) in a solution of pH $10$. ($E^{\circ}_{\mathrm{Cu^{2+}/Cu}}=+0.34\ \mathrm{V}$, $E^{\circ}_{\mathrm{Zn^{2+}/Zn}}=-0.76\ \mathrm{V}$.)
(a) $\mathrm{Cu^{2+}(aq)+2e^{-}\rightarrow Cu(s)}$, $n=2$: $$E=0.34-\frac{0.059}{2}\log\frac{1}{0.010}=0.34-0.0295\times2=0.281\ \mathrm{V}\approx0.28\ \mathrm{V}.$$
(b) $E=-0.76-0.0295\log\dfrac{1}{0.10}$ $=-0.76-0.0295$ $=-0.7895\ \mathrm{V}\approx-0.79\ \mathrm{V}$.
(c) $E=-0.059\,\mathrm{pH}=-0.059\times10=-0.59\ \mathrm{V}$.
Check Dilution makes every potential less positive: the oxidised form becomes scarcer, so reduction is less favoured. A tenfold dilution changes $E$ by $0.059/n$ volts: $0.0295\ \mathrm{V}$ for $n=2$.

For a whole cell, subtract the right and left electrode potentials as before. Both Nernst terms combine into one, involving the reaction quotient $Q$ of the cell reaction.

📐 Relation 2.5 — The Nernst equation for a cell

For a cell reaction $\ a\mathrm{A}+b\mathrm{B}\rightarrow c\mathrm{C}+d\mathrm{D}\ $ in which $n$ electrons are transferred,

$$E_{cell}=E^{\circ}_{cell}-\frac{2.303\,RT}{nF}\log Q,\qquad Q=\frac{[\mathrm{C}]^{c}[\mathrm{D}]^{d}}{[\mathrm{A}]^{a}[\mathrm{B}]^{b}},$$

with pure solids omitted from $Q$ and gases entered as partial pressures in bar. At $298\ \mathrm{K}$,

$$E_{cell}=E^{\circ}_{cell}-\frac{0.059}{n}\log Q.$$

Daniell cell: $\ E_{cell}=1.10-\dfrac{0.059}{2}\log\dfrac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cu^{2+}}]}$. Raising $[\mathrm{Cu^{2+}}]$ (a reactant) or lowering $[\mathrm{Zn^{2+}}]$ (a product) makes $Q$ smaller and the EMF larger, just as Le Chatelier's principle would suggest.

🧭 Problem-solving strategy — EMF at non-standard concentrations
  1. Identify the anode (left) and cathode (right), and find $E^{\circ}_{cell}=E^{\circ}_{R}-E^{\circ}_{L}$.
  2. Write the balanced cell reaction and read off $n$, the number of electrons transferred.
  3. Write $Q$: products over reactants, each to the power of its coefficient; leave out solids; gases as pressures in bar.
  4. Substitute into $E_{cell}=E^{\circ}_{cell}-\dfrac{0.059}{n}\log Q$, keeping track of the sign of $\log Q$.
  5. Check the direction: if the reactant ions are dilute and the product ions concentrated, $E_{cell}$ must come out smaller than $E^{\circ}_{cell}$.
✏️ Example 2 — A Daniell cell running down
After some use, a Daniell cell has $[\mathrm{Zn^{2+}}]=0.10\ \mathrm{M}$ and $[\mathrm{Cu^{2+}}]=0.0010\ \mathrm{M}$. Find its EMF at $298\ \mathrm{K}$.
Reaction $\mathrm{Zn(s)+Cu^{2+}(aq)\rightarrow Zn^{2+}(aq)+Cu(s)}$, $n=2$, $E^{\circ}_{cell}=1.10\ \mathrm{V}$.
Q $Q=\dfrac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cu^{2+}}]}=\dfrac{0.10}{0.0010}=100$, so $\log Q=2$.
Nernst $$E_{cell}=1.10-\frac{0.059}{2}\times2=1.10-0.059=1.041\ \mathrm{V}\approx1.04\ \mathrm{V}.$$
Check Product concentrated, reactant dilute: $E_{cell}\lt E^{\circ}_{cell}$ ✓. Notice how slowly the voltage falls: $Q$ had to rise a hundredfold to lose just $0.06\ \mathrm{V}$. That is why a battery's voltage stays nearly steady for most of its life.
✏️ Example 3 — A cell with unequal coefficients
Find the EMF at $298\ \mathrm{K}$ of $\mathrm{Ni(s)\,|\,Ni^{2+}(aq,\ 0.050\ M)\,\|}$ $\mathrm{Ag^{+}(aq,\ 0.010\ M)\,|\,Ag(s)}$, given $E^{\circ}_{\mathrm{Ni^{2+}/Ni}}=-0.25\ \mathrm{V}$ and $E^{\circ}_{\mathrm{Ag^{+}/Ag}}=+0.80\ \mathrm{V}$. ($\log 5=0.699$.)
Reaction $\mathrm{Ni(s)+2Ag^{+}(aq)\rightarrow Ni^{2+}(aq)+2Ag(s)}$, $n=2$, $E^{\circ}_{cell}=0.80-(-0.25)=1.05\ \mathrm{V}$.
Q The coefficient $2$ of $\mathrm{Ag^{+}}$ becomes a power: $$Q=\frac{[\mathrm{Ni^{2+}}]}{[\mathrm{Ag^{+}}]^{2}}=\frac{0.050}{(0.010)^{2}}=500,\qquad\log Q=\log5+2=2.699.$$
Nernst $E_{cell}=1.05-0.0295\times2.699$ $=1.05-0.0796$ $=0.970\ \mathrm{V}\approx0.97\ \mathrm{V}$.
Note Forgetting the square gives $Q=5$ and $E=1.03\ \mathrm{V}$, a typical lost mark.
✏️ Example 4 — Measuring pH with a cell; a concentration cell
(a) The cell $\mathrm{Zn(s)\,|\,Zn^{2+}(aq,\ 1\ M)\,\|}$ $\mathrm{H^{+}(aq,\ ?)\,|\,H_{2}(g,\ 1\ bar)\,|\,Pt(s)}$ has an EMF of $0.58\ \mathrm{V}$ at $298\ \mathrm{K}$. Find the pH of the acid solution.
(b) Find the EMF of the concentration cell $\mathrm{Cu(s)\,|\,Cu^{2+}(aq,\ 0.010\ M)\,\|}$ $\mathrm{Cu^{2+}(aq,\ 0.10\ M)\,|\,Cu(s)}$ at $298\ \mathrm{K}$.
(a) Set up $\mathrm{Zn(s)+2H^{+}(aq)\rightarrow Zn^{2+}(aq)+H_{2}(g)}$, $n=2$, $E^{\circ}_{cell}=0.76\ \mathrm{V}$, $Q=\dfrac{[\mathrm{Zn^{2+}}]\,p_{\mathrm{H_{2}}}}{[\mathrm{H^{+}}]^{2}}=\dfrac{1}{[\mathrm{H^{+}}]^{2}}$.
(a) Solve $\log Q=-2\log[\mathrm{H^{+}}]=2\,\mathrm{pH}$, so $E_{cell}=0.76-\dfrac{0.059}{2}\times2\,\mathrm{pH}=0.76-0.059\,\mathrm{pH}$. Then $0.58=0.76-0.059\,\mathrm{pH}$ gives $\mathrm{pH}=\dfrac{0.18}{0.059}=3.05\approx3.1$.
(b) Both electrodes are the same couple, so $E^{\circ}_{cell}=0$. The cell reaction is $\mathrm{Cu^{2+}(aq,\ 0.10\ M)\rightarrow Cu^{2+}(aq,\ 0.010\ M)}$, $n=2$: $$E_{cell}=0-\frac{0.059}{2}\log\frac{0.010}{0.10}=+0.0295\ \mathrm{V}\approx0.030\ \mathrm{V}.$$
Meaning A concentration cell runs because the dilute side "wants" to become more concentrated and vice versa. The more dilute electrode is the anode. The EMF falls to zero when the two concentrations become equal.
Interactive 1 — Nernst Lab: EMF, Q and Equilibrium

Choose a cell and set the concentrations of the ion made at the anode and the ion used up at the cathode. The graph is the Nernst line $E_{cell}=E^{\circ}_{cell}-\tfrac{0.059}{n}\log Q$: it crosses zero at $\log Q=\log K$. Press Play to let the cell discharge. Reactant ions are used up, $Q$ grows, the point slides down the line and the EMF falls until $Q=K$ and $E_{cell}=0$: a flat battery at equilibrium. The bars show how small the reactant concentration has to become.

−1.0−3.0
concentrations—
$Q$ · $\log Q$—
$\log K=nE^{\circ}_{cell}/0.059$—
$E_{cell}$—
$\Delta_{r}G=-nFE_{cell}$—

EMF, Gibbs Energy and the Equilibrium Constant

A galvanic cell is a machine that turns the Gibbs energy of a reaction into electrical work. If the cell reaction transfers $n$ moles of electrons, the charge that flows is $nF$ coulombs. Pushing a charge $q$ through a potential difference $E$ takes work $qE$. So the maximum electrical work a cell can do per mole of reaction, reached when it is discharged reversibly (very slowly, with almost no current), is $nFE_{cell}$. In Class 11 you learnt that the decrease in Gibbs energy is exactly the maximum useful (non-expansion) work a reaction can deliver. Putting the two together gives one of the most important relations in physical chemistry.

📐 Relation 2.6 — EMF and Gibbs energy
$$\Delta_{r}G=-nFE_{cell},\qquad\Delta_{r}G^{\circ}=-nFE^{\circ}_{cell}.$$

A spontaneous reaction ($\Delta_{r}G\lt0$) always gives a positive EMF. With $F$ in $\mathrm{C\,mol^{-1}}$ and $E$ in volts, $\Delta_{r}G$ comes out in $\mathrm{J\,mol^{-1}}$ (since $1\ \mathrm{C\times1\ V}=1\ \mathrm{J}$), per mole of the reaction as written.

Intensive and extensive. Write the Daniell reaction twice over, $\mathrm{2Zn(s)+2Cu^{2+}(aq)\rightarrow 2Zn^{2+}(aq)+2Cu(s)}$: then $n=4$ and $\Delta_{r}G^{\circ}$ doubles, but $E^{\circ}_{cell}=-\Delta_{r}G^{\circ}/nF$ stays $1.10\ \mathrm{V}$. The Gibbs energy depends on how much reacts; the EMF does not.

Now follow the cell as it discharges. $Q$ grows and $E_{cell}$ falls, until the reaction reaches equilibrium. At that point there is no driving force left: $\Delta_{r}G=0$, so $E_{cell}=0$, and $Q$ has become the equilibrium constant $K$. Putting $E_{cell}=0$ and $Q=K$ into the Nernst equation gives a direct bridge from a voltage to an equilibrium constant.

📐 Relation 2.7 — Standard EMF and the equilibrium constant
$$E^{\circ}_{cell}=\frac{2.303\,RT}{nF}\log K_{c},\qquad\text{at }298\ \mathrm{K}:\quad E^{\circ}_{cell}=\frac{0.059}{n}\log K_{c},\quad \log K_{c}=\frac{n\,E^{\circ}_{cell}}{0.059}.$$

Combining with Relation 2.6 gives $\Delta_{r}G^{\circ}=-2.303\,RT\log K_{c}$, the relation from Class 11 equilibrium. So the three quantities carry the same information:

$$E^{\circ}_{cell}\gt0\iff\Delta_{r}G^{\circ}\lt0\iff K_{c}\gt1.$$

Because $\log K$ is $n E^{\circ}_{cell}/0.059$, even a modest EMF means a huge $K$: with $n=2$, every $0.1\ \mathrm{V}$ multiplies $K$ by about $2500$ ($10^{0.2/0.059}=10^{3.39}$). A redox reaction with $E^{\circ}_{cell}$ above about $0.2\ \mathrm{V}$ goes practically to completion.

✏️ Example 5 — The Daniell cell: Gibbs energy and K
For $\mathrm{Zn(s)+Cu^{2+}(aq)\rightarrow Zn^{2+}(aq)+Cu(s)}$, $E^{\circ}_{cell}=1.10\ \mathrm{V}$. Find (a) $\Delta_{r}G^{\circ}$, (b) $K_{c}$ at $298\ \mathrm{K}$, and (c) the ratio $[\mathrm{Zn^{2+}}]/[\mathrm{Cu^{2+}}]$ when the cell is completely flat. ($F=96\,487\ \mathrm{C\,mol^{-1}}$.)
(a) $n=2$: $\Delta_{r}G^{\circ}=-2\times96\,487\ \mathrm{C\,mol^{-1}}\times1.10\ \mathrm{V}$ $=-212\,271\ \mathrm{J\,mol^{-1}}$ $\approx-212.3\ \mathrm{kJ\,mol^{-1}}$.
(b) $$\log K_{c}=\frac{2\times1.10}{0.059}=37.29,\qquad K_{c}=10^{37.29}=1.9\times10^{37}.$$
(c) A flat cell is at equilibrium, so $[\mathrm{Zn^{2+}}]/[\mathrm{Cu^{2+}}]=K_{c}=1.9\times10^{37}$. In practice every $\mathrm{Cu^{2+}}$ ion has been reduced: the reaction goes to completion.
Check $\Delta_{r}G^{\circ}=-2.303RT\log K_{c}$ $=-2.303\times8.314\times298\times37.29\ \mathrm{J\,mol^{-1}}$ $=-212.8\ \mathrm{kJ\,mol^{-1}}$, the same as (a) apart from the rounding of $0.059$.
✏️ Example 6 — How far does iron(III) oxidise iodide?
For $\mathrm{2Fe^{3+}(aq)+2I^{-}(aq)\rightarrow 2Fe^{2+}(aq)+I_{2}(s)}$, use $E^{\circ}_{\mathrm{Fe^{3+}/Fe^{2+}}}=+0.77\ \mathrm{V}$ and $E^{\circ}_{\mathrm{I_{2}/I^{-}}}=+0.54\ \mathrm{V}$ to find $E^{\circ}_{cell}$, $\Delta_{r}G^{\circ}$ and $K_{c}$ at $298\ \mathrm{K}$.
E° $\mathrm{Fe^{3+}}$ is reduced (cathode), $\mathrm{I^{-}}$ oxidised (anode): $E^{\circ}_{cell}=0.77-0.54=0.23\ \mathrm{V}$. Two electrons are transferred as written, so $n=2$.
ΔG° $\Delta_{r}G^{\circ}=-2\times96\,487\times0.23$ $=-44\,384\ \mathrm{J\,mol^{-1}}$ $\approx-44.4\ \mathrm{kJ\,mol^{-1}}$.
K $\log K_{c}=\dfrac{2\times0.23}{0.059}=7.80$, so $K_{c}=10^{7.80}=6.3\times10^{7}$.
Meaning Even a "small" $0.23\ \mathrm{V}$ gives $K\approx10^{8}$, so iron(III) converts iodide to iodine almost completely. (Written per mole of $\mathrm{Fe^{3+}}$, with $n=1$, $\Delta_{r}G^{\circ}$ and $\log K$ would both be halved, but $E^{\circ}_{cell}$ would still be $0.23\ \mathrm{V}$.)
⚠️ Common mistakes with Nernst, ΔG and K

1. $Q$ upside down. $Q$ is products over reactants of the cell reaction written with the left electrode as anode. For the Daniell cell it is $[\mathrm{Zn^{2+}}]/[\mathrm{Cu^{2+}}]$, not the reverse.

2. Powers left out. Coefficients become powers in $Q$: $[\mathrm{Ag^{+}}]^{2}$ in Example 3, $[\mathrm{H^{+}}]^{2}$ in Example 4.

3. Solids in $Q$. Zn, Cu, Ag and other pure solids do not appear in $Q$.

4. Wrong $n$. $n$ is the number of electrons in the balanced overall equation (6 for $\mathrm{2Al+3Ni^{2+}}$), not the charge on one ion.

5. $\ln$ and $\log$. $0.059\ \mathrm{V}$ goes with $\log_{10}$; with $\ln$ use $RT/F=0.0257\ \mathrm{V}$.

6. Units of $\Delta_{r}G^{\circ}$. $nFE$ gives joules per mole. Divide by $1000$ for $\mathrm{kJ\,mol^{-1}}$ and keep the minus sign.

🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — Combining half-cell potentials

In JEE Main, NEET and JEE Advanced 2026 (electrode potentials and their link with Gibbs energy); not required by the CBSE syllabus.

Two half-reactions may be added to give a third one, but their potentials may not simply be added, because $E^{\circ}$ is intensive. Add the Gibbs energies instead, which are extensive. Example: find $E^{\circ}_{\mathrm{Fe^{3+}/Fe}}$ from $E^{\circ}_{\mathrm{Fe^{3+}/Fe^{2+}}}=+0.77\ \mathrm{V}$ and $E^{\circ}_{\mathrm{Fe^{2+}/Fe}}=-0.44\ \mathrm{V}$.

$$\mathrm{Fe^{3+}(aq)+e^{-}\rightarrow Fe^{2+}(aq)}:\quad \Delta G_{1}^{\circ}=-1\cdot F\cdot0.77$$ $$\mathrm{Fe^{2+}(aq)+2e^{-}\rightarrow Fe(s)}:\quad \Delta G_{2}^{\circ}=-2\cdot F\cdot(-0.44)$$ $$\mathrm{Fe^{3+}(aq)+3e^{-}\rightarrow Fe(s)}:\quad -3FE^{\circ}=\Delta G_{1}^{\circ}+\Delta G_{2}^{\circ}=-F(0.77-0.88)$$ $$E^{\circ}_{\mathrm{Fe^{3+}/Fe}}=\frac{0.77-0.88}{3}=-0.037\ \mathrm{V}.$$

In general $E^{\circ}_{3}=\dfrac{n_{1}E^{\circ}_{1}+n_{2}E^{\circ}_{2}}{n_{1}+n_{2}}$. Adding the potentials directly would give the wrong answer, $+0.33\ \mathrm{V}$. (Adding two cell EMFs that share an electrode, as in Section 2.1, is different: there the electrons cancel.)

Interactive 2 — Three Faces of One Number: E°cell, ΔG° and K

Set a standard EMF and the number of electrons $n$. The graph plots $\log K=nE^{\circ}_{cell}/0.059$ for $n=1,2,3$ and $6$, with your $n$ in bold; the bar shows $\Delta_{r}G^{\circ}=-nFE^{\circ}_{cell}$. Press Play to sweep $E^{\circ}_{cell}$ from negative to positive and watch all three change sign together at $E^{\circ}_{cell}=0$, where $K=1$. Marked points are reactions from this chapter.

0.23
$E^{\circ}_{cell}$ · $n$—
$\Delta_{r}G^{\circ}$—
$\log K$—
$K$—

Exercises

I
Electrode potentials from the Nernst equation ($298\ \mathrm{K}$ unless stated; $E^{\circ}$ values from Section 2.1; $p_{\mathrm{H_{2}}}=1\ \mathrm{bar}$).
  • 1 Write the Nernst equation for the $\mathrm{Ag^{+}/Ag}$ electrode and find its potential in $0.010\ \mathrm{M}$ $\mathrm{Ag^{+}}$.
  • 2 Find the potential of a zinc electrode in $0.0010\ \mathrm{M}$ $\mathrm{Zn^{2+}}$.
  • 3 Find the potential of a hydrogen electrode in a solution of pH $3$.
  • 4 Find the potential of a hydrogen electrode in pure water at $298\ \mathrm{K}$.
  • 5 A platinum wire dips into a solution with $[\mathrm{Fe^{3+}}]=0.010\ \mathrm{M}$ and $[\mathrm{Fe^{2+}}]=0.10\ \mathrm{M}$. Find its potential.
  • 6 At what concentration of $\mathrm{Cu^{2+}}$ is the potential of a copper electrode $0.31\ \mathrm{V}$?
  • 7 Show that $\dfrac{2.303RT}{F}=0.059\ \mathrm{V}$ at $298\ \mathrm{K}$ ($R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$, $F=96\,487\ \mathrm{C\,mol^{-1}}$).
  • 8 Find $\dfrac{2.303RT}{F}$ at body temperature, $310\ \mathrm{K}$.
1 $E=E^{\circ}-0.059\log\dfrac{1}{[\mathrm{Ag^{+}}]}$ ($n=1$): $E=0.80-0.059\times2=0.682\approx0.68\ \mathrm{V}$.
2 $E=-0.76-0.0295\log1000$ $=-0.76-0.0885$ $=-0.8485\approx-0.85\ \mathrm{V}$.
3 $E=-0.059\times3=-0.177\approx-0.18\ \mathrm{V}$.
4 pH $=7$: $E=-0.059\times7=-0.413\approx-0.41\ \mathrm{V}$.
5 $\mathrm{Fe^{3+}(aq)+e^{-}\rightarrow Fe^{2+}(aq)}$, $n=1$: $E=0.77-0.059\log\dfrac{0.10}{0.010}=0.77-0.059=0.711\approx0.71\ \mathrm{V}$.
6 $0.31=0.34-0.0295\log\dfrac{1}{c}$, so $\log\dfrac{1}{c}=\dfrac{0.03}{0.0295}=1.017$ and $c=10^{-1.017}=0.096\ \mathrm{M}$.
7 $\dfrac{2.303\times8.314\times298}{96\,487}=\dfrac{5705.8}{96\,487}=0.05914\ \mathrm{V}\approx0.059\ \mathrm{V}$.
8 $\dfrac{2.303\times8.314\times310}{96\,487}=0.0615\ \mathrm{V}$, i.e. about $62\ \mathrm{mV}$ per tenfold change for $n=1$.
II
EMF of cells at $298\ \mathrm{K}$ (use $0.059\ \mathrm{V}$).
  • 1 A Daniell cell has $[\mathrm{Zn^{2+}}]=1.0\ \mathrm{M}$ and $[\mathrm{Cu^{2+}}]=0.010\ \mathrm{M}$. Find $E_{cell}$.
  • 2 A Daniell cell has $[\mathrm{Zn^{2+}}]=0.010\ \mathrm{M}$ and $[\mathrm{Cu^{2+}}]=1.0\ \mathrm{M}$. Find $E_{cell}$.
  • 3 $\mathrm{Mg(s)\,|\,Mg^{2+}(aq,\ 0.010\ M)\,\|}$ $\mathrm{Cu^{2+}(aq,\ 0.10\ M)\,|\,Cu(s)}$: find $E_{cell}$.
  • 4 $\mathrm{Fe(s)\,|\,Fe^{2+}(aq,\ 0.0010\ M)\,\|}$ $\mathrm{H^{+}(aq,\ 1\ M)\,|\,H_{2}(g,\ 1\ bar)\,|\,Pt(s)}$: find $E_{cell}$.
  • 5 $\mathrm{Al(s)\,|\,Al^{3+}(aq,\ 0.0010\ M)\,\|}$ $\mathrm{Ni^{2+}(aq,\ 0.50\ M)\,|\,Ni(s)}$: write the cell reaction, give $n$, and find $E_{cell}$ ($\log 8=0.903$).
  • 6 Find the EMF of the concentration cell $\mathrm{Ag(s)\,|\,Ag^{+}(aq,\ 0.0010\ M)\,\|}$ $\mathrm{Ag^{+}(aq,\ 0.10\ M)\,|\,Ag(s)}$.
  • 7 The cell $\mathrm{Pt(s)\,|\,H_{2}(g,\ 1\ bar)\,|\,H^{+}(aq)\,\|}$ $\mathrm{Cu^{2+}(aq,\ 1\ M)\,|\,Cu(s)}$ has $E_{cell}=0.48\ \mathrm{V}$. Find the pH of the solution in the hydrogen half-cell.
  • 8 At what ratio $[\mathrm{Zn^{2+}}]/[\mathrm{Cu^{2+}}]$ does the EMF of a Daniell cell fall to $1.00\ \mathrm{V}$?
Check: Nernst calculator for any cell at 298 K
$\log Q$—
$E_{cell}$—
$\log K$—
$\Delta_{r}G$ (kJ/mol)—
1 $Q=\dfrac{1.0}{0.010}=100$: $E=1.10-0.0295\times2=1.041\approx1.04\ \mathrm{V}$.
2 $Q=0.010$: $E=1.10+0.059=1.159\approx1.16\ \mathrm{V}$.
3 $E^{\circ}_{cell}=0.34-(-2.36)=2.70\ \mathrm{V}$; $Q=\dfrac{0.010}{0.10}=0.10$: $E=2.70+0.0295=2.7295\approx2.73\ \mathrm{V}$.
4 $\mathrm{Fe(s)+2H^{+}(aq)\rightarrow Fe^{2+}(aq)+H_{2}(g)}$, $E^{\circ}_{cell}=0.44\ \mathrm{V}$, $Q=\dfrac{0.0010\times1}{1^{2}}=10^{-3}$: $E=0.44+0.0295\times3=0.5285\approx0.53\ \mathrm{V}$.
5 $\mathrm{2Al(s)+3Ni^{2+}(aq)\rightarrow 2Al^{3+}(aq)+3Ni(s)}$, $n=6$, $E^{\circ}_{cell}=-0.25-(-1.66)=1.41\ \mathrm{V}$. $Q=\dfrac{(0.0010)^{2}}{(0.50)^{3}}=8.0\times10^{-6}$, $\log Q=-5.097$. $E=1.41-\dfrac{0.059}{6}(-5.097)=1.41+0.050=1.460\approx1.46\ \mathrm{V}$.
6 $E^{\circ}_{cell}=0$, $n=1$, $Q=\dfrac{0.0010}{0.10}=0.010$: $E=-0.059\log0.010=0.118\approx0.12\ \mathrm{V}$.
7 $\mathrm{H_{2}(g)+Cu^{2+}(aq)\rightarrow 2H^{+}(aq)+Cu(s)}$, $n=2$, $Q=[\mathrm{H^{+}}]^{2}$: $E=0.34-0.0295\log[\mathrm{H^{+}}]^{2}=0.34+0.059\,\mathrm{pH}$. So $\mathrm{pH}=\dfrac{0.48-0.34}{0.059}=2.37\approx2.4$.
8 $1.00=1.10-0.0295\log Q$, so $\log Q=\dfrac{0.10}{0.0295}=3.39$ and $Q=10^{3.39}=2.5\times10^{3}$.
III
Gibbs energy and equilibrium constants at $298\ \mathrm{K}$ ($F=96\,487\ \mathrm{C\,mol^{-1}}$, $R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$).
  • 1 Find $\Delta_{r}G^{\circ}$ for $\mathrm{Zn(s)+2Ag^{+}(aq)\rightarrow Zn^{2+}(aq)+2Ag(s)}$, $E^{\circ}_{cell}=1.56\ \mathrm{V}$.
  • 2 Find $\Delta_{r}G^{\circ}$ for $\mathrm{Fe(s)+Cu^{2+}(aq)\rightarrow Fe^{2+}(aq)+Cu(s)}$, $E^{\circ}_{cell}=0.78\ \mathrm{V}$.
  • 3 Find $K_{c}$ for $\mathrm{Zn(s)+Fe^{2+}(aq)\rightleftharpoons Zn^{2+}(aq)+Fe(s)}$, $E^{\circ}_{cell}=0.32\ \mathrm{V}$.
  • 4 Find $K_{c}$ for $\mathrm{Cl_{2}(g)+2Fe^{2+}(aq)}$ $\mathrm{\rightleftharpoons 2Cl^{-}(aq)+2Fe^{3+}(aq)}$, $E^{\circ}_{cell}=0.59\ \mathrm{V}$.
  • 5 A one-electron redox reaction has $K_{c}=1.0\times10^{-5}$. Find $E^{\circ}_{cell}$ and say whether the forward reaction is favoured.
  • 6 A two-electron cell reaction has $\Delta_{r}G^{\circ}=-193\ \mathrm{kJ\,mol^{-1}}$. Find $E^{\circ}_{cell}$.
  • 7 For the Daniell cell of Exercise II.1 ($Q=100$, $E_{cell}=1.041\ \mathrm{V}$), find $\Delta_{r}G$ from $-nFE_{cell}$, and check it with $\Delta_{r}G=\Delta_{r}G^{\circ}+2.303RT\log Q$.
  • 8 The reaction in III.1 is rewritten per mole of silver: $\mathrm{\tfrac12 Zn(s)+Ag^{+}(aq)\rightarrow\tfrac12 Zn^{2+}(aq)+Ag(s)}$. What are $n$, $E^{\circ}_{cell}$, $\Delta_{r}G^{\circ}$ and $\log K_{c}$ now?
1 $n=2$: $\Delta_{r}G^{\circ}=-2\times96\,487\times1.56$ $=-301\,039\ \mathrm{J\,mol^{-1}}$ $\approx-301.0\ \mathrm{kJ\,mol^{-1}}$.
2 $-2\times96\,487\times0.78=-150\,520\ \mathrm{J\,mol^{-1}}\approx-150.5\ \mathrm{kJ\,mol^{-1}}$.
3 $\log K_{c}=\dfrac{2\times0.32}{0.059}=10.85$, $K_{c}=7.0\times10^{10}$.
4 $\log K_{c}=\dfrac{2\times0.59}{0.059}=20.0$, $K_{c}=1.0\times10^{20}$.
5 $E^{\circ}_{cell}=\dfrac{0.059}{1}\log(1.0\times10^{-5})=-0.295\ \mathrm{V}\approx-0.30\ \mathrm{V}$. Negative, $K\lt1$: the reverse reaction is favoured.
6 $E^{\circ}_{cell}=-\dfrac{\Delta_{r}G^{\circ}}{nF}=\dfrac{193\,000}{2\times96\,487}=1.00\ \mathrm{V}$.
7 $\Delta_{r}G=-2\times96\,487\times1.041=-200.9\ \mathrm{kJ\,mol^{-1}}$. Check: $\Delta_{r}G^{\circ}=-212.27\ \mathrm{kJ\,mol^{-1}}$ and $2.303RT\log Q=2.303\times8.314\times298\times2=11.41\ \mathrm{kJ\,mol^{-1}}$, so $\Delta_{r}G=-212.27+11.41=-200.9\ \mathrm{kJ\,mol^{-1}}$ ✓.
8 $n=1$; $E^{\circ}_{cell}=1.56\ \mathrm{V}$ (unchanged, intensive); $\Delta_{r}G^{\circ}=-96\,487\times1.56=-150.5\ \mathrm{kJ\,mol^{-1}}$ (halved); $\log K_{c}=\dfrac{1.56}{0.059}=26.4$ (halved, so the new $K$ is the square root of the old one, $10^{52.9}$).
IV
Applying the Nernst equation.
  • 1 Biology. A nerve cell has $[\mathrm{K^{+}}]=140\ \mathrm{mM}$ inside and $5.0\ \mathrm{mM}$ outside. Treating the membrane as a concentration cell for $\mathrm{K^{+}}$ ($n=1$) at $310\ \mathrm{K}$, estimate the potential difference across it in millivolts. ($\log28=1.447$; use your answer to I.8.)
  • 2 Explain, using $Q$ and $K$, why a battery "goes flat" although it still contains reactants.
  • 3 Environment. Iron in rainwater of pH $4.0$ that already holds $[\mathrm{Fe^{2+}}]=1.0\times10^{-3}\ \mathrm{M}$: find $E_{cell}$ for $\mathrm{Fe(s)+2H^{+}(aq)\rightarrow Fe^{2+}(aq)+H_{2}(g)}$ with $p_{\mathrm{H_{2}}}=1\ \mathrm{bar}$. Does iron still dissolve?
  • 4 Two copper rods dip into $1.0\ \mathrm{M}$ and $0.0010\ \mathrm{M}$ $\mathrm{CuSO_{4}}$, joined by a salt bridge. Which rod is the anode? Find the initial EMF, and the EMF when the concentrations have become equal.
1 $E=0.0615\log\dfrac{140}{5.0}=0.0615\times1.447=0.0890\ \mathrm{V}\approx89\ \mathrm{mV}$ (the inside is negative). Measured resting potentials of nerve cells are about $70\ \mathrm{mV}$, close to this estimate, because the membrane is mostly permeable to $\mathrm{K^{+}}$.
2 As the cell works, products build up and reactants are used up, so $Q$ rises towards $K$ and $E_{cell}=\tfrac{0.059}{n}\log\tfrac{K}{Q}$ falls. When $Q=K$ the reaction is at equilibrium, $\Delta_{r}G=0$ and $E_{cell}=0$, even though large amounts of both reactants and products may remain.
3 $Q=\dfrac{[\mathrm{Fe^{2+}}]\,p_{\mathrm{H_{2}}}}{[\mathrm{H^{+}}]^{2}}=\dfrac{10^{-3}}{(10^{-4})^{2}}=10^{5}$; $E=0.44-0.0295\times5=0.2925\approx0.29\ \mathrm{V}$. Still positive, so iron still dissolves, though with less driving force than in $1\ \mathrm{M}$ acid.
4 The rod in the dilute solution is the anode (Cu dissolves there, raising the low concentration). $E=0.0295\log\dfrac{1.0}{0.0010}=0.0295\times3=0.0885\approx0.089\ \mathrm{V}$. When the two concentrations are equal, $Q=1$ and $E=0$.
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks Calculate the EMF of the cell $\mathrm{Al(s)\,|\,Al^{3+}(aq,\ 0.010\ M)\,\|}$ $\mathrm{Fe^{2+}(aq,\ 0.10\ M)\,|\,Fe(s)}$ at $298\ \mathrm{K}$. Given $E^{\circ}_{\mathrm{Al^{3+}/Al}}=-1.66\ \mathrm{V}$, $E^{\circ}_{\mathrm{Fe^{2+}/Fe}}=-0.44\ \mathrm{V}$.

Reaction $\mathrm{2Al(s)+3Fe^{2+}(aq)\rightarrow 2Al^{3+}(aq)+3Fe(s)}$, $n=6$; $E^{\circ}_{cell}=-0.44-(-1.66)=1.22\ \mathrm{V}$.
Q $Q=\dfrac{[\mathrm{Al^{3+}}]^{2}}{[\mathrm{Fe^{2+}}]^{3}}=\dfrac{(0.010)^{2}}{(0.10)^{3}}=\dfrac{10^{-4}}{10^{-3}}=0.10$, $\log Q=-1$.
Answer $E_{cell}=1.22-\dfrac{0.059}{6}(-1)=1.22+0.0098=1.2298\approx1.23\ \mathrm{V}$.
Marking tip 1 mark for the balanced equation with $n=6$, 1 for $Q$ with the correct powers, 1 for the answer with unit.

E2. JEE Main Numerical value. In a Daniell cell ($E^{\circ}_{cell}=1.10\ \mathrm{V}$) at $298\ \mathrm{K}$, the EMF becomes zero when $[\mathrm{Zn^{2+}}]/[\mathrm{Cu^{2+}}]=10^{x}$. Taking $\tfrac{2.303RT}{F}=0.059\ \mathrm{V}$, the value of $x$ to the nearest integer is ______.

Idea $E_{cell}=0$ means equilibrium, so the ratio equals $K_{c}$.
Compute $x=\log K_{c}=\dfrac{nE^{\circ}_{cell}}{0.059}=\dfrac{2\times1.10}{0.059}=37.29$.
Answer $\mathbf{37}$.

E3. NEET For a cell reaction involving a two-electron change, $E^{\circ}_{cell}=0.50\ \mathrm{V}$ at $298\ \mathrm{K}$. Taking $F=96\,500\ \mathrm{C\,mol^{-1}}$, $\Delta_{r}G^{\circ}$ is (A) $-48.25\ \mathrm{kJ\,mol^{-1}}$  (B) $-96.5\ \mathrm{kJ\,mol^{-1}}$  (C) $+96.5\ \mathrm{kJ\,mol^{-1}}$  (D) $-193\ \mathrm{kJ\,mol^{-1}}$

Compute $\Delta_{r}G^{\circ}=-nFE^{\circ}_{cell}=-2\times96\,500\times0.50=-96\,500\ \mathrm{J\,mol^{-1}}$.
Answer (B) $-96.5\ \mathrm{kJ\,mol^{-1}}$. (A) forgets $n$; (C) has the wrong sign; (D) doubles twice.

E4. JEE Advanced The cell $\mathrm{Pt(s)\,|\,H_{2}(g,\ 1\ bar)\,|\,HA(aq,\ 0.10\ M)\,\|}$ $\mathrm{H^{+}(aq,\ 1\ M)\,|\,H_{2}(g,\ 1\ bar)\,|\,Pt(s)}$ has an EMF of $0.177\ \mathrm{V}$ at $298\ \mathrm{K}$, where HA is a weak monobasic acid. Find the pH of the HA solution and the acid dissociation constant $K_{a}$ of HA.

Cell reaction Left: $\mathrm{H_{2}(g)\rightarrow 2H^{+}(aq,\ L)+2e^{-}}$; right: $\mathrm{2H^{+}(aq,\ R)+2e^{-}\rightarrow H_{2}(g)}$. Overall $\mathrm{2H^{+}(aq,\ R)\rightarrow 2H^{+}(aq,\ L)}$, $n=2$, $E^{\circ}_{cell}=0$.
Nernst $E_{cell}=-\dfrac{0.059}{2}\log\dfrac{[\mathrm{H^{+}}]_{L}^{2}}{1^{2}}=-0.059\log[\mathrm{H^{+}}]_{L}=0.059\,\mathrm{pH}_{L}$. So $\mathrm{pH}_{L}=\dfrac{0.177}{0.059}=3.00$ and $[\mathrm{H^{+}}]=1.0\times10^{-3}\ \mathrm{M}$.
$K_{a}$ $\mathrm{HA(aq)\rightleftharpoons H^{+}(aq)+A^{-}(aq)}$: $K_{a}=\dfrac{[\mathrm{H^{+}}][\mathrm{A^{-}}]}{[\mathrm{HA}]}=\dfrac{(1.0\times10^{-3})^{2}}{0.10-0.001}=1.0\times10^{-5}$.
Answer pH $=3.00$, $K_{a}\approx1.0\times10^{-5}$.

E5. JEE Main NEET Which change increases the EMF of a working Daniell cell? (A) adding aqueous ammonia to the copper half-cell, which binds $\mathrm{Cu^{2+}}$ as a complex ion  (B) diluting the zinc sulphate solution  (C) using a larger zinc rod  (D) diluting the copper(II) sulphate solution

Idea $E_{cell}=1.10-0.0295\log\dfrac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cu^{2+}}]}$: the EMF rises if $Q$ falls.
Options (A) and (D) lower $[\mathrm{Cu^{2+}}]$, raising $Q$: EMF falls. (C) Solids do not appear in $Q$: no change. (B) lowers $[\mathrm{Zn^{2+}}]$, so $Q$ falls.
Answer (B).

E6. Board Assertion–reason. Assertion (A): The EMF of a galvanic cell becomes zero when the cell reaction reaches equilibrium. Reason (R): At equilibrium $\Delta_{r}G=0$. (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.

Check Both are true, and $E_{cell}=-\Delta_{r}G/nF$ links them: $\Delta_{r}G=0$ forces $E_{cell}=0$.
Answer (A).
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. A pH meter is a voltmeter that relies on the Nernst equation. How is it calibrated, how close are real electrodes to the ideal "59 mV per pH unit", and why must the temperature be known?

Why it matters. pH is measured millions of times a day: in hospital blood-gas analysers, in water-treatment plants, in dairies, breweries and swimming pools, and on research ships tracking ocean acidification. An error of $0.1$ pH unit in blood ($7.40$ is normal) can mean a wrong diagnosis.

What to do.

  1. The ideal slope of a pH electrode is $\dfrac{2.303RT}{F}$ volts per pH unit. Compute it in $\mathrm{mV}$ at $0$, $25$, $37$ and $50\ ^{\circ}\mathrm{C}$ ($F=96\,487\ \mathrm{C\,mol^{-1}}$).
  2. A meter at $25\ ^{\circ}\mathrm{C}$ reads $0.0\ \mathrm{mV}$ in a pH $7.00$ buffer and $+174.0\ \mathrm{mV}$ in a pH $4.00$ buffer. Find the actual slope and express it as a percentage of the ideal value.
  3. The same electrode then reads $+101.0\ \mathrm{mV}$ in a sample of rainwater. Find the pH of the rain using your calibration. What pH would you report if you had wrongly used the ideal slope instead?
  4. Decide whether the rain counts as "acid rain" (pH below about $5.6$, the value for pure water in equilibrium with atmospheric $\mathrm{CO_{2}}$).

Tools: Calculator; the check-lab under Exercises II; Interactive 1 for the slope $0.059/n$.

Hand in: Your slope table, the calibrated and uncalibrated pH of the rain, and two sentences on why laboratories calibrate with two buffers before each use.

Going further. Ocean surface pH has fallen from about $8.2$ to about $8.1$ since the industrial revolution. Find how many millivolts that is for an ideal electrode at $15\ ^{\circ}\mathrm{C}$, and read how oceanographers achieve the precision needed to see such small changes.

Instructor note: ideal slopes $54.2$, $59.2$, $61.5$ and $64.1\ \mathrm{mV}$ per pH unit at $0$, $25$, $37$ and $50\ ^{\circ}\mathrm{C}$. Calibrated slope $=174.0/3.00=58.0\ \mathrm{mV}$ per pH, $98.0\%$ of $59.2$. Rain: $\mathrm{pH}=7.00-101.0/58.0=5.26$; with the ideal slope, $7.00-101.0/59.2=5.29$. Either way the rain is mildly acidic (below $5.6$). Going further: $0.1\times57.2\approx5.7\ \mathrm{mV}$ at $15\ ^{\circ}\mathrm{C}$.

🧪 Self-check quiz

0/5 correct

Q1. At $298\ \mathrm{K}$ the factor $\dfrac{2.303RT}{F}$ in the Nernst equation is about

Q2. When the reaction in a galvanic cell reaches equilibrium,

Q3. In a Daniell cell at $298\ \mathrm{K}$, $[\mathrm{Cu^{2+}}]$ is made ten times smaller while $[\mathrm{Zn^{2+}}]$ is unchanged. The EMF

Q4. For a cell reaction with $E^{\circ}_{cell}\gt0$,

Q5. The potential of a hydrogen electrode ($p_{\mathrm{H_{2}}}=1\ \mathrm{bar}$) dipping into a solution of pH $2$ at $298\ \mathrm{K}$ is

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