ChemistryISC Class 12 · Electrochemistry

Nernst Equation, Gibbs Energy and Equilibrium Constant: ISC Class 12

Updated 15 October 2026 · Fact-checked

The Nernst equation gives a cell's emf at any concentration: at 298 K, E = E° − (0.0591/n) log Q. The emf is linked to Gibbs energy by ΔG = −nFE, so a positive emf means a spontaneous reaction. At equilibrium E = 0, which gives log K = nE°/0.0591 at 298 K.

Notice that as the cell runs, Q rises to K, emf falls to zero, and ΔG = −nFE reaches zero, so the reaction stops at equilibrium.

Illustration drawn for QuestionBuddy.

Understand Nernst Equation, Gibbs Energy and Equilibrium Constant

Standard emf. In a galvanic cell the electrode with the lower (more negative) standard reduction potential is the anode (oxidation, written on the left); the one with the higher potential is the cathode (reduction, on the right). The standard emf is $E_{ce l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{an o d e}^{\circ} = E_{r i g h t}^{\circ} - E_{l e f t}^{\circ}$ . For Zn | Zn²⁺ ‖ Cu²⁺ | Cu, $E^{\circ} = 0.34 - (- 0.76) = 1.10$ V. Standard means 1 M solutions, 1 bar gases and 298 K.

Concentration matters. Real cells rarely have 1 M solutions. For an electrode $M^{n +} + n e^{-} \to M$ , Nernst showed that $E = E^{\circ} - \frac{R T}{n F} ln \frac{1}{[ M ^{n +} ]}$ . At 298 K, with $R = 8.314$ J K⁻¹ mol⁻¹ and $F = 96500$ C mol⁻¹, $\frac{2.303 R T}{F} = 0.0591$ V, so $E = E^{\circ} - \frac{0.0591}{n} lo g \frac{1}{[ M ^{n +} ]}$ . A more dilute ion solution makes its electrode potential less positive.

For a whole cell with reaction $a A + b B \to c C + d D$ , the Nernst equation is

$E_{ce l l} = E_{ce l l}^{\circ} - \frac{0.0591}{n} lo g \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}}$ at 298 K,

where $n$ is the number of electrons transferred in the balanced equation and pure solids and liquids are left out of the quotient $Q$ . For the Daniell cell, $E = 1.10 - \frac{0.0591}{2} lo g \frac{[ Z n ^{2 +} ]}{[ C u ^{2 +} ]}$ . Raising the cathode ion concentration or lowering the anode ion concentration raises the emf.

Emf and Gibbs energy. A cell does electrical work by pushing $n F$ coulombs through a potential difference $E$ . The maximum useful work equals the decrease in Gibbs energy: $Δ G = - n F E_{ce l l}$ , and under standard conditions $Δ G^{\circ} = - n F E_{ce l l}^{\circ}$ . A positive emf gives a negative ΔG, so the cell reaction is spontaneous; a negative emf means the reaction runs the other way. Emf is an intensive property (it does not change if the equation is doubled), while ΔG is extensive (doubling the equation doubles $n$ and ΔG).

Emf and the equilibrium constant. As a cell discharges, $Q$ rises and $E$ falls. When $E = 0$ the reaction is at equilibrium and $Q = K_{c}$ . Putting $E = 0$ in the Nernst equation gives $E_{ce l l}^{\circ} = \frac{2.303 R T}{n F} lo g K_{c}$ , which at 298 K is

$lo g K_{c} = \frac{n E _{ce l l}^{\circ}}{0.0591}$ .

Combined with $Δ G^{\circ} = - n F E^{\circ}$ , this gives $Δ G^{\circ} = - 2.303 R T lo g K_{c}$ . A cell with $E^{\circ}$ of about 1 V has $K$ near $1 0^{34}$ : such reactions go essentially to completion.

Uses. The three relations let you predict whether a metal displaces another, whether an oxidising agent can oxidise an ion (for example whether Fe³⁺ can oxidise I⁻), find an unknown concentration from a measured emf, and find equilibrium constants that would be hard to measure directly.

Nernst Equation, Gibbs Energy and Equilibrium Constant: formulas and key facts

Standard emf: $E_{ce l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{an o d e}^{\circ} = E_{r i g h t}^{\circ} - E_{l e f t}^{\circ}$; Use standard reduction potentials for both; do not change the sign of the anode value before subtracting.
Nernst equation for an electrode: $E_{M^{n +} / M} = E_{M^{n +} / M}^{\circ} - \frac{0.0591}{n} lo g \frac{1}{[ M ^{n +} ]}$; At 298 K; general form uses $\frac{2.303 R T}{n F}$ .
Nernst equation for a cell: $E_{ce l l} = E_{ce l l}^{\circ} - \frac{0.0591}{n} lo g Q$; $Q$ = products over reactants from the balanced cell reaction; solids omitted.
Gibbs energy and emf: $Δ G = - n F E_{ce l l}, Δ G^{\circ} = - n F E_{ce l l}^{\circ}$; $F = 96500$ C mol⁻¹; ΔG in J. Maximum electrical work $= - Δ G$ .
Equilibrium constant: $lo g K_{c} = \frac{n E _{ce l l}^{\circ}}{0.0591} (298 K), Δ G^{\circ} = - 2.303 R T lo g K_{c}$; From $E = 0$ and $Q = K_{c}$ at equilibrium.

How to solve Nernst Equation, Gibbs Energy and Equilibrium Constant questions

ISC builds 2-, 3- and 5-mark questions out of the same chain: half-reactions, balanced cell reaction, $E^{\circ}$ , Nernst, then ΔG or K. Follow the chain and each link earns a mark.

1Identify anode and cathode from the cell notation (left is anode) or from the reduction potentials (lower one is anode).
2Write both half-reactions, multiply to equal electrons, add, and read off $n$ .
3Find $E_{ce l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{an o d e}^{\circ}$ ; for a working cell it should be positive.
4Write $Q$ from the balanced equation (products over reactants, powers = coefficients, solids left out).
5Substitute in $E = E^{\circ} - \frac{0.0591}{n} lo g Q$ ; work out $lo g Q$ on its own line.
6For energy, use $Δ G = - n F E$ with the emf the question asks for ( $E$ for ΔG, $E^{\circ}$ for ΔG°) and keep the negative sign; give the answer in J or kJ.
7For $K_{c}$ , use $lo g K_{c} = \frac{n E ^{\circ}}{0.0591}$ , then take the antilog and write it in standard form.

Nernst Equation, Gibbs Energy and Equilibrium Constant in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Concentrations shift a cell's voltage

Suppose a school science-fair team builds a Daniell cell with a zinc strip in 0.50 M ZnSO₄ and a copper strip in 0.0050 M CuSO₄, joined by a salt bridge, at 298 K. They want to light a small LED that needs at least 2.0 V.

Calculate the emf of one cell and find the least number of such cells to connect in series.

Try it first, then show the solutionSolution

Cell reaction: $Z n + C u^{2 +} \to Z n^{2 +} + C u$ , $n = 2$ , $E^{\circ} = 1.10$ V.
$E = 1.10 - \frac{0.0591}{2} lo g \frac{0.50}{0.0050} = 1.10 - 0.0591 = 1.041$ V.
One cell gives 1.04 V, less than 2.0 V; two in series give $2 \times 1.041 = 2.08$ V.
So two cells are enough. Using more concentrated CuSO₄ would raise each cell's emf slightly.

Answer: Each cell gives about 1.04 V; two cells in series (about 2.08 V) light the LED.

Memory hook: More ions at the cathode or fewer at the anode push the emf up; the Nernst equation tells you by how much.

Tricks to remember Nernst Equation, Gibbs Energy and Equilibrium Constant

Short memory aids. Say each one aloud once, then use it on a question the same day.

Right minus left
$E_{ce l l}^{\circ} = E_{R}^{\circ} - E_{L}^{\circ}$ : in the cell diagram, Reduction is on the Right.
Use it when: Getting the sign of $E^{\circ}$ right every time.
Products over reactants, like K
$Q$ is built exactly like $K_{c}$ : products on top, reactants below, solids ignored. For the Daniell cell that is $[Z n^{2 +}] / [C u^{2 +}]$ .
Use it when: Writing the log term in the Nernst equation.
Plus E, minus G
Positive emf, negative ΔG, spontaneous. The minus sign in $Δ G = - n F E$ is the whole story.
Use it when: Explaining why a working cell must have a positive emf.

Nernst Equation, Gibbs Energy and Equilibrium Constant: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Sign check · 1 mark · MCQBoard level

For a cell reaction that occurs spontaneously under standard conditions:
(a) $E_{ce l l}^{\circ} < 0$ and $Δ G^{\circ} > 0$
(b) $E_{ce l l}^{\circ} > 0$ and $Δ G^{\circ} < 0$
(c) $E_{ce l l}^{\circ} > 0$ and $K_{c} < 1$
(d) $E_{ce l l}^{\circ} = 0$ and $K_{c} = 0$

Try it first, then show the solutionSolution

$Δ G^{\circ} = - n F E^{\circ}$ : spontaneous needs $Δ G^{\circ} < 0$ , so $E^{\circ} > 0$ .
Then $lo g K_{c} = \frac{n E ^{\circ}}{0.0591} > 0$ , so $K_{c} > 1$ ; (c) is wrong.

Answer: (b).

Example 2: Electrode potential at low concentration · 2 marks · Short answerBoard level

Calculate the electrode potential of a zinc electrode dipped in 0.001 M ZnSO₄ solution at 298 K. ( $E_{Z n^{2 +} / Z n}^{\circ} = - 0.76$ V)

Try it first, then show the solutionSolution

$Z n^{2 +} + 2 e^{-} \to Z n$ , $n = 2$ (½ mark).
$E = E^{\circ} - \frac{0.0591}{2} lo g \frac{1}{[ Z n ^{2 +} ]} = - 0.76 - 0.02955 lo g 1000$ (1 mark).
$E = - 0.76 - 0.0887 = - 0.849$ V (½ mark).

Answer: $E \approx - 0.849$ V.

Example 3: Emf and ΔG of an iron-silver cell · 3 marks · Short answerHarder board level

For the cell Fe | Fe²⁺(0.010 M) ‖ Ag⁺(0.050 M) | Ag at 298 K: (a) write the cell reaction, (b) calculate the emf, (c) calculate ΔG. ( $E_{F e^{2 +} / F e}^{\circ} = - 0.44$ V, $E_{A g^{+} / A g}^{\circ} = + 0.80$ V, $F = 96500$ C mol⁻¹)

Try it first, then show the solutionSolution

(a) $F e (s) + 2 A g^{+} (a q) \to F e^{2 +} (a q) + 2 A g (s)$ , $n = 2$ (½ mark).
$E^{\circ} = 0.80 - (- 0.44) = 1.24$ V; $Q = \frac{[ F e ^{2 +} ]}{[ A g ^{+} ] ^{2}} = \frac{0.010}{( 0.050 ) ^{2}} = 4.0$ (½ mark).
(b) $E = 1.24 - \frac{0.0591}{2} lo g 4.0 = 1.24 - 0.0178 = 1.222$ V (1 mark).
(c) $Δ G = - n F E = - 2 \times 96500 \times 1.222 = - 2.36 \times 1 0^{5}$ J (1 mark).

Answer: (a) Fe + 2Ag⁺ → Fe²⁺ + 2Ag. (b) $E \approx 1.222$ V. (c) $Δ G \approx - 236$ kJ mol⁻¹.

Example 4: ΔG° and K for a nickel-silver cell · 3 marks · Short answerHarder board level

Calculate $Δ G^{\circ}$ and the equilibrium constant at 298 K for the reaction $N i (s) + 2 A g^{+} (a q) ⇌ N i^{2 +} (a q) + 2 A g (s)$ . ( $E_{N i^{2 +} / N i}^{\circ} = - 0.25$ V, $E_{A g^{+} / A g}^{\circ} = + 0.80$ V)

Try it first, then show the solutionSolution

$E_{ce l l}^{\circ} = 0.80 - (- 0.25) = 1.05$ V; $n = 2$ (½ mark).
$Δ G^{\circ} = - 2 \times 96500 \times 1.05 = - 202650$ J $= - 202.7$ kJ (1 mark).
$lo g K_{c} = \frac{2 \times 1.05}{0.0591} = 35.53$ (1 mark).
$K_{c} = 1 0^{35.53} \approx 3.4 \times 1 0^{35}$ (½ mark): the reaction goes practically to completion.

Answer: $Δ G^{\circ} \approx - 202.7$ kJ mol⁻¹; $K_{c} \approx 3.4 \times 1 0^{35}$ .

Example 5: Unknown concentration from a measured emf · 5 marks · Long answerHOTS / competency

The cell Zn | Zn²⁺(x M) ‖ Cu²⁺(0.10 M) | Cu has an emf of 1.13 V at 298 K. ( $E_{ce l l}^{\circ} = 1.10$ V)
(a) Write the Nernst equation for the cell and calculate x. [3]
(b) Calculate ΔG for the cell reaction under these conditions. [1]
(c) Explain what happens to the emf as the cell is used, and what its value is when the cell is 'dead'. [1]

Try it first, then show the solutionSolution

(a) $Z n + C u^{2 +} \to Z n^{2 +} + C u$ , $n = 2$ ; $E = E^{\circ} - \frac{0.0591}{2} lo g \frac{[ Z n ^{2 +} ]}{[ C u ^{2 +} ]}$ (1 mark).
$1.13 = 1.10 - 0.02955 lo g \frac{x}{0.10}$ , so $lo g \frac{x}{0.10} = - \frac{0.03}{0.02955} = - 1.015$ (1 mark).
$\frac{x}{0.10} = 0.0966$ , so $x = 9.7 \times 1 0^{- 3}$ M (1 mark). The emf is above $E^{\circ}$ because the anode solution is dilute.
(b) $Δ G = - 2 \times 96500 \times 1.13 = - 2.18 \times 1 0^{5}$ J (1 mark).
(c) As the cell works, [Zn²⁺] rises and [Cu²⁺] falls, so $Q$ rises and $E$ falls. When equilibrium is reached, $E = 0$ and no current flows (1 mark).

Answer: (a) $x \approx 9.7 \times 1 0^{- 3}$ M. (b) $Δ G \approx - 218$ kJ mol⁻¹. (c) The emf falls steadily and becomes zero at equilibrium.

Where students lose marks in Nernst Equation, Gibbs Energy and Equilibrium Constant

The slips examiners see most often, and what to write instead.

Mistake: Stopping at $E_{ce l l}^{\circ}$ when work is asked
ISC examiners noted that candidates calculated only $E_{ce l l}^{\circ}$ and did not go on to the maximum work (2024).
Do this: Maximum work $= - Δ G^{\circ} = n F E^{\circ}$ . Finish the chain and state the unit.
Usually costs: 1 mark of 2.
Mistake: Dropping the negative sign in ΔG
Examiners reported ΔG° written without the negative sign, or with a wrong formula, in 2024.
Wrong working
$Δ G^{\circ} = n F E^{\circ} = 2 \times 96500 \times 1.05$
Right working
$Δ G^{\circ} = - n F E^{\circ} = - 2 \times 96500 \times 1.05 = - 202650$ J
Do this: Write $Δ G = - n F E$ as a separate line before substituting.
Usually costs: 1 mark.
Mistake: Confusing ΔG with ΔG°
Examiners found ΔG° calculated where ΔG (with the actual emf) was asked.
Do this: Use $E_{ce l l}$ from the Nernst equation for ΔG; use $E^{\circ}$ only for ΔG°.
Usually costs: 1 mark.
Mistake: Wrong quotient or wrong unknown in the Nernst equation
Examiners noted wrong substitutions, and $E_{ce l l}$ calculated when $E_{ce l l}^{\circ}$ was asked (2025).
Wrong working
$E = E^{\circ} - \frac{0.0591}{2} lo g \frac{[ C u ^{2 +} ]}{[ N i ^{2 +} ]}$ for Ni | Ni²⁺ ‖ Cu²⁺ | Cu
Right working
$E = E^{\circ} - \frac{0.0591}{2} lo g \frac{[ N i ^{2 +} ]}{[ C u ^{2 +} ]}$
Do this: Write the balanced cell reaction first; $Q$ is products over reactants. Then rearrange for the quantity asked.
Usually costs: 1 to 2 marks.
Mistake: Unbalanced or half-only cell reaction
Examiners recorded unbalanced Cu-Ag cell reactions and answers giving only separate half-reactions.
Do this: Equalise electrons (Cu + 2Ag⁺ → Cu²⁺ + 2Ag) and add; $n$ comes from this equation.
Usually costs: 1 mark, and every later step that uses $n$.

Exam tips

A working galvanic cell always has a positive $E_{ce l l}^{\circ}$ ; a negative answer means you swapped anode and cathode.
Show $lo g Q$ as a number on its own line; examiners can then give method credit even if the final value slips.
Keep four significant figures in intermediate steps and round only the final emf (to three decimal places in volts).
Use the constant the paper gives (0.0591 or 0.059); answers may differ slightly in the last figure, which is accepted.
For $K_{c}$ , write the answer in standard form; for very large K, add a line saying the reaction goes to completion.

Practice questions on Nernst Equation, Gibbs Energy and Equilibrium Constant

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · Very short answerBoard level
Complete the relation: the standard Gibbs energy change of a cell reaction is related to its standard emf by $Δ G^{\circ}$ = ______.
Hint
It involves $n$ and the Faraday constant.
Answer
$Δ G^{\circ} = - n F E_{ce l l}^{\circ}$ .
Q2 · 2 marks · Short answerBoard level
Write the cell reaction and the Nernst equation at 298 K for the cell Mg | Mg²⁺(C₁) ‖ Ag⁺(C₂) | Ag.
Hint
Two silver ions take the two electrons from one magnesium atom.
Answer
$M g + 2 A g^{+} \to M g^{2 +} + 2 A g$ , $n = 2$ . $E_{ce l l} = E_{ce l l}^{\circ} - \frac{0.0591}{2} lo g \frac{C _{1}}{C _{2}^{2}}$ .
Q3 · 2 marks · Short answerHarder board level
For the cell Cr | Cr³⁺ ‖ Fe²⁺ | Fe under standard conditions, calculate $E_{ce l l}^{\circ}$ and $Δ G^{\circ}$ . ( $E^{\circ}$ : Cr³⁺/Cr = −0.74 V, Fe²⁺/Fe = −0.44 V)
Hint
Balance electrons: 2Cr gives 6 electrons.
Answer
$2 C r + 3 F e^{2 +} \to 2 C r^{3 +} + 3 F e$ , $n = 6$ . $E^{\circ} = - 0.44 - (- 0.74) = 0.30$ V. $Δ G^{\circ} = - 6 \times 96500 \times 0.30 = - 173700$ J ( $- 173.7$ kJ).
Q4 · 3 marks · Short answerHarder board level
Calculate the emf at 298 K of the cell Pb | Pb²⁺(0.10 M) ‖ H⁺(0.010 M) | H₂(1 bar) | Pt. ( $E_{P b^{2 +} / P b}^{\circ} = - 0.13$ V)
Hint
$Q = \frac{[ P b ^{2 +} ] p _{H_{2}}}{[ H ^{+} ] ^{2}}$ .
Answer
$P b + 2 H^{+} \to P b^{2 +} + H_{2}$ , $n = 2$ , $E^{\circ} = 0.13$ V. $Q = \frac{0.10}{( 0.010 ) ^{2}} = 1000$ . $E = 0.13 - 0.02955 \times 3 = 0.041$ V.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): If a cell reaction equation is multiplied by 2, $E_{ce l l}^{\circ}$ stays the same but $Δ G^{\circ}$ doubles.
Reason (R): $E_{ce l l}^{\circ}$ is an intensive property, whereas $Δ G^{\circ} = - n F E^{\circ}$ depends on the number of electrons transferred.
(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.
Hint
Doubling the equation doubles $n$ .
Answer
(a).
Q6 · 3 marks · Short answerHarder board level
A cell reaction with $n = 2$ has $Δ G^{\circ} = - 50.0$ kJ mol⁻¹ at 298 K. Calculate $E_{ce l l}^{\circ}$ and the equilibrium constant. ( $R = 8.314$ J K⁻¹ mol⁻¹)
Hint
Use $Δ G^{\circ} = - n F E^{\circ}$ , then $Δ G^{\circ} = - 2.303 R T lo g K$ .
Answer
$E^{\circ} = \frac{50 000}{2 \times 96500} = 0.259$ V. $lo g K = \frac{50 000}{2.303 \times 8.314 \times 298} = 8.76$ , so $K \approx 5.8 \times 1 0^{8}$ .
Q7 · 3 marks · Short answerHOTS / competency
Using $E^{\circ}$ values Fe³⁺/Fe²⁺ = +0.77 V, I₂/I⁻ = +0.54 V and Br₂/Br⁻ = +1.09 V, predict whether Fe³⁺ ions can oxidise (a) I⁻ to I₂ and (b) Br⁻ to Br₂ under standard conditions. Support each answer with $E_{ce l l}^{\circ}$ , and give $Δ G^{\circ}$ for the feasible reaction.
Hint
Fe³⁺ is reduced at the cathode; the halide is oxidised at the anode.
Answer
(a) $2 F e^{3 +} + 2 I^{-} \to 2 F e^{2 +} + I_{2}$ : $E^{\circ} = 0.77 - 0.54 = + 0.23$ V, feasible; $Δ G^{\circ} = - 2 \times 96500 \times 0.23 = - 44390$ J. (b) $E^{\circ} = 0.77 - 1.09 = - 0.32$ V, so Fe³⁺ cannot oxidise Br⁻.
Q8 · 3 marks · Short answerHOTS / competency
For the cell Ni | Ni²⁺(1.0 M) ‖ Cu²⁺(x M) | Cu at 298 K, find the value of x at which the emf becomes zero. ( $E^{\circ}$ : Ni²⁺/Ni = −0.25 V, Cu²⁺/Cu = +0.34 V) What does the answer tell you about how far the reaction goes?
Hint
Set $E = 0$ : $E^{\circ} = \frac{0.0591}{2} lo g \frac{[ N i ^{2 +} ]}{[ C u ^{2 +} ]}$ .
Answer
$E^{\circ} = 0.59$ V. $0 = 0.59 - 0.02955 lo g \frac{1.0}{x}$ , so $lo g \frac{1}{x} = 19.97$ and $x \approx 1.1 \times 1 0^{- 20}$ M. The emf falls to zero only when almost all Cu²⁺ has been reduced: the reaction goes practically to completion.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Value of $\frac{2.303 R T}{F}$ at 298 K?Show answer
0.0591 V.
2.Relation between $E_{ce l l}^{\circ}$ and $K_{c}$ at 298 K?Show answer
$lo g K_{c} = \frac{n E _{ce l l}^{\circ}}{0.0591}$ .
3.What is the emf of a cell at equilibrium?Show answer
Zero.
4.Why must the emf of a spontaneous cell reaction be positive?Show answer
$Δ G = - n F E$ ; a positive $E$ gives a negative ΔG, the condition for spontaneity.
5.Is emf intensive or extensive?Show answer
Intensive; ΔG is extensive.
6.Write the Nernst equation for the Daniell cell at 298 K.Show answer
$E = 1.10 - \frac{0.0591}{2} lo g \frac{[ Z n ^{2 +} ]}{[ C u ^{2 +} ]}$ V.

Nernst Equation, Gibbs Energy and Equilibrium Constant: frequently asked questions

What is the Nernst equation used for in ISC Class 12?

To find the electrode potential or cell emf at concentrations other than 1 M, to find an unknown concentration from a measured emf, and, with E = 0, to find the equilibrium constant.

How do you calculate the maximum work from a galvanic cell?

Maximum electrical work equals -\Delta G = nFE. Under standard conditions use E^\circ_{cell}, for example 2 \times 96500 \times 1.56 J for a cell with n = 2 and E^\circ = 1.56 V.

Why is solid metal not included in Q?

The concentration (activity) of a pure solid is taken as unity, so it does not appear in the reaction quotient.

What is the relation between ΔG° and the equilibrium constant?

\Delta G^\circ = -2.303RT\log K_c, which follows from \Delta G^\circ = -nFE^\circ and \log K_c = \frac{nE^\circ}{0.0591}.

Can a cell have an emf larger than its standard emf?

Yes. If Q < 1 (a concentrated cathode solution or a dilute anode solution), \log Q is negative and E > E^\circ.