ChemistryCBSE Class 12 · Electrochemistry

Nernst Equation, Gibbs Energy and Equilibrium Constant

Updated 15 October 2026 · Fact-checked

The Nernst equation gives a cell's potential at any concentration: at 298 K, $E_{cell} = E^\circ_{cell} - \frac{0.059}{n}\log Q$. The same $E^\circ_{cell}$ gives the standard Gibbs energy, $\Delta_rG^\circ = -nFE^\circ_{cell}$, and the equilibrium constant, $\log K_c = \frac{nE^\circ_{cell}}{0.059}$. A positive $E_{cell}$ means a spontaneous reaction.

Understand Nernst Equation, Gibbs Energy and Equilibrium Constant

Standard electrode potentials ( $E^{\circ}$ ) are measured with every ion at 1 mol L⁻¹, every gas at 1 bar and 298 K. Real cells rarely have such concentrations, and as a cell runs the concentrations keep changing. The Nernst equation tells you the potential at the actual concentrations.

For one electrode. For the reduction $M^{n +} (a q) + n e^{-} \to M (s)$ , $E_{(M^{n +} / M)} = E_{(M^{n +} / M)}^{\circ} - \frac{R T}{n F} ln \frac{1}{[ M ^{n +} ]}$ . The concentration of a pure solid is taken as 1. Converting to base-10 logarithms at 298 K, $\frac{2.303 R T}{F} = 0.059$ V, so $E = E^{\circ} - \frac{0.059}{n} lo g \frac{1}{[ M ^{n +} ]}$ . A more concentrated ion solution makes the reduction potential higher.

For a cell. For the general reaction $a A + b B \to c C + d D$ with $n$ electrons transferred, $E_{ce l l} = E_{ce l l}^{\circ} - \frac{0.059}{n} lo g \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}}$ at 298 K. In the Daniell cell, $Z n + C u^{2 +} \to Z n^{2 +} + C u$ , this becomes $E_{ce l l} = 1.10 - \frac{0.059}{2} lo g \frac{[ Z n ^{2 +} ]}{[ C u ^{2 +} ]}$ V. Always find $E_{ce l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{an o d e}^{\circ}$ (both as reduction potentials), write the balanced cell reaction, read $n$ from it, and build $Q$ with products on top.

Equilibrium constant. As the cell discharges, $Q$ grows and $E_{ce l l}$ falls. When $E_{ce l l} = 0$ the reaction is at equilibrium and $Q = K_{c}$ . Putting these in the Nernst equation gives $E_{ce l l}^{\circ} = \frac{2.303 R T}{n F} lo g K_{c}$ , or $lo g K_{c} = \frac{n E _{ce l l}^{\circ}}{0.059}$ at 298 K. Because $E^{\circ}$ sits in an exponent, even 1 V gives enormous $K_{c}$ values, around $1 0^{34}$ for a two-electron reaction.

Gibbs energy. Electrical work done by a cell equals charge × potential difference. For $n$ mol of electrons the charge is $n F$ , so $Δ_{r} G = - n F E_{ce l l}$ and, under standard conditions, $Δ_{r} G^{\circ} = - n F E_{ce l l}^{\circ}$ . A positive cell potential gives a negative Gibbs energy: the reaction is spontaneous. $E_{ce l l}$ is an intensive property, but $Δ_{r} G$ is extensive: doubling the equation doubles $n$ and $Δ_{r} G$ while $E_{ce l l}^{\circ}$ stays the same. Combining the two results gives $Δ_{r} G^{\circ} = - 2.303 R T lo g K_{c}$ , linking electrochemistry with thermodynamics.

Useful constants: $F = 96487$ C mol⁻¹ (take 96500 unless told otherwise), $R = 8.314$ J K⁻¹ mol⁻¹. Give $Δ_{r} G^{\circ}$ in J mol⁻¹ or kJ mol⁻¹ with its sign.

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

Nernst equation for an electrode (298 K): $E_{(M^{n +} / M)} = E_{(M^{n +} / M)}^{\circ} - \frac{0.059}{n} lo g \frac{1}{[ M ^{n +} ]}$; General form uses $\frac{R T}{n F} ln$ ; 0.059 V $= \frac{2.303 R T}{F}$ at 298 K.
Nernst equation for a cell (298 K): $E_{ce l l} = E_{ce l l}^{\circ} - \frac{0.059}{n} lo g Q, Q = \frac{[ products ]}{[ reactants ]}$; Pure solids and liquids are left out of $Q$ ; gases enter as partial pressures in bar.
Standard cell potential: $E_{ce l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{an o d e}^{\circ} = E_{r i g h t}^{\circ} - E_{l e f t}^{\circ}$; Both values are standard reduction potentials; do not multiply $E^{\circ}$ by stoichiometric coefficients.
Equilibrium constant: $lo g K_{c} = \frac{n E _{ce l l}^{\circ}}{0.059}$; At 298 K; from $E_{ce l l} = 0$ when $Q = K_{c}$ .
Gibbs energy: $Δ_{r} G^{\circ} = - n F E_{ce l l}^{\circ}, Δ_{r} G^{\circ} = - 2.303 R T lo g K_{c}$; $F = 96500$ C mol⁻¹; answer in J mol⁻¹ then convert to kJ mol⁻¹.
Hydrogen electrode: $E_{H^{+} / H_{2}} = - 0.059 pH (p_{H_{2}} = 1 bar)$; From $E = 0 - \frac{0.059}{1} lo g \frac{1}{[ H ^{+} ]}$ .

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

Most 3-mark electrochemistry numericals in the board paper are Nernst or $Δ G$ / $K$ questions. Use the same five steps every time.

1From the cell notation, identify the anode (left, oxidation) and cathode (right, reduction). Write both half-reactions and the balanced overall reaction.
2Read $n$ , the number of electrons transferred in the balanced reaction (the lowest common multiple of the two half-reactions).
3Find $E_{ce l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{an o d e}^{\circ}$ from reduction potentials.
4Write $Q$ from the balanced equation, products over reactants, with concentrations raised to their coefficients; leave out solids.
5Substitute in $E_{ce l l} = E_{ce l l}^{\circ} - \frac{0.059}{n} lo g Q$ and evaluate the log carefully (log of a number below 1 is negative).
6For $Δ_{r} G^{\circ}$ use $- n F E_{ce l l}^{\circ}$ ; for $K_{c}$ use $lo g K_{c} = \frac{n E _{ce l l}^{\circ}}{0.059}$ and take the antilog.
7Check signs: a positive $E_{ce l l}$ must go with a negative $Δ_{r} G$ and $K_{c} > 1$ .

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.

Cell potential depends only logarithmically on concentration

Suppose your school science club builds a Daniell cell for an exhibition but has only 0.01 M copper sulphate instead of 1 M, while the zinc sulphate is 1 M. A member worries that the voltage will drop to a hundredth.

Using $E_{ce l l}^{\circ} = 1.10$ V, calculate the cell potential at 298 K and explain whether the worry is justified.

Try it first, then show the solutionSolution

Reaction: $Z n (s) + C u^{2 +} (a q) \to Z n^{2 +} (a q) + C u (s)$ , $n = 2$ .
$Q = \frac{[ Z n ^{2 +} ]}{[ C u ^{2 +} ]} = \frac{1}{0.01} = 100$ , so $lo g Q = 2$ .
$E_{ce l l} = 1.10 - \frac{0.059}{2} \times 2 = 1.10 - 0.059 = 1.041$ V.
A hundred-fold change in concentration changes the voltage by only 0.059 V because concentration enters through a logarithm.

Answer: About 1.04 V; the voltage drops by only about 5%, so the worry is not justified.

Memory hook: Ten times the concentration, only 0.059/n volts: concentration hides inside a log.

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.

Minus, point zero five nine over n, log Q
Read it as a sentence: 'E equals E-nought minus 0.059 over n log Q'. Q is always products over reactants, just like an equilibrium constant.
Use it when: Writing the equation correctly under exam pressure.
Positive E, negative G, big K
Three faces of one fact: $E^{\circ} > 0 \Leftrightarrow Δ G^{\circ} < 0 \Leftrightarrow K > 1$ .
Use it when: Instant check on signs.
Right minus left
$E_{ce l l}^{\circ} = E_{R}^{\circ} - E_{L}^{\circ}$ : in cell notation the cathode is on the right.
Use it when: Getting $E_{ce l l}^{\circ}$ from a cell diagram.

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: Spontaneity from E · 1 mark · MCQBoard level

For a cell reaction to be spontaneous under standard conditions:
(a) $E_{ce l l}^{\circ} > 0$ and $Δ_{r} G^{\circ} > 0$
(b) $E_{ce l l}^{\circ} > 0$ and $Δ_{r} G^{\circ} < 0$
(c) $E_{ce l l}^{\circ} < 0$ and $Δ_{r} G^{\circ} < 0$
(d) $E_{ce l l}^{\circ} = 0$ and $Δ_{r} G^{\circ} < 0$

Try it first, then show the solutionSolution

$Δ_{r} G^{\circ} = - n F E_{ce l l}^{\circ}$ ; $n$ and $F$ are positive.
So a positive $E_{ce l l}^{\circ}$ gives a negative $Δ_{r} G^{\circ}$ , the condition for spontaneity.

Answer: (b).

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

Calculate the electrode potential of a copper electrode dipped in 0.0010 M $C u S O_{4}$ solution at 298 K. ( $E_{C u^{2 +} / C u}^{\circ} = + 0.34$ V)

Try it first, then show the solutionSolution

$C u^{2 +} + 2 e^{-} \to C u$ , $n = 2$ ; $E = E^{\circ} - \frac{0.059}{2} lo g \frac{1}{[ C u ^{2 +} ]}$ (1 mark).
$E = 0.34 - 0.0295 lo g 1000 = 0.34 - 0.0885 = 0.2515 \approx 0.252$ V (1 mark).

Answer: 0.252 V (lower than $E^{\circ}$ because $[C u^{2 +}]$ is below 1 M).

Example 3: EMF of a cell at non-standard concentrations · 3 marks · Short answerHarder board level

Calculate the emf of the cell at 298 K:
$Z n (s) ∣ Z n^{2 +} (0.10 M) ∣∣ A g^{+} (0.010 M) ∣ A g (s)$
( $E_{Z n^{2 +} / Z n}^{\circ} = - 0.76$ V, $E_{A g^{+} / A g}^{\circ} = + 0.80$ V)

Try it first, then show the solutionSolution

Cell reaction: $Z n (s) + 2 A g^{+} (a q) \to Z n^{2 +} (a q) + 2 A g (s)$ , $n = 2$ (½ mark).
$E_{ce l l}^{\circ} = 0.80 - (- 0.76) = 1.56$ V (½ mark).
$Q = \frac{[ Z n ^{2 +} ]}{[ A g ^{+} ] ^{2}} = \frac{0.10}{( 0.010 ) ^{2}} = 1000$ , so $lo g Q = 3$ (1 mark).
$E_{ce l l} = 1.56 - \frac{0.059}{2} \times 3 = 1.56 - 0.0885 = 1.47$ V (1 mark).

Answer: $E_{ce l l} \approx 1.47$ V.

Example 4: Gibbs energy and equilibrium constant · 3 marks · Short answerHarder board level

For the reaction $N i (s) + 2 A g^{+} (a q) \to N i^{2 +} (a q) + 2 A g (s)$ , calculate (a) $Δ_{r} G^{\circ}$ and (b) the equilibrium constant at 298 K. ( $E_{N i^{2 +} / N i}^{\circ} = - 0.25$ V, $E_{A g^{+} / A g}^{\circ} = + 0.80$ V, $F = 96500$ C mol⁻¹)

Try it first, then show the solutionSolution

$E_{ce l l}^{\circ} = 0.80 - (- 0.25) = 1.05$ V; $n = 2$ (½ mark).
(a) $Δ_{r} G^{\circ} = - 2 \times 96500 \times 1.05 = - 202650$ J mol⁻¹ $= - 202.65$ kJ mol⁻¹ (1 mark).
(b) $lo g K_{c} = \frac{2 \times 1.05}{0.059} = 35.59$ (1 mark).
$K_{c} = antilog (35.59) \approx 3.9 \times 1 0^{35}$ (½ mark).

Answer: (a) $- 202.65$ kJ mol⁻¹. (b) $K_{c} \approx 3.9 \times 1 0^{35}$ .

Example 5: Working backwards to a concentration · 3 marks · Short answerHOTS / competency

The cell $Z n (s) ∣ Z n^{2 +} (x M) ∣∣ C u^{2 +} (1.0 M) ∣ C u (s)$ has an emf of 1.159 V at 298 K. $E_{ce l l}^{\circ} = 1.10$ V. (a) Find $x$ . (b) At what ratio $[Z n^{2 +}] / [C u^{2 +}]$ will this cell stop producing current? What does that ratio represent?

Try it first, then show the solutionSolution

$1.159 = 1.10 - \frac{0.059}{2} lo g \frac{x}{1.0}$ , so $lo g x = - \frac{0.059}{0.0295} = - 2$ (1 mark).
$x = 0.010$ M. A lower $[Z n^{2 +}]$ pushes the reaction forward, which is why $E_{ce l l} > E_{ce l l}^{\circ}$ (½ mark).
(b) Current stops when $E_{ce l l} = 0$ : $lo g \frac{[ Z n ^{2 +} ]}{[ C u ^{2 +} ]} = \frac{1.10}{0.0295} = 37.3$ , ratio $\approx 1.9 \times 1 0^{37}$ (1 mark).
This ratio is the equilibrium constant $K_{c}$ of the cell reaction (½ mark).

Answer: (a) $x = 0.010$ M. (b) About $1.9 \times 1 0^{37}$ , which is $K_{c}$ for the Daniell cell reaction.

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

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

Mistake: Multiplying $E^{\circ}$ by the stoichiometric coefficient
Students treat $E^{\circ}$ like an enthalpy.
Wrong working
$E_{ce l l}^{\circ} = 2 (0.80) - (- 0.76) = 2.36$ V for $Z n + 2 A g^{+}$
Right working
$E_{ce l l}^{\circ} = 0.80 - (- 0.76) = 1.56$ V
Do this: Electrode potential is intensive. The coefficient appears only as a power in $Q$ and through $n$ .
Usually costs: Most of a 3-mark numerical.
Mistake: Wrong $n$ or wrong $Q$
Students write $n = 1$ for $A g^{+}$ or forget to square $[A g^{+}]$ .
Wrong working
$Q = \frac{0.10}{0.010} = 10$ , $n = 1$
Right working
$Q = \frac{0.10}{( 0.010 ) ^{2}} = 1000$ , $n = 2$
Do this: Balance the full cell reaction first, then read $n$ and the powers in $Q$ from it.
Usually costs: 1-2 marks.
Mistake: Sign slips with logs of numbers below 1
$lo g 0.1 = - 1$ is dropped as +1.
Wrong working
$E = 0.78 - 0.0295 (1) = 0.75$ V
Right working
$E = 0.78 - 0.0295 (- 1) = 0.81$ V
Do this: Write $lo g Q$ as a separate line with its sign before substituting.
Usually costs: 1 mark.
Mistake: Unit errors in $Δ_{r} G^{\circ}$
The answer in joules is labelled kJ.
Wrong working
$Δ_{r} G^{\circ} = - 202650$ kJ mol⁻¹
Right working
$- 202650$ J mol⁻¹ $= - 202.65$ kJ mol⁻¹
Do this: $n F E$ with $F$ in C mol⁻¹ and $E$ in V gives joules; divide by 1000 for kJ.
Usually costs: ½ mark.

Exam tips

Write the cell reaction and $n$ on the first line; examiners award a step mark for each.
Use 0.059 V unless the question gives 0.0591 V; use the value given if any.
Show $lo g Q$ separately, then substitute; most lost marks come from sign errors in the log.
State the conclusion for $Δ_{r} G^{\circ}$ : 'negative, so the reaction is spontaneous'.
For $K_{c}$ , give the antilog in scientific notation, such as $3.9 \times 1 0^{35}$ .

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 · MCQBoard level
For the electrode reaction $M^{n +} (a q) + n e^{-} \to M (s)$ at 298 K, the Nernst equation is:
(a) $E = E^{\circ} + \frac{0.059}{n} lo g \frac{1}{[ M ^{n +} ]}$
(b) $E = E^{\circ} - \frac{0.059}{n} lo g \frac{1}{[ M ^{n +} ]}$
(c) $E = E^{\circ} - \frac{0.059}{n} lo g [M^{n +}]$
(d) $E = E^{\circ} - 0.059 n lo g \frac{1}{[ M ^{n +} ]}$
Hint
Q for a reduction is products over reactants: solid over ion.
Answer
(b).
Q2 · 2 marks · Very short answerBoard level
Write the Nernst equation and calculate the emf of the cell at 298 K:
$F e (s) ∣ F e^{2 +} (0.010 M) ∣∣ C u^{2 +} (0.10 M) ∣ C u (s)$
( $E_{F e^{2 +} / F e}^{\circ} = - 0.44$ V, $E_{C u^{2 +} / C u}^{\circ} = + 0.34$ V)
Hint
$n = 2$ , $Q = 0.1$ .
Answer
$E_{ce l l} = E_{ce l l}^{\circ} - \frac{0.059}{2} lo g \frac{[ F e ^{2 +} ]}{[ C u ^{2 +} ]}$ . $E_{ce l l}^{\circ} = 0.78$ V; $lo g (0.1) = - 1$ ; $E_{ce l l} = 0.78 + 0.0295 = 0.81$ V (0.8095 V).
Q3 · 2 marks · Very short answerHarder board level
Calculate the potential of a hydrogen electrode in contact with a solution of pH 4 at 298 K, with hydrogen gas at 1 bar.
Hint
$H^{+} + e^{-} \to \frac{1}{2} H_{2}$ , $E^{\circ} = 0$ .
Answer
$E = 0 - 0.059 lo g \frac{1}{[ H ^{+} ]} = - 0.059 \times pH = - 0.059 \times 4 = - 0.236$ V.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): $E_{ce l l}^{\circ}$ is an intensive property, but $Δ_{r} G^{\circ}$ of the cell reaction is an extensive property.
Reason (R): $Δ_{r} G^{\circ} = - n F E_{ce l l}^{\circ}$ , and $n$ doubles when the cell reaction is multiplied by 2.
(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
What happens to $E_{ce l l}^{\circ}$ when you double the equation?
Answer
(a). $E_{ce l l}^{\circ}$ does not change on doubling, but $n$ and hence $Δ_{r} G^{\circ}$ double.
Q5 · 2 marks · Very short answerHarder board level
Calculate the standard Gibbs energy change for the reaction $M g (s) + 2 A g^{+} (a q) \to M g^{2 +} (a q) + 2 A g (s)$ . ( $E_{M g^{2 +} / M g}^{\circ} = - 2.37$ V, $E_{A g^{+} / A g}^{\circ} = + 0.80$ V, $F = 96500$ C mol⁻¹)
Hint
$E_{ce l l}^{\circ} = 3.17$ V, $n = 2$ .
Answer
$Δ_{r} G^{\circ} = - 2 \times 96500 \times 3.17 = - 611810$ J mol⁻¹ $\approx - 611.8$ kJ mol⁻¹.
Q6 · 3 marks · Short answerHarder board level
Calculate the equilibrium constant at 298 K for $F e (s) + C u^{2 +} (a q) ⇌ F e^{2 +} (a q) + C u (s)$ . ( $E_{F e^{2 +} / F e}^{\circ} = - 0.44$ V, $E_{C u^{2 +} / C u}^{\circ} = + 0.34$ V) What does the size of $K_{c}$ tell you about the reaction?
Hint
$lo g K_{c} = \frac{n E ^{\circ}}{0.059}$ .
Answer
$E_{ce l l}^{\circ} = 0.78$ V; $lo g K_{c} = \frac{2 \times 0.78}{0.059} = 26.44$ ; $K_{c} \approx 2.8 \times 1 0^{26}$ . The reaction goes practically to completion: iron displaces copper almost entirely.
Q7 · 3 marks · Short answerHOTS / competency
A concentration cell is set up: $C u (s) ∣ C u^{2 +} (0.0010 M) ∣∣ C u^{2 +} (0.10 M) ∣ C u (s)$ . (a) What is $E_{ce l l}^{\circ}$ ? (b) Calculate $E_{ce l l}$ at 298 K. (c) Which electrode loses mass as the cell works, and why does the emf fall to zero eventually?
Hint
Both electrodes are the same couple; only concentrations differ.
Answer
(a) 0 V. (b) Overall: $C u^{2 +} (0.10) \to C u^{2 +} (0.0010)$ ; $E_{ce l l} = 0 - \frac{0.059}{2} lo g \frac{0.0010}{0.10} = + 0.059$ V. (c) The anode in the dilute solution dissolves (loses mass); as it runs the dilute side gets more concentrated and the concentrated side more dilute, and the emf becomes zero when they are equal.
Q8 · 4 marks · Case-basedHOTS / competency
Read and answer.
Suppose students set up a Daniell cell with 1.0 M $Z n S O_{4}$ and 1.0 M $C u S O_{4}$ of equal volumes and connect it to a small LED. Its initial emf is 1.10 V. They notice that the emf slowly falls while the LED is on.
(a) Write the Nernst equation for this cell at 298 K. (1 mark)
(b) Calculate the emf when $[Z n^{2 +}]$ has risen to 1.8 M and $[C u^{2 +}]$ has fallen to 0.20 M. (2 marks)
(c) Explain why the emf keeps falling and what its value will be when the LED finally goes off for good. (1 mark)
Hint
$Q = 9$ ; $lo g 9 = 0.954$ .
Answer
(a) $E_{ce l l} = 1.10 - \frac{0.059}{2} lo g \frac{[ Z n ^{2 +} ]}{[ C u ^{2 +} ]}$ . (b) $E_{ce l l} = 1.10 - 0.0295 \times 0.954 = 1.072$ V. (c) As $Z n^{2 +}$ builds up and $C u^{2 +}$ is used up, $Q$ increases and $E_{ce l l}$ decreases; it reaches 0 V when $Q$ becomes equal to $K_{c}$ , i.e. at equilibrium.

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
0.059 V.
2.Relation between $Δ_{r} G^{\circ}$ and $E_{ce l l}^{\circ}$ ?Show answer
$Δ_{r} G^{\circ} = - n F E_{ce l l}^{\circ}$ .
3.What is $E_{ce l l}$ at equilibrium?Show answer
Zero; then $Q = K_{c}$ .
4.Formula for $lo g K_{c}$ at 298 K?Show answer
$lo g K_{c} = \frac{n E _{ce l l}^{\circ}}{0.059}$ .
5.Potential of a hydrogen electrode at pH 2?Show answer
$- 0.118$ V.
6.Which is extensive: $E_{ce l l}$ or $Δ_{r} G$ ?Show answer
$Δ_{r} G$ .

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

What is the Nernst equation in Class 12 Chemistry?

It relates the potential of an electrode or cell to the concentrations of the species involved: at 298 K, E_{cell} = E^\circ_{cell} - \frac{0.059}{n}\log Q.

How is the equilibrium constant related to the standard cell potential?

At equilibrium E_{cell} = 0 and Q = K_c, so E^\circ_{cell} = \frac{0.059}{n}\log K_c at 298 K.

Why is $\Delta_rG^\circ$ negative for a working galvanic cell?

The cell does electrical work, so its E^\circ_{cell} is positive, and \Delta_rG^\circ = -nFE^\circ_{cell} is negative, which means the reaction is spontaneous.

Should I use 96500 or 96487 for the Faraday constant?

Use the value given in the question. If none is given, 96500 C mol⁻¹ is accepted.

Are solids included in Q in the Nernst equation?

No. Pure solids and pure liquids have unit activity and are left out; only ion concentrations and gas pressures appear.