Under what condition is $E_{Cell} = 0$ or $\Delta_r G = 0$?

Under what condition is ECell = 0 or Δr G = 0?

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E_{Cell} = 0

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Concept understanding — Cell Representation Nernst Equation

Cell Representation and the Nernst Equation: From Intuition to Precision

Imagine you have a Daniell cell — a zinc rod in zinc sulphate solution connected by a salt bridge to a copper rod in copper sulphate solution. You know it produces a voltage. But what happens if you dilute the copper sulphate solution? Or if you change the temperature? The voltage changes. The Nernst equation is the tool that tells you exactly how much it changes.

The Intuition First

A battery works because the two half-cells "want" to react — zinc wants to lose electrons, copper ions want to gain them. This "want" is measured as a tendency, or potential. But the strength of that tendency depends on how crowded the ions are.

Think of it like this: If you have a room full of people who all want to leave (like zinc ions wanting to form), the push to get out is stronger when the room is packed. If the room is nearly empty, the push is weaker. Similarly, for copper ions wanting to enter the metal (gain electrons), the pull is stronger when there are many copper ions around, and weaker when there are few.

The Nernst equation quantifies this: the actual cell potential depends on the concentrations (or activities) of the ions involved.

The Precise Statement

For a general cell reaction:

aA + bB \rightarrow cC + dD

The cell potential $E$ under non-standard conditions is given by:

E = E^\circ - \frac{RT}{nF} \ln Q

Where:

$E$ = cell potential under the given conditions (in volts)
$E^\circ$ = standard cell potential (when all reactants/products are at 1 M, 1 atm, 25°C)
$R$ = universal gas constant (8.314 J/mol·K)
$T$ = temperature in Kelvin
$n$ = number of moles of electrons transferred in the balanced half-reactions
$F$ = Faraday constant (96,485 C/mol)
$Q$ = reaction quotient = $\frac{[C]^c [D]^d}{[A]^a [B]^b}$ (using concentrations for dilute solutions)

At 25°C (298 K), the equation simplifies to a very practical form:

E = E^\circ - \frac{0.0591}{n} \log_{10} Q

The 0.0591 comes from $\frac{2.303RT}{F}$ at 298 K. Notice it uses $\log_{10}$ (common log), not natural log.

Cell Representation: How We Write It

In electrochemistry, we represent a cell with a shorthand notation. For the Daniell cell:

\text{Zn(s)} \mid \text{Zn}^{2+}(\text{aq}) \parallel \text{Cu}^{2+}(\text{aq}) \mid \text{Cu(s)}

The single vertical line $\mid$ represents a phase boundary (solid electrode | solution). The double line $\parallel$ represents the salt bridge.

The anode (oxidation) is written on the left, the cathode (reduction) on the right. Electrons flow from left to right in the external circuit.

Applying the Nernst Equation to a Cell Representation

For the Daniell cell, the half-reactions are:

Anode (oxidation): $\text{Zn(s)} \rightarrow \text{Zn}^{2+}(\text{aq}) + 2e^-$
Cathode (reduction): $\text{Cu}^{2+}(\text{aq}) + 2e^- \rightarrow \text{Cu(s)}$

Overall: $\text{Zn(s)} + \text{Cu}^{2+}(\text{aq}) \rightarrow \text{Zn}^{2+}(\text{aq}) + \text{Cu(s)}$

Here $n = 2$ (two electrons transferred). The reaction quotient is:

Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}

So the Nernst equation becomes:

E = E^\circ - \frac{0.0591}{2} \log_{10} \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}

Note

Solids (Zn, Cu) do not appear in $Q$ because their concentrations are constant (activity = 1).

A Worked Example

Suppose you have a Daniell cell where $[\text{Zn}^{2+}] = 0.1$ M and $[\text{Cu}^{2+}] = 1.0$ M at 25°C. $E^\circ$ for the cell is 1.10 V.

$E = 1.10 - \frac{0.0591}{2} \log_{10} \frac{0.1}{1.0}$ …

Why this formula?

Cell Representation & the Nernst Equation: Why It Works

The Core Question

Why does a cell's voltage change when concentrations change? The Nernst equation answers this — but the reason lies in the link between chemical free energy and electrical work.

1. The Fundamental Link: Gibbs Free Energy & Cell Potential

A galvanic cell does electrical work. The maximum useful work a cell can do equals the change in Gibbs free energy ( $\Delta G$ ):

\Delta G = -nFE_{\text{cell}}

Where:

$n$ = moles of electrons transferred
$F$ = Faraday constant ( $96,485 \, \text{C mol}^{-1}$ )
$E_{\text{cell}}$ = cell potential (volts)

Why negative? A spontaneous reaction has $\Delta G < 0$ and $E_{\text{cell}} > 0$ — the negative sign makes this consistent.

2. The Chemical Side: $\Delta G$ Depends on Concentration

For a general redox reaction:

aA + bB \rightarrow cC + dD

The Gibbs free energy under non-standard conditions is:

\Delta G = \Delta G^\circ + RT \ln Q

Where $Q$ is the reaction quotient:

Q = \frac{[C]^c [D]^d}{[A]^a [B]^b}

Why this form? It comes from the relationship between chemical potential and concentration — the entropy of mixing drives concentration dependence.

3. Combining Both Sides: The Derivation

Set the electrical work equal to the chemical free energy change:

-nFE_{\text{cell}} = -nFE^\circ_{\text{cell}} + RT \ln Q

Divide both sides by $-nF$ :

E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln Q

This is the Nernst equation.

4. The "Why" in Plain Terms

Concept	Physical Meaning
$E^\circ_{\text{cell}}$	Voltage when all species are at 1 M (standard state)
$-\frac{RT}{nF} \ln Q$	Correction factor — adjusts voltage for real concentrations
$Q$	Tells you how far the reaction is from equilibrium

Key insight: When $Q = K$ (equilibrium), $E_{\text{cell}} = 0$ — the battery is dead because no net reaction occurs.

5. The Common Form (log base 10)

At $25^\circ \text{C}$ ( $298 \, \text{K}$ ):

\frac{RT}{F} \ln 10 \approx 0.0592 \, \text{V}

So:

E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0592}{n} \log_{10} Q

Why convert to log? Exam convenience — most concentration values are powers of 10.

--- …

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The cell potential $E_{Cell}$ and the Gibbs free energy change $\Delta_r G$ become zero when the cell reaction reaches equilibrium — that is, when the concentrations of reactants and products are such that the reaction quotient $Q$ equals the equilibrium constant $K$ .

The Nernst equation is the bridge between electrochemistry and thermodynamics. For a cell reaction, it tells us how the voltage changes as the reaction proceeds. The key idea is simple: a battery runs because the reaction is not yet balanced — there’s a driving force. Once the reaction reaches equilibrium, that driving force vanishes, and both $E_{Cell}$ and $\Delta_r G$ drop to zero.

Let’s see why, step by step.

Start with the Nernst equation for a cell. For a general cell reaction:

aA + bB \rightarrow cC + dD

the cell potential at any point is:

E_{Cell} = E^\circ_{Cell} - \frac{RT}{nF} \ln Q

where $Q = \frac{[C]^c [D]^d}{[A]^a [B]^b}$ is the reaction quotient, $n$ is the number of electrons transferred, $F$ is Faraday’s constant, $R$ is the gas constant, and $T$ is the temperature.

Recall the thermodynamic link. The Gibbs free energy change for the cell reaction is:

\Delta_r G = -nF E_{Cell}

This is a direct relation — if $E_{Cell} = 0$ , then $\Delta_r G = 0$ , and vice versa. So the condition is the same for both.

What happens as the reaction proceeds?

As the cell discharges, reactants are consumed and products build up. The value of $Q$ increases. The Nernst equation shows that $E_{Cell}$ decreases as $Q$ grows. This continues until the system can go no further — that is, until chemical equilibrium is reached.
At equilibrium, $Q$ becomes $K$ .

At equilibrium, the reaction quotient equals the equilibrium constant: $Q = K$ . Plugging this into the Nernst equation:

E_{Cell} = E^\circ_{Cell} - \frac{RT}{nF} \ln K

But we also know from thermodynamics that at equilibrium:

\Delta_r G^\circ = -RT \ln K

And since $\Delta_r G^\circ = -nF E^\circ_{Cell}$ , we get:

E^\circ_{Cell} = \frac{RT}{nF} \ln K

Substituting this back:

$E_{Cell} = \frac{RT}{nF} \ln K - \frac{RT}{nF} \ln K = 0$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
$2\text{Cr}(s) + 3\text{Cd}^{2+}(aq) \rightarrow 2\text{Cr}^{3+}(aq) + 3\text{Cd}(s)$
›Reveal solutionSolution
The standard Gibbs free energy change is found from the cell potential via $\Delta_r G^\Theta = -nFE^\Theta_\text{cell}$ .

With $E^\Theta_\text{cell} = +0.34\ \text{V}$ and $n = 6$ , we get $\Delta_r G^\Theta = -196.86\ \text{kJ mol}^{-1}$ , so the correct option is (A).

The key idea is that a galvanic cell converts chemical energy into electrical work. The standard Gibbs free energy change $\Delta_r G^\Theta$ is directly related to the maximum electrical work the cell can do, which is given by the product of the charge transferred and the cell potential.

We are given the overall reaction:

$2\text{Cr}(s) + 3\text{Cd}^{2+}(aq) \rightarrow 2\text{Cr}^{3+}(aq) + 3\text{Cd}(s)$

To find $\Delta_r G^\Theta$ , we need the standard cell potential $E^\Theta_\text{cell}$ and the number of electrons transferred $n$ .

Identify the half-reactions and their standard potentials

Reduction of cadmium: $\text{Cd}^{2+} + 2e^- \rightarrow \text{Cd}(s)$ , $E^\circ = -0.40\ \text{V}$

Reduction of chromium: $\text{Cr}^{3+} + 3e^- \rightarrow \text{Cr}(s)$ , $E^\circ = -0.74\ \text{V}$

In the overall reaction, $\text{Cr}$ is oxidized (it loses electrons) and $\text{Cd}^{2+}$ is reduced. So the cell potential is:

$E^\Theta_\text{cell} = E^\Theta_\text{cathode} - E^\Theta_\text{anode}$

The cathode is where reduction occurs: $\text{Cd}^{2+}|\text{Cd}$ with $E^\circ = -0.40\ \text{V}$

The anode is where oxidation occurs: $\text{Cr}|\text{Cr}^{3+}$ with $E^\circ = -0.74\ \text{V}$ (but we use the reduction potential as given, then subtract).

Thus:

$E^\Theta_\text{cell} = (-0.40) - (-0.74) = +0.34\ \text{V}$

Determine the number of electrons transferred ( $n$ )

From the half-reactions:

$\text{Cd}^{2+} + 2e^- \rightarrow \text{Cd}$ (each Cd²⁺ takes 2 electrons)

$\text{Cr} \rightarrow \text{Cr}^{3+} + 3e^-$ (each Cr loses 3 electrons)

To balance the overall reaction, we need the least common multiple of 2 and 3, which is 6.

Multiply the cadmium half-reaction by 3: $3\text{Cd}^{2+} + 6e^- \rightarrow 3\text{Cd}$

Multiply the chromium half-reaction by 2: $2\text{Cr} \rightarrow 2\text{Cr}^{3+} + 6e^-$

So $n = 6$ moles of electrons transferred per mole of reaction as written.

Apply the relationship between $\Delta_r G^\Theta$ and $E^\Theta_\text{cell}$ …
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TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
$^+$
›Reveal solutionSolution
The cell potential depends on the reaction quotient $Q$ via the Nernst equation; increasing $[Ag^+]$ makes $Q$ smaller, which raises $E$ above $E^\circ$ . The correct choice is (A).

The key concept here is the Nernst equation, which tells us how the cell potential changes when concentrations deviate from standard conditions (1 M for solutes, 1 atm for gases, pure solids ignored). For the reaction
$\text{Ni}(s) + 2\text{Ag}^+(aq) \rightarrow \text{Ni}^{2+}(aq) + 2\text{Ag}(s)$
the standard potential is $E^\circ = 1.05\ \text{V}$ . The student wants $E > E^\circ$ . Since $E^\circ$ is fixed, we must adjust the reaction quotient $Q$ to make $E$ larger.

Write the Nernst equation for this cell At 25°C, the Nernst equation is

$E = E^\circ - \frac{0.0592}{n} \log Q$
where $n$ is the number of electrons transferred. Here, Ni loses 2 electrons and each Ag⁺ gains 1, so $n = 2$ .

The reaction quotient is
$Q = \frac{[\text{Ni}^{2+}]}{[\text{Ag}^+]^2}$
(Solids like Ni and Ag do not appear in $Q$ .)

How to make $E > E^\circ$ ? Since $E = E^\circ - \frac{0.0592}{2} \log Q$ , we need $E > E^\circ$ , which means

$-\frac{0.0592}{2} \log Q > 0 \quad \Rightarrow \quad \log Q < 0 \quad \Rightarrow \quad Q < 1$
So the cell potential exceeds the standard value when the reaction quotient is less than 1.

What does $Q < 1$ mean in terms of concentrations?

$Q = \frac{[\text{Ni}^{2+}]}{[\text{Ag}^+]^2} < 1 \quad \Rightarrow \quad [\text{Ni}^{2+}] < [\text{Ag}^+]^2$
Under standard conditions, both concentrations are 1 M, so $Q = 1$ and $E = E^\circ$ . To make $Q < 1$ , we can either decrease $[\text{Ni}^{2+}]$ or increase $[\text{Ag}^+]$ .

Evaluate the options

(A) Increase the concentration of Ag⁺(aq) → This makes the denominator larger, so $Q$ becomes smaller → $E$ increases. ✓ …
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TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
$\ce{Fe^{2+}(aq) + Ag^+(aq) -> Fe^{3+}(aq) + Ag(s)}$
›Reveal solutionSolution
The standard cell potential is $E^\Theta_{\text{cell}} = 0.03\ \text{V}$ , giving $\Delta_r G^\Theta = -nFE^\Theta_{\text{cell}} = -2.895\ \text{kJ mol}^{-1}$ and $\log K_c = \frac{nE^\Theta_{\text{cell}}}{0.059} \approx 0.508$ , so the correct pair is option (C).

Concept and intuition:

In a galvanic cell, the spontaneous reaction drives electrons from the anode (oxidation) to the cathode (reduction). The standard cell potential $E^\Theta_{\text{cell}}$ is the difference between the reduction potentials of the two half-cells. From $E^\Theta_{\text{cell}}$ we directly get the Gibbs free energy change via $\Delta_r G^\Theta = -nFE^\Theta_{\text{cell}}$ , and the equilibrium constant via the Nernst equation at standard conditions: $\log K_c = \frac{nE^\Theta_{\text{cell}}}{0.059}$ at 298 K. The sign of $\Delta_r G^\Theta$ must be negative for a spontaneous reaction, which immediately eliminates options with positive $\Delta_r G^\Theta$ .

Step-by-step reasoning:

Identify the half-reactions and their standard potentials.

Reduction: $\ce{Ag^+ + e^- -> Ag(s)}$ , $E^\Theta = +0.80\ \text{V}$ (cathode).

Oxidation: $\ce{Fe^{2+} -> Fe^{3+} + e^-}$ , but we have the reduction potential for $\ce{Fe^{3+}/Fe^{2+}} = +0.77\ \text{V}$ . For oxidation, we reverse the sign: $E^\Theta_{\text{ox}} = -0.77\ \text{V}$ . The cell potential is $E^\Theta_{\text{cell}} = E^\Theta_{\text{cathode}} - E^\Theta_{\text{anode}} = 0.80 - 0.77 = 0.03\ \text{V}$ .

Determine the number of electrons transferred ( $n$ ).

The balanced reaction $\ce{Fe^{2+} + Ag^+ -> Fe^{3+} + Ag}$ involves one electron transfer: $\ce{Fe^{2+} -> Fe^{3+} + e^-}$ and $\ce{Ag^+ + e^- -> Ag}$ . So $n = 1$ .

Calculate $\Delta_r G^\Theta$ .

$\Delta_r G^\Theta = -nFE^\Theta_{\text{cell}} = -1 \times 96500\ \text{C mol}^{-1} \times 0.03\ \text{V}$ .

Since $1\ \text{C}\cdot\text{V} = 1\ \text{J}$ , we get $\Delta_r G^\Theta = -2895\ \text{J mol}^{-1} = -2.895\ \text{kJ mol}^{-1}$ .

Calculate $\log K_c$ .

At 298 K, the Nernst equation simplifies to $E^\Theta_{\text{cell}} = \frac{0.059}{n} \log K_c$ (using $R = 8.3$ , $F = 96500$ , and $\ln \to \log_{10}$ factor 2.303). …
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TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
$^{3+}$
›Reveal solutionSolution

In the given spontaneous cell $\text{Zn}^{2+}/\text{Zn}$ is the cathode, so $E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}$ gives $E^\circ_{\text{M}^{3+}/\text{M}} = -0.76-0.90 = -1.66$ V.

The spontaneous reaction
$2\text{M}(s) + 3\text{Zn}^{2+}(aq) \rightarrow 2\text{M}^{3+}(aq) + 3\text{Zn}(s),\qquad E^\circ_{cell}=+0.90\text{ V}$
has $\text{Zn}^{2+}$ reduced (cathode) and M oxidised (anode). Therefore

$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = E^\circ_{\text{Zn}^{2+}/\text{Zn}} - E^\circ_{\text{M}^{3+}/\text{M}}$ …
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TG EAPCET 2023Set ap-2023-05-10-FN1 markMCQ
$E^\circ_{\ce{Cu^{2+}/Cu}}$
›Reveal solutionSolution
The standard potential for the two-electron reduction $\ce{Cu^{2+} -> Cu}$ is not the sum of the given potentials — it is the weighted average of the Gibbs free energy changes. The correct value is $0.336\ \text{V}$ , option (D).

The trap here is tempting: just add $0.153\ \text{V}$ and $0.520\ \text{V}$ to get $0.673\ \text{V}$ . That would be the answer if potentials were additive — but they are not. Electrode potentials are intensive properties; they depend on the number of electrons transferred. What is additive is the Gibbs free energy change, $\Delta G^\circ = -nFE^\circ$ .

So to find $E^\circ$ for $\ce{Cu^{2+} + 2e^- -> Cu}$ , we must work through the free energies.

Write the half-reactions with their $n$ and $\Delta G^\circ$ .

For $\ce{Cu^{2+} + e^- -> Cu^+}$ :

$n_1 = 1$ , $E_1^\circ = +0.153\ \text{V}$

$\Delta G_1^\circ = -n_1 F E_1^\circ = -F(0.153)$

For $\ce{Cu^+ + e^- -> Cu}$ :

$n_2 = 1$ , $E_2^\circ = +0.520\ \text{V}$

$\Delta G_2^\circ = -F(0.520)$

Add the two steps to get the overall reaction.

$\ce{Cu^{2+} + e^- -> Cu^+}$

$\ce{Cu^+ + e^- -> Cu}$

Sum: $\ce{Cu^{2+} + 2e^- -> Cu}$

The overall $\Delta G^\circ$ is the sum:

$\Delta G^\circ_{\text{total}} = \Delta G_1^\circ + \Delta G_2^\circ = -F(0.153 + 0.520) = -F(0.673)$

Relate total $\Delta G^\circ$ to the overall $E^\circ$ .

For the overall reaction, $n = 2$ . So:

$\Delta G^\circ_{\text{total}} = -n F E^\circ_{\ce{Cu^{2+}/Cu}} = -2F E^\circ_{\ce{Cu^{2+}/Cu}}$

Equate:

$-2F E^\circ_{\ce{Cu^{2+}/Cu}} = -F(0.673)$

Cancel $-F$ (non-zero):

$2 E^\circ_{\ce{Cu^{2+}/Cu}} = 0.673$ …
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TG EAPCET 2023Set ap-2023-05-10-AN1 markMCQ
$^{2+}$
›Reveal solutionSolution
The standard Gibbs energy change is directly related to the cell potential by $\Delta G^\circ = -nFE^\circ_{\text{cell}}$ . For this reaction, $n = 2$ electrons transferred, so $\Delta G^\circ = -2 \times 96500 \times 0.51 = -98430\ \text{J mol}^{-1} = -98.43\ \text{kJ mol}^{-1}$ , matching option (D).

The key idea is that the standard Gibbs free energy change ( $\Delta G^\circ$ ) for a spontaneous electrochemical cell reaction is negative and proportional to the cell potential. The relationship is $\Delta G^\circ = -nFE^\circ_{\text{cell}}$ , where $n$ is the number of moles of electrons transferred in the balanced reaction, and $F$ is Faraday’s constant (charge per mole of electrons). This formula comes from the fact that electrical work done by the cell equals the charge transferred times the potential difference, and at constant temperature and pressure, that work equals the decrease in Gibbs free energy.

Determine the number of electrons transferred ( $n$ ).

The half-reactions are:

Oxidation: $\text{Zn(s)} \rightarrow \text{Zn}^{2+}(\text{aq}) + 2e^-$

Reduction: $\text{Ni}^{2+}(\text{aq}) + 2e^- \rightarrow \text{Ni(s)}$ Each zinc atom loses two electrons, and each nickel ion gains two electrons. So $n = 2$ .

Apply the formula $\Delta G^\circ = -nFE^\circ_{\text{cell}}$ .

Given:

$E^\circ_{\text{cell}} = 0.51\ \text{V}$

$F = 96500\ \text{C mol}^{-1}$

Therefore:

$\Delta G^\circ = -2 \times 96500 \times 0.51$

Calculate step by step. First, $2 \times 96500 = 193000$ . Then, $193000 \times 0.51 = 98430$ . …
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TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
$25~^\circ \mathrm{C}$
›Reveal solutionSolution
Converting $\Delta U$ to $\Delta H$ using the $P\Delta V$ work of the expanding gas, then applying $\Delta G = \Delta H - T\Delta S$ , gives $\Delta G \approx -2.67~\mathrm{k.Cal}$ , so the correct option is (A).

Concept. We're given $\Delta U$ and $\Delta S$ for a reaction where a liquid converts into gas. Since gas is produced, the reaction does $P\Delta V$ work on the surroundings, so $\Delta H \ne \Delta U$ . We first find $\Delta H$ , then apply $\Delta G = \Delta H - T\Delta S$ .

Relation between $\Delta H$ and $\Delta U$ . At constant pressure, $\Delta H = \Delta U + P\Delta V = \Delta U + \Delta n_g RT$ , where $\Delta n_g$ is the change in moles of gas.

Find $\Delta n_g$ . Reaction: $X_2O_4(l) \rightarrow 2XO_2(g)$ . The reactant is a liquid (0 mol gas); the product is 2 mol gas. So $\Delta n_g = 2$ .

Compute $P\Delta V$ . With $R = 2~\mathrm{Cal\,K^{-1}mol^{-1}}$ and $T = 25^\circ\mathrm{C} = 298~\mathrm{K}$ :

$P\Delta V = \Delta n_g RT = 2 \times 2 \times 298 = 1192~\mathrm{Cal} = 1.192~\mathrm{k.Cal}$

Compute $\Delta H$ .

$\Delta H = \Delta U + P\Delta V = 2.1 + 1.192 = 3.292~\mathrm{k.Cal}$

Compute $T\Delta S$ . Converting $\Delta S = 20~\mathrm{Cal/K}$ to k.Cal/K: $\Delta S = 0.020~\mathrm{k.Cal/K}$ . $T\Delta S = 298 \times 0.020 = 5.96~\mathrm{k.Cal}$ …
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←Previous questionQ19 — Can E^°Cell or Δr G^° for cell reaction ever be equal to zero?→Next questionQ21 — What does the negative sign in the expression E^°Zn²⁺/Zn = -0.76V mean…

Q.Under what condition is $E_{Cell} = 0$ or $\Delta_r G = 0$ ?

Concept understanding — Cell Representation Nernst Equation

Cell Representation and the Nernst Equation: From Intuition to Precision

The Intuition First

The Precise Statement

Cell Representation: How We Write It

Applying the Nernst Equation to a Cell Representation

A Worked Example

Why this formula?

Cell Representation & the Nernst Equation: Why It Works

The Core Question

1. The Fundamental Link: Gibbs Free Energy & Cell Potential

2. The Chemical Side: $\Delta G$ Depends on Concentration

3. Combining Both Sides: The Derivation

4. The "Why" in Plain Terms

5. The Common Form (log base 10)

Method: Nernst Equation Analysis for Equilibrium Condition

Concept Foundation

Step-by-Step Solution

Common Mistakes & How to Avoid Them

Mistake 1: Confusing $E_{Cell}$ (actual) with $E^\circ_{Cell}$ (standard)

Mistake 2: Thinking the reaction "stops" in the sense of no longer occurring at all

Mistake 3: Forgetting the direct link between $E_{Cell}$ and $\Delta_r G$

Q.Under what condition is ECell=0E_{Cell} = 0ECell​=0 or ΔrG=0\Delta_r G = 0Δr​G=0?

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Q.Under what condition is $E_{Cell} = 0$ or $\Delta_r G = 0$ ?

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