Write the Nernst equation for the cell reaction in the Daniel cell. How will the ECell be affected when concen

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E_{Cell}

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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 Nernst equation for the Daniel cell is $E_{cell} = E^\circ_{cell} - \frac{0.059}{2} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$ at 298 K. Increasing $[Zn^{2+}]$ increases the log term, which decreases $E_{cell}$ .

The Daniel cell is the classic example of a galvanic cell — zinc and copper electrodes in their respective sulfate solutions, connected by a salt bridge. The cell reaction is:

Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)

The driving force for this reaction is the difference in reduction potentials. Zinc is more reactive (easier to oxidise), so it acts as the anode. Copper ions are more easily reduced, so copper is the cathode.

The Nernst equation lets us calculate the actual cell potential under non-standard conditions — when concentrations aren't 1 M. It adjusts the standard potential $E^\circ_{cell}$ by a term that depends on the reaction quotient $Q$ .

aA + bB \rightarrow cC + dD

Now let's apply this to the Daniel cell step by step.

Identify $n$ , the number of electrons transferred.

In the balanced reaction $Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu$ , each zinc atom loses 2 electrons, and each copper ion gains 2 electrons. So $n = 2$ .
Write the reaction quotient $Q$ .

For the reaction as written, $Q = \frac{[Zn^{2+}]}{[Cu^{2+}]}$ . Solids (Zn and Cu) have activity = 1, so they don't appear.
Plug into the Nernst equation.

At 298 K:

E_{cell} = E^\circ_{cell} - \frac{0.059}{2} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}

This is the required Nernst equation for the Daniel cell.

Now analyse the effect of increasing $[Zn^{2+}]$ .

Look at the log term: $\log \frac{[Zn^{2+}]}{[Cu^{2+}]}$ . If $[Zn^{2+}]$ increases while $[Cu^{2+}]$ stays the same, the fraction becomes larger, so $\log$ becomes larger (less negative, or more positive).

Since this log term is subtracted from $E^\circ_{cell}$ , a larger log term means a smaller $E_{cell}$ . …

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Common Mistakes: Cell Representation & Nernst Equation (Daniel Cell)

Mistake 1: Writing the Nernst equation with wrong sign or form

The error:

Students often write:

E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log \frac{[\text{Anode}]}{[\text{Cathode}]}

...but then flip the ratio or forget the sign.

Why it happens:

They memorise the formula without understanding that the reaction quotient $Q$ is always products over reactants — and in a Daniel cell, the spontaneous reaction is:

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

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

How to avoid:

Always write the balanced cell reaction first, then:

E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log Q

where $Q = \frac{[\text{products}]}{[\text{reactants}]}$ (excluding solids). For Daniel cell:

E_{cell} = E^\circ_{cell} - \frac{0.0591}{2} \log \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}

Mistake 2: Confusing anode and cathode in the ratio

The error:

Writing $\log \frac{[\text{Cu}^{2+}]}{[\text{Zn}^{2+}]}$ instead of the correct $\log \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}$ .

Why it happens:

Students think "cathode over anode" because reduction happens at cathode — but the Nernst equation uses the reaction quotient, not electrode labels.

How to avoid:

Remember: Zn is oxidised (anode, loses electrons) → Zn²⁺ appears in products. Cu²⁺ is reduced (cathode) → appears in reactants. So:

Products: Zn²⁺ (numerator)
Reactants: Cu²⁺ (denominator)

Mistake 3: Forgetting $n = 2$ in the Daniel cell

The error:

Using $n = 1$ or writing $0.0591$ without dividing by $n$ .

Why it happens:

Rushing — the Daniel cell involves transfer of 2 electrons:

\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-

\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}

How to avoid:

Always count electrons from the balanced half-reactions. For Daniel cell, $n = 2$ always.

Mistake 4: Misinterpreting the effect of increasing [Zn²⁺]

The error:

Saying "E_cell increases" or "no change" when [Zn²⁺] is increased.

Why it happens:

Students think "more ions = more reaction" without checking the Nernst equation.

The correct reasoning:

From the Nernst equation:

E_{cell} = E^\circ_{cell} - \frac{0.0591}{2} \log \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}

If [Zn²⁺] increases:

The fraction $\frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}$ increases
$\log$ of a larger number is more positive
Subtracting a larger positive number → $E_{cell}$ decreases

How to avoid:

Always plug into the equation qualitatively:

More products → larger $Q$ → more negative $\log$ term → lower $E_{cell}$

--- …

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

AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
$Zn^{2+}$
›Reveal solutionSolution
The cell is a Daniell cell (Zn|Zn²⁺||Cu²⁺|Cu) with an external opposing voltage $E_{\text{ext}} < 1.1\ \text{V}$ . Since the opposing voltage is less than the cell’s standard emf (1.1 V), the cell still drives electrons from Zn (anode) to Cu (cathode) — so Statement I is correct. As the cell operates, Zn dissolves (weight decreases) and Cu deposits (weight increases) — so Statement II is also correct. Hence the correct option is (A).

Concept and Intuition

This is a Daniell cell — the classic Zn–Cu galvanic cell. Normally, without any external source, electrons flow spontaneously from the Zn electrode (anode, oxidation) to the Cu electrode (cathode, reduction). The standard cell potential is about 1.1 V.

Here, an external voltage source $E_{\text{ext}}$ is inserted in the external circuit, but it is less than 1.1 V. That means the external source is trying to oppose the cell’s natural electron flow, but it is too weak to reverse it. The cell still operates as a galvanic cell (not an electrolytic cell), because the net driving force remains from Zn to Cu.

The Nernst equation helps us understand that even if concentrations change slightly, the cell potential remains positive as long as the reaction quotient $Q$ is less than the equilibrium constant $K$ . Since $E_{\text{ext}} < E_{\text{cell}}$ , the cell continues to discharge.

Step-by-step reasoning

Identify the spontaneous direction In a Zn–Cu cell, the standard reduction potentials are:

$\begin{aligned} \text{Cu}^{2+} + 2e^- &\rightarrow \text{Cu} \quad E^\circ = +0.34\ \text{V} \\ \text{Zn}^{2+} + 2e^- &\rightarrow \text{Zn} \quad E^\circ = -0.76\ \text{V} \end{aligned}$
The cell reaction is:
$\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}$
with $E^\circ_{\text{cell}} = 0.34 - (-0.76) = 1.10\ \text{V}$ .

Electrons flow from Zn (anode) to Cu (cathode) in the external circuit.

Effect of the external source $E_{\text{ext}} < 1.1\ \text{V}$

The external source is connected opposing the cell’s natural polarity. However, because its voltage is less than the cell’s emf, the net potential difference still drives electrons from Zn to Cu.

Tip

Think of it like two batteries in series opposing: the larger one (the cell) wins, so current flows in its direction. Here the cell is the “larger” battery.

Evaluate Statement I: “Electrons flow from Cu electrode to Zn electrode.”

The statement claims the opposite of the spontaneous direction. But as argued, the external source is too weak to reverse the flow. Therefore electrons still flow from Zn to Cu, not from Cu to Zn.

Statement I is false as written — wait, careful: The statement says “from Cu electrode to Zn electrode”. That is the reverse of the actual direction. So Statement I is incorrect. …
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AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
$Zn^{2+}$
›Reveal solutionSolution
In a standard Zn–Cu galvanic cell, electrons flow from the Zn anode (oxidation) to the Cu cathode (reduction), so conventional current flows from Cu to Zn. The Zn electrode loses mass (Zn → Zn²⁺) and the Cu electrode gains mass (Cu²⁺ → Cu). Therefore Statement I is correct, Statement II is incorrect, and the answer is (C).

Concept and Intuition: The Cell Representation and the Nernst Equation

This question tests your understanding of the direction of current and the mass changes at the electrodes in a working galvanic cell. The key is to remember that in a galvanic (voltaic) cell, the spontaneous redox reaction drives electrons through the external circuit. By convention, current flows opposite to electron flow (from positive to negative). The electrode where oxidation occurs (anode) loses mass; the electrode where reduction occurs (cathode) gains mass. The external opposing voltage $E_{\text{ext}} < 1.1\ \text{V}$ tells us the cell is still operating spontaneously (the cell’s emf is about 1.1 V for Zn–Cu under standard conditions).

Step-by-step reasoning

Identify the spontaneous reaction In a Zn–Cu cell, zinc is more reactive (higher reduction potential for Zn²⁺/Zn is –0.76 V, for Cu²⁺/Cu is +0.34 V). The spontaneous reaction is:

$\text{Zn(s)} + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu(s)}$
Zinc is oxidized (loses electrons), copper ions are reduced (gain electrons).

Determine electron flow and conventional current

Oxidation at the Zn electrode (anode): $\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$ Electrons leave the Zn electrode and travel through the external circuit toward the Cu electrode.

Reduction at the Cu electrode (cathode): $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$ Electrons arrive at the Cu electrode.

Conventional current is defined as the flow of positive charge, opposite to electron flow. So electrons flow from Zn → Cu, meaning conventional current flows from Cu → Zn.

Statement I says: “Current flows from Cu electrode to Zn electrode.” This matches the conventional current direction. Statement I is correct.

Analyze mass changes at the electrodes

At the Zn anode: Zn metal is converted to Zn²⁺ ions, which go into solution. The solid Zn electrode loses mass.

At the Cu cathode: Cu²⁺ ions from solution gain electrons and deposit as solid Cu on the electrode. The Cu electrode gains mass.

Statement II says: “With increase in time the mass of Zn electrode increases and mass of Cu electrode decreases.” This is the exact opposite of what happens. Statement II is incorrect. …
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AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
$\Delta G^\circ$
›Reveal solutionSolution
This applies $\Delta G^\circ = -nFE^\circ_{cell}$ to the Ca/Au3+ redox couple; the standard free energy change works out to $-2.53\times10^3$ kJ.

Concept and Intuition

The standard free energy change of a redox reaction is related to the cell's standard EMF by $\Delta G^\circ = -nFE^\circ_{cell}$ , where $n$ is the total number of electrons transferred as balanced by the overall equation, and $F$ is Faraday's constant. A positive $E^\circ_{cell}$ (spontaneous reaction) corresponds to a negative $\Delta G^\circ$ .

Step-by-Step Solution

Identify the half-reactions: reduction at cathode, $Au^{3+} + 3e^- \rightarrow Au$ , $E^\circ = +1.50$ V; oxidation at anode, $Ca \rightarrow Ca^{2+} + 2e^-$ , $E^\circ_{Ca^{2+}/Ca} = -2.87$ V.

Balance electrons: multiply the Au half-reaction by 2 and the Ca half-reaction by 3, giving $n = 6$ electrons transferred overall, matching the given equation $3Ca + 2Au^{3+} \rightarrow 3Ca^{2+} + 2Au$ .

Compute $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 1.50 - (-2.87) = 4.37$ V.

Compute $\Delta G^\circ = -nFE^\circ_{cell} = -(6)(96500\ \text{C/mol})(4.37\ \text{V})$ . …
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AP EAPCET 2022Set ap-2022-07-11-FN1 markMCQ
$\Delta G^0$
›Reveal solutionSolution
This tests the relation $\Delta G^\circ=-nFE^\circ_{cell}$ , with the key subtlety of finding the correct $n$ (total electrons transferred, found via LCM of the two half-cell electron counts).

Concept and Intuition

The standard Gibbs energy change of a cell reaction is directly proportional to the number of moles of electrons transferred in the balanced overall reaction and the cell potential. Since the anode gives 3 electrons per P atom and the cathode needs 2 electrons per Q $^{2+}$ ion, the reaction must be balanced so both half-reactions transfer the same total number of electrons — found via the LCM of 3 and 2, which is 6.

Step-by-Step Solution

Anode (oxidation): $P \to P^{3+} + 3e^-$ ; Cathode (reduction): $Q^{2+} + 2e^- \to Q$ . …
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AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
$Q. The energy conversion involved in a galvanic cell is (A) Chemical energy to mechanical energy (B) Chemical energy to electrical energy (C) Electrical energy to chemical energy (D) Electrical energy to thermal energy$
›Reveal solutionSolution
A galvanic cell is the device that turns a spontaneous chemical (redox) reaction into usable electrical energy — the opposite of electrolysis.

Concept and Intuition

In a galvanic cell, a spontaneous ( $\Delta G < 0$ ) redox reaction is split into separate oxidation (anode) and reduction (cathode) half-reactions connected by an external circuit, so the electron transfer that would otherwise happen directly (releasing energy as heat) instead flows through a wire and does electrical work. This is the reverse energy conversion of an electrolytic cell, which consumes electrical energy to drive a non-spontaneous chemical reaction.

Step-by-Step Solution

Identify the device: a galvanic/voltaic cell (e.g. Daniell cell) generates electricity from a spontaneous redox reaction.

The reaction proceeds because it is thermodynamically favourable (releases chemical energy). …
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AP EAPCET 2021Set ap-2021-09-07-FN1 markMCQ
$Cr/Cr^{3+}\ //\ Cd^{2+}/Cd$
›Reveal solutionSolution
Compute $E^\circ_{cell}$ from the given half-cell potentials, balance electrons for the overall cell reaction to get $n$ , then use $\Delta G^\circ = -nFE^\circ_{cell}$ .

Concept and Intuition

In the cell notation $Cr/Cr^{3+}\,//\,Cd^{2+}/Cd$ , the left electrode (written first) is the anode (oxidation), and the right electrode is the cathode (reduction). The cell potential is always cathode minus anode: $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$ . Once $E^\circ_{cell}$ is known, $\Delta G^\circ$ follows from the fundamental relation between electrical work and free energy, $\Delta G^\circ = -nFE^\circ_{cell}$ , where $n$ is the total number of electrons transferred in the balanced overall cell reaction (found by matching the electron counts of the two half-reactions via their LCM).

Step-by-Step Solution

Identify electrodes: anode = $Cr^{3+}/Cr$ ( $E^\circ=-0.74\ V$ , oxidation: $Cr\to Cr^{3+}+3e^-$ ), cathode = $Cd^{2+}/Cd$ ( $E^\circ=-0.40\ V$ , reduction: $Cd^{2+}+2e^-\to Cd$ ).

Compute cell potential: $E^\circ_{cell} = E^\circ_{cathode}-E^\circ_{anode} = (-0.40)-(-0.74) = 0.34\ V$ (positive, confirming the reaction is spontaneous as written -- consistent with it being a genuine galvanic cell). …
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AP EAPCET 2021Set eng-2021-10-05-FN1 markMCQ
$X\,|\,X^+(x\ M)\,||\,Y^+(y\ M)\,|\,Y$
›Reveal solutionSolution
By the standard cell-notation convention (anode | ... || ... | cathode) and a positive measured emf, the spontaneous cell reaction is $X + Y^+ \to X^+ + Y$ .

Concept and Intuition

Electrochemical cell notation has a fixed convention: the electrode written on the left is the anode, where oxidation occurs, and the electrode on the right is the cathode, where reduction occurs — the double vertical line "||" represents the salt bridge separating the two half-cells. A positive emf value for the cell as written confirms the reaction is indeed spontaneous in the direction implied by this left-to-right convention (anode oxidation feeding electrons to cathode reduction).

Step-by-Step Solution

Left half-cell, $X\,|\,X^+(x\,M)$ : this is the anode — oxidation occurs here, $X \to X^+ + e^-$ .

Right half-cell, $Y^+(y\,M)\,|\,Y$ : this is the cathode — reduction occurs here, $Y^+ + e^- \to Y$ .

Combine the two half-reactions (balancing the single electron transferred in each): $X + Y^+ \to X^+ + Y$ . …
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Q.Write the Nernst equation for the cell reaction in the Daniel cell. How will the $E_{Cell}$ be affected when concentration of $Zn^{2+}$ ions is increased?

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 for Cell Potential

Steps:

Common Mistakes: Cell Representation & Nernst Equation (Daniel Cell)

Mistake 1: Writing the Nernst equation with wrong sign or form

Mistake 2: Confusing anode and cathode in the ratio

Mistake 3: Forgetting $n = 2$ in the Daniel cell

Mistake 4: Misinterpreting the effect of increasing [Zn²⁺]

Concept and Intuition

Step-by-step reasoning

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Q.Write the Nernst equation for the cell reaction in the Daniel cell. How will the ECellE_{Cell}ECell​ be affected when concentration of Zn2+Zn^{2+}Zn2+ ions is increased?

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Q.Write the Nernst equation for the cell reaction in the Daniel cell. How will the $E_{Cell}$ be affected when concentration of $Zn^{2+}$ ions is increased?

Unlock everything free for 14 days