Write de Broglie's wave equation and explain terms in it.

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Q. Write de Broglie's wave equation and explain terms in it.

Karnataka PUCKarnataka School Examination and Assessment Board (1st PUC) 2026Subjective· 2mImportance★★★★★

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Concept understanding — Standard Reduction Potential

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. …

Why this formula?

Standard Reduction Potential: Why the Formula Holds

Let's build this from first principles — understanding why before memorising what.

1. The Core Idea: A Half-Cell's "Tendency to Gain Electrons"

A standard reduction potential ( $E^\circ$ ) measures how strongly a species wants to gain electrons (be reduced) under standard conditions (1 M concentration, 1 atm pressure, 25°C).

But why can't we measure this directly? Because every reduction must be paired with an oxidation — you can't have electrons flowing without a complete circuit.

2. The Formula: $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$

Why subtraction, not addition?

Consider a Daniell cell:

Zn | Zn²⁺ (1 M) || Cu²⁺ (1 M) | Cu

Experimentally, we measure the cell potential as +1.10 V.

Now, we define the standard hydrogen electrode (SHE) as exactly 0.00 V:

2\text{H}^+ + 2e^- \rightarrow \text{H}_2 \quad E^\circ = 0.00 \text{ V}

The reasoning step-by-step:

We can only measure differences — like measuring height difference between two points.
If we connect the SHE to the copper half-cell, we measure +0.34 V (Cu²⁺ is reduced).
If we connect the SHE to the zinc half-cell, we measure −0.76 V (Zn²⁺ is reduced less readily than H⁺).

Now, the cell potential is the difference in their tendencies:

E^\circ_{\text{cell}} = E^\circ_{\text{Cu}} - E^\circ_{\text{Zn}} = (+0.34) - (-0.76) = +1.10 \text{ V}

Key insight: The formula uses subtraction because we're comparing two half-cells against the same reference (SHE). The cathode is where reduction happens (higher $E^\circ$ ), the anode is where oxidation happens (lower $E^\circ$ ).

3. The Nernst Equation: Why $E = E^\circ - \frac{RT}{nF} \ln Q$

This is the thermodynamic derivation — the real "why."

From Gibbs free energy:

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

For an electrochemical cell:

\Delta G = -nFE \quad \text{and} \quad \Delta G^\circ = -nFE^\circ

Substituting:

-nFE = -nFE^\circ + RT \ln Q

Rearranging:

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

Why this makes physical sense:

$RT \ln Q$ represents the entropy penalty of non-standard concentrations
$nF$ converts charge to energy (Faraday's constant × number of electrons)
The minus sign means: as products accumulate ( $Q$ increases), the cell potential drops — the reaction is approaching equilibrium

At equilibrium ( $Q = K$ ), $E = 0$ — the battery is "dead."

--- …

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[!TLDR]

λ = h / (m v)

Method

de Broglie proposed that every moving particle has an associated wavelength λ = h/mv (equivalently λ=h/p), where λ is the de Broglie wavelength, h is Planck's constant, m is the …

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

CBSE 2026Set ANNUAL1 markMCQ
$Q. A transition element X is placed in the first series of transition elements. It has the following characteristics:$
(i) It is a coinage metal.
(ii) It has positive reduction potential.
(iii) It does not react with HCl but reacts with nitric acid. The element 'X' is likely to be
(a) Zn
(b) Cu
(c) Au
(d) Ag
›Reveal solutionSolution
Among the first-series transition metals, copper uniquely fits all three clues: it is a coinage metal, has a positive standard reduction potential, and dissolves only in oxidising acids like HNO3, not in HCl.

Coinage metals are Cu, Ag, and Au - but the question specifies the element is in the FIRST series of transition elements (3d series), which rules out Ag (4d series) and Au (5d series), leaving only Cu.

Cu has E°(Cu2+/Cu) = +0.34 V, a positive reduction potential, meaning it is less reactive than hydrogen and will not displace H2 from dilute HCl (a non-oxidising acid). …
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CBSE 2025Set ANNUAL1 markMCQ
$KMnO_4$
(a) reduces permanganate to $Mn^{2+}$
(b) oxidises oxalic acid to $CO_2$ and $H_2O$
(c) gets oxidised by oxalic acid to chlorine
(d) furnishes $H^+$ ions in addition to those from oxalic acid.
›Reveal solutionSolution

KMnO₄ is a strong enough oxidiser to attack the chloride ion of HCl itself, converting it to chlorine gas and introducing a side reaction that spoils the titration — this rules out using HCl as the acidifying agent.

$KMnO_4$ is a powerful oxidising agent. If dilute HCl were used to acidify the oxalic acid solution for this titration, the chloride ion ( $Cl^-$ ) itself would be oxidised by $KMnO_4$ to chlorine gas, as an unwanted side reaction:

$2KMnO_4 + 16HCl \rightarrow 2KCl + 2MnCl_2 + 5Cl_2\uparrow + 8H_2O$

…

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Q.Write de Broglie's wave equation and explain terms in it.

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