The electrode potential of a magnesium electrode varies with the concentration of Mg2+ ions according to EMg²⁺

NCERT Exemplar · Q2

E_{Mg^{2+}\mid Mg} = E^{\circ}_{Mg^{2+}\mid Mg} - \frac{0.059}{2}\log\frac{1}{[Mg^{2+}]}

(i) A straight line of positive slope that rises from the lower-left toward the upper-right, cutting the E-axis at a negative (below-origin) intercept.

(ii) A straight line of positive slope but drawn entirely in the positive-E region, cutting the E-axis at a positive (above-origin) intercept.

(iii) An upward-curving (concave-up) curve in which E rises more and more steeply as

\log[Mg^{2+}]

increases.

(iv) A straight line of negative slope that falls from the upper-left toward the lower-right, so that E decreases as

\log[Mg^{2+}]

increases.

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Concept understanding — Nernst Equation

The Nernst Equation: Why Batteries Don't Always Give Their Rated Voltage

Imagine you have a fresh AA battery. It says 1.5 V on the side. But if you measure it with a voltmeter, you might get 1.58 V when it's new, and 1.2 V when it's almost dead. Why does the voltage change? The Nernst equation is the tool that tells you exactly why.

The Core Idea: Concentration Drives Voltage

Every electrochemical cell works because of a chemical reaction that wants to happen. But here's the key: how badly the reaction wants to happen depends on how much of each chemical is present.

Think of it like a slope. A steep hill gives you more energy when you roll down. A shallow hill gives you less. In a battery, the "hill" is the difference in concentration (or more precisely, activity) of ions between the two electrodes. When the battery is fresh, the hill is steep — lots of reactants, few products. As the battery runs, reactants get used up, products build up, the hill flattens, and the voltage drops.

The Nernst equation is the mathematical formula that calculates the exact voltage for any given set of concentrations.

The Precise Statement

For a general electrochemical reaction:

aA + bB \rightarrow cC + dD

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

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

Where:

$E$ = cell potential under the given conditions (what you actually measure)
$E^\circ$ = standard cell potential (the voltage when all reactants and products are at 1 M concentration, 1 atm pressure, 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 reaction
$F$ = Faraday constant (96,485 C/mol)
$Q$ = reaction quotient = $\frac{[C]^c [D]^d}{[A]^a [B]^b}$ (using concentrations for now)

At 25°C (298 K), the constants combine into a simpler form:

E = E^\circ - \frac{0.0592}{n} \log_{10} Q

Note

The 0.0592 comes from $\frac{2.303RT}{F}$ at 298 K. The 2.303 converts natural log to base-10 log, which is more convenient for calculations.

What It Actually Means

The equation has three parts:

$E^\circ$ — the "ideal" voltage when everything is at standard conditions. This is what you'd get in a textbook table.
$\frac{RT}{nF}$ — a scaling factor. It tells you how sensitive the voltage is to concentration changes. More electrons transferred ( $n$ ) means less sensitivity.
$\ln Q$ — the "concentration penalty". When $Q$ is small (lots of reactants, few products), $\ln Q$ is negative, so $E$ is higher than $E^\circ$ . When $Q$ is large (products building up), $\ln Q$ is positive, so $E$ drops below $E^\circ$ .

A Concrete Example

Consider the Daniell cell: $\text{Zn} | \text{Zn}^{2+} || \text{Cu}^{2+} | \text{Cu}$

The reaction is: $\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}$

$E^\circ = 1.10 \text{ V}$ , $n = 2$

If $[\text{Cu}^{2+}] = 0.1 \text{ M}$ and $[\text{Zn}^{2+}] = 1.0 \text{ M}$ :

Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \frac{1.0}{0.1} = 10

E = 1.10 - \frac{0.0592}{2} \log_{10}(10) = 1.10 - 0.0296 \times 1 = 1.07 \text{ V}

The voltage dropped by 0.03 V because the copper ion concentration is lower than standard.

Watch out

A common mistake: forgetting that $Q$ uses the concentrations of aqueous species and gases (as partial pressures), but not pure solids or liquids. In the Daniell cell, solid Zn and Cu don't appear in $Q$ .

Why It Matters

The Nernst equation isn't just for batteries. It explains:

Why a pH meter works (it measures the voltage across a membrane sensitive to H⁺ concentration)
How nerve cells maintain their resting potential (concentration gradients of Na⁺ and K⁺ across the cell membrane)
Why corrosion happens faster in salt water (the Nernst equation shows that lower ion concentrations can make metals more reactive)

The Takeaway

The Nernst equation is the bridge between thermodynamics (how much energy a reaction could release) and real-world conditions (what's actually in the beaker). It tells you that voltage isn't fixed — it's a dynamic quantity that responds to what's happening inside the cell.

The Nernst equation is one of the most heavily tested formulas in the NCERT/CBSE Class 12 Chemistry Electrochemistry chapter, and ‘Nernst equation derivation’ or ‘Nernst equation numericals’ appear repeatedly in board important-questions lists and JEE Main/NEET chemistry papers. Being comfortable with this equation is essential for solving cell-potential problems in competitive exams.

Rewriting the Nernst expression shows E depends linearly on $\log[Mg^{2+}]$ with a positive slope. Because Mg has a negative standard electrode potential, the line rises from a negative intercept. So the correct graph is a straight line going up from lower-left to upper-right (option A / graph (i)).

Concept

The potential of a single electrode follows the Nernst equation. For the half-reaction $Mg^{2+} + 2e^- \rightarrow Mg$ the given form is

E_{Mg^{2+}\mid Mg} = E^{\circ}_{Mg^{2+}\mid Mg} - \frac{0.059}{2}\log\frac{1}{[Mg^{2+}]}.

Why this form

A graph is easiest to read when the equation is in the straight-line form $y = mx + c$ . Here $y = E$ , $x = \log[Mg^{2+}]$ .

Steps

Use the log identity $\log\frac{1}{[Mg^{2+}]} = -\log[Mg^{2+}]$ .
Substitute:

E = E^{\circ} - \frac{0.059}{2}\big(-\log[Mg^{2+}]\big) = E^{\circ} + \frac{0.059}{2}\log[Mg^{2+}].

Compare with $y = mx + c$ : slope $m = +\frac{0.059}{2} = +0.0295$ (positive), intercept $c = E^{\circ}_{Mg^{2+}\mid Mg}$ .
So E increases linearly as $\log[Mg^{2+}]$ increases — a rising straight line.
The standard reduction potential of magnesium is negative ( $E^{\circ} \approx -2.37\ \text{V}$ ), so the intercept lies below the origin.

Eliminating the distractors

B — a rising straight line, but drawn with a positive intercept; it cannot represent Mg, whose $E^{\circ}$ is negative.
C — a curve; the relation is linear, not curved, so this is wrong.
D — a falling straight line (negative slope); the slope here is positive, so this is wrong.

✓Final answer

0.0295

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Concept

Why this form

Steps

Eliminating the distractors

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