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

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

Showing the 12 most recent of 28 on this concept.

AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
$\mathrm{Pt, H_2(g) \mid H^+(10^{-8}\,M) \parallel H^+(0.01\,M) \mid H_2(g), Pt}$
›Reveal solutionSolution
This tests the Nernst equation applied to a hydrogen concentration cell; the answer is +0.36 V.

Concept and Intuition

A concentration cell has identical electrodes (here, $H_2$ / $Pt$ ) differing only in electrolyte concentration. No standard cell potential exists ( $E^\circ_{cell}=0$ ); the driving force is entirely the concentration gradient, and the cell spontaneously moves ions/electrons to equalise it — the compartment with the higher $[H^+]$ acts as the cathode (reduction is favoured where $H^+$ is more concentrated).

Step-by-Step Solution

Electrode reaction (as a reduction): $H^+ + e^- \to \tfrac12 H_2(g)$ , with $E^\circ = 0$ V and $n=1$ .

Nernst equation for each half-cell (taking $P_{H_2}=1$ atm): $E = E^\circ - \dfrac{2.303RT}{F}\log\dfrac{1}{[H^+]} = 0.06\log[H^+]$ .

Left electrode (written first, so it is the anode): $[H^+]=10^{-8}$ , $E_{anode}=0.06\times(-8)=-0.48$ V.

Right electrode (cathode): $[H^+]=0.01=10^{-2}$ , $E_{cathode}=0.06\times(-2)=-0.12$ V.

$E_{cell}=E_{cathode}-E_{anode}=-0.12-(-0.48)=+0.36$ V.

Equivalently, directly: $E_{cell}=0.06\log\dfrac{[H^+]_{cathode}}{[H^+]_{anode}}=0.06\log\dfrac{10^{-2}}{10^{-8}}=0.06\times6=+0.36$ V.

Common Mistakes

Forgetting that in a concentration cell $E^\circ_{cell}=0$ , and trying to look up a non-existent standard potential.

Mixing up which electrode is the cathode — it must be the one with the higher $H^+$ concentration (favours reduction).

Sign errors when substituting negative logarithms of very small concentrations.

✓Final answer
$The correct option is (A) — +0.36.$

ANSWER: A
AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
$2M^{3+}(aq) + 2I^-(aq) \rightarrow 2M^{2+}(aq) + I_2(s)$
›Reveal solutionSolution
Apply the Nernst-equilibrium relation $E^0_{cell}=\dfrac{2.303RT}{nF}\log K_c$ with $n=2$ to get $E^0_{cell}\approx0.235\ V$ .

Concept and Intuition

At equilibrium, the standard cell potential is directly linked to the equilibrium constant of the overall cell reaction through $\Delta G^0=-RT\ln K_c=-nFE^0_{cell}$ , which rearranges to $E^0_{cell}=\dfrac{2.303RT}{nF}\log K_c$ . Getting $n$ right (the number of electrons in the balanced equation) is the crucial first step.

Step-by-Step Solution

Balance electrons: $M^{3+}+e^-\rightarrow M^{2+}$ occurs twice (2 electrons), and $2I^-\rightarrow I_2+2e^-$ supplies 2 electrons — so $n=2$ .

$E^0_{cell} = \dfrac{2.303RT}{nF}\log K_c = \dfrac{2.303\times8.3\times298}{2\times96500}\times7.98$ .

Numerator: $2.303\times8.3=19.1149$ ; $19.1149\times298\approx5696.2$ .

Denominator: $2\times96500=193000$ .

$\dfrac{5696.2}{193000}\approx0.02951$ .

$E^0_{cell}\approx0.02951\times7.98\approx0.2355\ V \approx 0.235\ V$ .

Common Mistakes

Using $n=1$ (forgetting to double the electron count for the balanced 2M³⁺ equation).

Mixing up $\ln$ and $\log$ (the $2.303$ factor already converts $\ln$ to $\log_{10}$ , so don't apply it twice).

✓Final answer
$The correct option is (C) — 0.235.$

ANSWER: C
AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
$E_1, E_2$
›Reveal solutionSolution
This tests using the Nernst equation to compare emfs of Daniell-type cells at different ion concentrations. The order comes out $E_3 > E_2 > E_1$ .

Concept and Intuition

For the cell reaction $Zn + Cu^{2+} \to Zn^{2+} + Cu$ , increasing $[Cu^{2+}]$ (reactant) or decreasing $[Zn^{2+}]$ (product) both push the reaction further forward (Le Chatelier), raising the emf above $E°$ ; the reverse concentrations lower it below $E°$ .

Step-by-Step Solution

Write the Nernst equation for this two-electron process ( $n=2$ ): $E_{cell} = E°_{cell} - \dfrac{0.0591}{2}\log\dfrac{[Zn^{2+}]}{[Cu^{2+}]}$ .

Cell I: $[Zn^{2+}]=1M$ , $[Cu^{2+}]=0.1M$ . $\log\dfrac{1}{0.1} = \log 10 = 1$ . So $E_1 = E° - \dfrac{0.0591}{2}(1) = E° - 0.0296$ (below standard).

Cell II: $[Zn^{2+}]=1M$ , $[Cu^{2+}]=1M$ . $\log\dfrac{1}{1} = 0$ . So $E_2 = E°$ (standard conditions).

Cell III: $[Zn^{2+}]=0.1M$ , $[Cu^{2+}]=1M$ . $\log\dfrac{0.1}{1} = -1$ . So $E_3 = E° - \dfrac{0.0591}{2}(-1) = E° + 0.0296$ (above standard).

Comparing: $E_3 (E°+0.0296) > E_2 (E°) > E_1 (E°-0.0296)$ .

Common Mistakes

Getting the sign of the log term backwards (forgetting the Nernst equation subtracts the log term, so a positive log term lowers E, not raises it).

Mixing up which concentration ratio corresponds to which cell.

✓Final answer
$E_3 > E_2 > E_1$

ANSWER: A
AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
$E_{cell}$
›Reveal solutionSolution
Apply the Nernst equation to the given galvanic cell to relate the measured $E_{cell}$ to the standard $E°_{cell}$ . The answer is 0.76 V.

Concept and Intuition

In cell notation, the left electrode is the anode (oxidation) and the right is the cathode (reduction). Here Fe is oxidised to $Fe^{2+}$ and $Cu^{2+}$ is reduced to Cu, so the overall spontaneous cell reaction is $Fe + Cu^{2+} \to Fe^{2+} + Cu$ , transferring $n=2$ electrons. The Nernst equation corrects the standard potential for the actual (non-standard) ion concentrations present.

Step-by-Step Solution

Overall reaction: $Fe(s) + Cu^{2+}(aq) \rightarrow Fe^{2+}(aq) + Cu(s)$ , with reaction quotient $Q = \dfrac{[Fe^{2+}]}{[Cu^{2+}]}$ (solids are omitted).

Nernst equation: $E_{cell} = E°_{cell} - \dfrac{0.06}{n}\log Q = E°_{cell} - \dfrac{0.06}{2}\log\dfrac{[Fe^{2+}]}{[Cu^{2+}]}$ .

Substitute concentrations: $Q = \dfrac{0.001}{0.1} = 0.01$ , so $\log Q = \log(10^{-2}) = -2$ .

$E_{cell} = E°_{cell} - 0.03\times(-2) = E°_{cell} + 0.06$ .

Given $E_{cell} = 0.82$ V: $0.82 = E°_{cell} + 0.06 \Rightarrow E°_{cell} = 0.76$ V.

Common Mistakes

Writing $Q$ upside down (as $[Cu^{2+}]/[Fe^{2+}]$ ), which flips the sign of the correction term.

Using $n=1$ instead of the correct $n=2$ electrons for this reaction.

Sign errors when subtracting a negative log term.

✓Final answer
$The correct option is (C) — 0.76 V.$

ANSWER: C
AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
$M(s)\,|\,M^{2+}(xM)\,\|\,H^+(0.02M)\,|\,H_2(g),Pt(s)$
›Reveal solutionSolution
Applying the Nernst equation to the given cell with $E^o_{cell}=0.14\ V$ and $E_{cell}=0.077\ V$ gives $x = 0.05\ M$ .

Concept and Intuition

The cell has $M$ metal as the anode (oxidation: $M \rightarrow M^{2+} + 2e^-$ ) and the standard hydrogen electrode as the cathode (reduction: $2H^+ + 2e^- \rightarrow H_2$ ). The overall reaction and its Nernst equation let us relate the measured cell potential to the unknown concentration $x$ of $M^{2+}$ .

Step-by-Step Solution

Cell reaction: $M(s) + 2H^+(aq) \rightarrow M^{2+}(aq) + H_2(g)$ , with $n = 2$ electrons transferred.

$E^o_{cell} = E^o_{cathode} - E^o_{anode} = E^o_{H^+/H_2} - E^o_{M^{2+}/M} = 0 - (-0.14) = 0.14\ V$ .

Nernst equation: $E_{cell} = E^o_{cell} - \dfrac{0.06}{n}\log Q$ , where $Q = \dfrac{[M^{2+}]\,p_{H_2}}{[H^+]^2} = \dfrac{x \times 1}{(0.02)^2} = \dfrac{x}{4\times10^{-4}}$ .

Substitute: $0.077 = 0.14 - \dfrac{0.06}{2}\log\dfrac{x}{4\times10^{-4}} = 0.14 - 0.03\log\dfrac{x}{4\times10^{-4}}$ .

$0.03\log\dfrac{x}{4\times10^{-4}} = 0.14 - 0.077 = 0.063 \Rightarrow \log\dfrac{x}{4\times10^{-4}} = 2.1$ .

$\log x - \log(4\times10^{-4}) = 2.1 \Rightarrow \log x = 2.1 + \log(4\times10^{-4}) = 2.1 + (0.602-4) = 2.1 - 3.398 = -1.298$ .

Writing $-1.298$ in the characteristic-mantissa form used by the given antilog table, $-1.298 = \bar{2}.7$ (i.e. $-2+0.702 \approx -2+0.70$ ), and using the supplied value $antilog(\bar{2}.7) = 0.05$ , we get $x \approx 0.05$ .

Common Mistakes

Getting the sign of $E^o_{cell}$ wrong by mixing up which electrode is the anode and which is the cathode.

Forgetting to square $[H^+]$ in $Q$ since two moles of $H^+$ appear in the balanced equation.

Misreading the bar notation in the antilog table (e.g. $\bar{2}.7$ means $-2+0.7$ , not $-2.7$ ).

✓Final answer
$The correct option is (A) — 0.05.$

ANSWER: A
AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
$E°$
›Reveal solutionSolution
The cell potential is the difference between the standard reduction potentials of the cathode and anode; here it is positive 0.23 V, and using the Nernst equation at equilibrium gives $\log K_C \approx 7.79$ , so the correct option is (C).

We start with the reaction:

$2Fe^{3+}(aq) + 2I^-(aq) \to 2Fe^{2+}(aq) + I_2(s)$

The key is to see this as a galvanic cell — a spontaneous redox reaction that can produce electricity. The Nernst equation connects the cell potential to the equilibrium constant, so once we find $E^\circ_{\text{cell}}$ , we can directly compute $\log K_C$ .

1. Identify the half-reactions and their standard potentials

Reduction half (cathode):

$Fe^{3+} + e^- \to Fe^{2+}$

$E^\circ = +0.77\ \text{V}$

Oxidation half (anode):

$2I^- \to I_2 + 2e^-$

The standard reduction potential for $I_2/I^-$ is given as $0.54\ \text{V}$ , so the oxidation potential is $-0.54\ \text{V}$ .

Watch out

A common mistake is to forget that the cell potential is $E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ when both are given as reduction potentials. Here, the anode is the $I^-/I_2$ couple being oxidized, so we subtract its reduction potential.

2. Calculate the standard cell potential
$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.77 - 0.54 = 0.23\ \text{V}$
So $X = 0.23\ \text{V}$ . This eliminates options (B) and (D) immediately.

3. Relate $E^\circ_{\text{cell}}$ to the equilibrium constant

The Nernst equation at equilibrium ( $Q = K_C$ ) and standard conditions gives:
$E^\circ_{\text{cell}} = \frac{RT}{nF} \ln K_C$
Convert to base-10 log:
$E^\circ_{\text{cell}} = \frac{2.303 RT}{nF} \log K_C$
At $T = 298\ \text{K}$ , $R = 8.3\ \text{J K}^{-1}\text{mol}^{-1}$ , $F = 96500\ \text{C mol}^{-1}$ :
$\frac{2.303 RT}{F} = \frac{2.303 \times 8.3 \times 298}{96500} \approx 0.059\ \text{V}$
Tip

The value $0.059\ \text{V}$ at 298 K is a handy constant to remember: it’s the factor that converts $\log K$ to volts per electron.

4. Determine $n$ , the number of electrons transferred

From the balanced reaction:

$2Fe^{3+} + 2e^- \to 2Fe^{2+}$ (2 electrons gained)

$2I^- \to I_2 + 2e^-$ (2 electrons lost)

So $n = 2$ .

5. Solve for $\log K_C$
$0.23 = \frac{0.059}{2} \log K_C \quad \Rightarrow \quad 0.23 = 0.0295 \log K_C$ $\log K_C = \frac{0.23}{0.0295} \approx 7.7966 \approx 7.80$
This matches option (C): $0.23\ \text{V}$ and $7.79$ (rounding difference is negligible).
$\displaystyle E^\circ_{\text{cell}} = \frac{0.059}{n} \log K_C$

✓Final answer
$The correct option is (C) . ANSWER: C$
AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
$\log K_c$
›Reveal solutionSolution
The Nernst equation at equilibrium gives $\log K_c = nE^\circ_{cell}/0.06$ ; identifying the cathode/anode and electron count $n=2$ gives $\log K_c = 15.33$ .

Concept and Intuition

At equilibrium, the cell's Gibbs energy relation $\Delta G^\circ = -nFE^\circ_{cell} = -RT\ln K$ combines with $\dfrac{2.303RT}{F}=0.06$ V to give the compact working formula $\log K_c = \dfrac{nE^\circ_{cell}}{0.06}$ . The larger the cell EMF, the more the reaction favours products, i.e., the larger $K_c$ .

Step-by-Step Solution

Identify the two half-reactions from the overall reaction $Cu(s)+2Ag^+(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)$ :

Reduction (cathode): $Ag^+ + e^- \rightarrow Ag$ , $E^\circ = 0.80$ V

Oxidation (anode): $Cu \rightarrow Cu^{2+}+2e^-$ , $E^\circ_{Cu^{2+}/Cu} = 0.34$ V

$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.80 - 0.34 = 0.46$ V.

Number of electrons transferred in the balanced equation: $n=2$ .

$\log K_c = \dfrac{nE^\circ_{cell}}{0.06} = \dfrac{2\times 0.46}{0.06} = \dfrac{0.92}{0.06} = 15.33$ .

Common Mistakes

Subtracting the electrode potentials in the wrong order (anode $-$ cathode), which flips the sign.

Using $n=1$ instead of the actual 2 electrons transferred in the balanced cell reaction.

✓Final answer
$The correct option is (C) — 15.33.$

ANSWER: C
AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
$2M^{3+}(aq) + 2I^-(aq) \rightarrow 2M^{2+}(aq) + I_2(s)$
›Reveal solutionSolution
Standard Nernst-equation relation between cell EMF and equilibrium constant: $\log K_c = \dfrac{nFE^{\circ}_{cell}}{2.303RT}$ . Answer: 7.96.

Concept and Intuition

At equilibrium, the Gibbs free energy relation $\Delta G^{\circ}=-RT\ln K = -nFE^{\circ}_{cell}$ links a cell's standard EMF to the equilibrium constant of the underlying redox reaction. Rearranging and converting to base-10 log gives $\log K_c = \dfrac{nFE^{\circ}_{cell}}{2.303RT}$ .

Step-by-Step Solution

Reaction: $2M^{3+}+2I^- \rightarrow 2M^{2+}+I_2$ . Each $M^{3+}\rightarrow M^{2+}$ step transfers 1 electron, and there are 2 such reductions plus the corresponding oxidation of $2I^-\rightarrow I_2$ (also a 2-electron process) — so $n=2$ .

Compute $\dfrac{2.303RT}{F} = \dfrac{2.303\times8.3\times298}{96500}$ .

Numerator: $2.303\times8.3=19.115$ ; $19.115\times298=5696.3$ .

Divide by $F=96500$ : $5696.3/96500 = 0.05903$ V.

$\log K_c = \dfrac{nE^{\circ}_{cell}}{0.05903} = \dfrac{2\times0.235}{0.05903} = \dfrac{0.47}{0.05903} \approx 7.96$ .

Common Mistakes

Using $n=1$ instead of $n=2$ (miscounting the electrons transferred from the balanced equation).

Using the commonly memorised constant 0.0591/0.059 without recomputing it from the given R, F, T values, which can introduce small rounding mismatches with the intended answer.

✓Final answer
$The correct option is (B) — 7.96.$

ANSWER: B
AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
$E_{cell}$
›Reveal solutionSolution
This tests applying the Nernst equation to a galvanic cell with non-standard ion concentrations to find the actual cell potential.

Concept and Intuition

The cell notation $\mathrm{Zn(s)|Zn^{2+}(0.01\,M)||Ni^{2+}(0.1\,M)|Ni(s)}$ tells us Zn is the anode (oxidation) and Ni is the cathode (reduction), since by convention the anode is written on the left. The standard cell potential is found from the standard reduction potentials, and then the Nernst equation corrects for the actual (non-1M) ion concentrations present.

Step-by-Step Solution

Identify half-reactions: anode (oxidation): $Zn \to Zn^{2+}+2e^-$ ; cathode (reduction): $Ni^{2+}+2e^- \to Ni$ .

Standard cell potential: $E^{\ominus}_{cell} = E^{\ominus}_{cathode}-E^{\ominus}_{anode} = (-0.25)-(-0.76) = 0.51\ \mathrm{V}$ .

Overall reaction: $Zn(s)+Ni^{2+}(aq)\to Zn^{2+}(aq)+Ni(s)$ , with $n=2$ electrons transferred.

Nernst equation: $E_{cell}=E^{\ominus}_{cell}-\dfrac{0.06}{n}\log\dfrac{[Zn^{2+}]}{[Ni^{2+}]}$ (products of oxidation over reactant of reduction, since Zn²⁺ is a product formed and Ni²⁺ is consumed).

Substitute: $E_{cell}=0.51-\dfrac{0.06}{2}\log\dfrac{0.01}{0.1} = 0.51-0.03\log(0.1) = 0.51-0.03(-1)=0.51+0.03=0.54\ \mathrm{V}$ .

Common Mistakes

Swapping the ratio inside the log (using $[Ni^{2+}]/[Zn^{2+}]$ instead of $[Zn^{2+}]/[Ni^{2+}]$ ), which flips the sign of the correction term.

Sign error when subtracting a negative anode potential from a negative cathode potential.

✓Final answer
$The correct option is (D) — 0.54.$

ANSWER: D
AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
$\mathrm{H^+(aq) + e^- \longrightarrow \dfrac{1}{2}H_2(1\ bar)}$
›Reveal solutionSolution
This is a direct Nernst-equation application to the standard hydrogen electrode at non-standard pH; the electrode potential comes out to $-0.6\ V$ .

Concept and Intuition

The hydrogen electrode's potential depends on $\mathrm{[H^+]}$ (equivalently, pH) through the Nernst equation. At pH = 0 (standard, $[\mathrm{H^+}]=1\ M$ ) the electrode potential is exactly $0\ V$ by definition. At any other pH it shifts, becoming more negative as pH increases (as $\mathrm{[H^+]}$ decreases).

Step-by-Step Solution

Half-reaction: $\mathrm{H^+(aq) + e^- \rightarrow \tfrac{1}{2}H_2(1\ bar)}$ , with $n = 1$ electron and $E^\circ = 0\ V$ (by convention).

Nernst equation: $E = E^\circ - \dfrac{0.06}{n}\log\dfrac{P_{H_2}^{1/2}}{[\mathrm{H^+}]}$ . With $P_{H_2}=1\ bar$ , this simplifies to $E = E^\circ - 0.06\log\dfrac{1}{[\mathrm{H^+}]}$ .

Since $\log\dfrac{1}{[\mathrm{H^+}]} = -\log[\mathrm{H^+}] = \mathrm{pH}$ , we get $E = E^\circ - 0.06 \times \mathrm{pH}$ .

Substituting $E^\circ = 0$ and $\mathrm{pH} = 10.0$ : $E = 0 - 0.06 \times 10 = -0.6\ V$ .

Common Mistakes

Getting the sign of the log term wrong and reporting $+0.6\ V$ .

Forgetting that at pH other than 0, the SHE potential is no longer exactly zero.

✓Final answer
$-0.6$

ANSWER: A
AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
$\mathrm{M(s) \mid M^{2+}(x\ M) \parallel Zn^{2+}(y\ M) \mid Zn(s)}$
›Reveal solutionSolution
Using the Nernst equation at equilibrium ( $E_{cell}=0$ ) with the given $\log$ ratio gives $E^\ominus_{cell}=1.6$ V, and subtracting from $E^\ominus_{Zn^{2+}/Zn}=-0.76$ V gives $E^\ominus_{M^{2+}/M}=-2.36$ V.

Concept and Intuition

In the cell notation $\mathrm{M(s)\mid M^{2+}\parallel Zn^{2+}\mid Zn(s)}$ , the left electrode is always the anode (oxidation) and the right is the cathode (reduction) by convention. At equilibrium, no net current flows and the cell EMF is zero — this is exactly the condition used to relate $E^\ominus_{cell}$ to the reaction quotient via the Nernst equation.

Step-by-Step Solution

Half reactions: anode $\mathrm{M \to M^{2+} + 2e^-}$ ; cathode $\mathrm{Zn^{2+}+2e^- \to Zn}$ ; overall $\mathrm{M + Zn^{2+} \to M^{2+} + Zn}$ , with $n=2$ .

Nernst equation: $E_{cell}=E^\ominus_{cell}-\dfrac{0.06}{2}\log\dfrac{[M^{2+}]}{[Zn^{2+}]}$ .

At equilibrium, $E_{cell}=0$ , so $E^\ominus_{cell}=\dfrac{0.06}{2}\log\dfrac{[M^{2+}]}{[Zn^{2+}]}=0.03\times53.33=1.6\ \text{V}$ .

Also $E^\ominus_{cell}=E^\ominus_{cathode}-E^\ominus_{anode}=E^\ominus_{Zn^{2+}/Zn}-E^\ominus_{M^{2+}/M}$ .

So $1.6 = (-0.76) - E^\ominus_{M^{2+}/M} \Rightarrow E^\ominus_{M^{2+}/M} = -0.76-1.6=-2.36\ \text{V}$ .

Common Mistakes

Mixing up which electrode is anode/cathode from the cell notation, or forgetting to divide by $n=2$ before multiplying by 0.06.

✓Final answer
$The correct option is (C) — -2.36.$

ANSWER: C
AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
$\mathrm{Mg(s)\,|\,Mg^{2+}(x\,M)\,||\,Zn^{2+}(y\,M)\,|\,Zn(s)}$
›Reveal solutionSolution
At equilibrium the cell voltage is zero, so the Nernst equation directly gives $\log([\mathrm{Mg^{2+}}]/[\mathrm{Zn^{2+}}]) = E^\circ_{cell}/(0.06/n) = 1.60/0.03 = 53.33$ .

Concept and Intuition

In the cell $\mathrm{Mg(s)\,|\,Mg^{2+}\,||\,Zn^{2+}\,|\,Zn(s)}$ , Mg is oxidised at the anode and $\mathrm{Zn^{2+}}$ is reduced at the cathode. When a cell "reaches equilibrium," its net driving force is exhausted — the measured cell potential $E_{cell}$ becomes zero (this is the condition that also defines $K_{eq}$ for the overall cell reaction via $E^\circ_{cell}=\dfrac{0.06}{n}\log K$ ).

Step-by-Step Solution

Cell reaction: $\mathrm{Mg(s) + Zn^{2+}(aq) \to Mg^{2+}(aq) + Zn(s)}$ , with $n=2$ electrons transferred.

$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = E^\circ_{Zn^{2+}/Zn} - E^\circ_{Mg^{2+}/Mg} = (-0.76) - (-2.36) = 1.60$ V.

Nernst equation: $E_{cell} = E^\circ_{cell} - \dfrac{0.06}{n}\log Q$ , where $Q = \dfrac{[\mathrm{Mg^{2+}}]}{[\mathrm{Zn^{2+}}]}$ (Mg and Zn solids don't appear).

At equilibrium, $E_{cell}=0$ : $0 = 1.60 - \dfrac{0.06}{2}\log\dfrac{[\mathrm{Mg^{2+}}]}{[\mathrm{Zn^{2+}}]}$ .

$\dfrac{0.06}{2}\log Q = 1.60 \Rightarrow 0.03 \log Q = 1.60 \Rightarrow \log Q = \dfrac{1.60}{0.03} = 53.33$ .

Common Mistakes

Forgetting to divide by $n=2$ when using $0.06/n$ (using 0.06 directly would give a wrong factor-of-2 answer, 26.67).

Mixing up which electrode is the cathode/anode and getting the sign of $E^\circ_{cell}$ wrong.

✓Final answer
$The correct option is (A) — 53.33.$

ANSWER: A

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