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.

COMEDK 2026Set 2026-M1 markMCQ
$25^{\circ} \mathrm{C}$
›Reveal solutionSolution
The electrode potential of a chromium electrode in a chromic sulphate solution becomes less negative (increases) when the solution is diluted, because the Nernst equation shows that decreasing $\mathrm{Cr}^{3+}$ concentration raises the potential; the calculated increase is $+19.7\ \mathrm{mV}$ , corresponding to option (B).

The key concept here is the Nernst equation, which relates the electrode potential of a half-cell to the concentration of the ion involved. For a metal/metal-ion electrode like $\mathrm{Cr}^{3+}/\mathrm{Cr}$ , the reduction reaction is:
$\mathrm{Cr}^{3+} + 3e^- \rightarrow \mathrm{Cr(s)}$
The Nernst equation at $25^\circ\mathrm{C}$ is:
$E = E^0 + \frac{0.059}{n} \log [\mathrm{Cr}^{3+}]$
where $n = 3$ (number of electrons transferred). Notice the sign: because the standard potential $E^0$ is for the reduction, and we are using the form $E = E^0 + \frac{0.059}{n}\log[\text{oxidised}]$ , this is correct for a reduction half-cell.

When we dilute the solution 10 times, $[\mathrm{Cr}^{3+}]$ becomes $\frac{1}{10}$ of its original value. The logarithm of a smaller number is more negative, so $E$ becomes less negative (i.e., increases). The question asks for the change in potential, so we compute $\Delta E = E_{\text{new}} - E_{\text{old}}$ .

Let’s work through it step by step.

Write the Nernst equation for the initial and final states. Let the initial concentration be $c$ . Then:

$E_{\text{initial}} = E^0 + \frac{0.059}{3} \log c$
After 10‑fold dilution, $c_{\text{new}} = \frac{c}{10}$ , so:
$E_{\text{final}} = E^0 + \frac{0.059}{3} \log\left(\frac{c}{10}\right)$

Find the change $\Delta E = E_{\text{final}} - E_{\text{initial}}$ . Subtract:

$\Delta E = \left[E^0 + \frac{0.059}{3} \log\left(\frac{c}{10}\right)\right] - \left[E^0 + \frac{0.059}{3} \log c\right]$
The $E^0$ terms cancel, leaving:
$\Delta E = \frac{0.059}{3} \left[ \log\left(\frac{c}{10}\right) - \log c \right]$

Simplify the logarithm difference. Using $\log(a/b) = \log a - \log b$ :

$\log\left(\frac{c}{10}\right) - \log c = \log c - \log 10 - \log c = -\log 10$
Since $\log_{10} 10 = 1$ , this is simply $-1$ .

Compute the numerical value.

$\Delta E = \frac{0.059}{3} \times (-1) = -\frac{0.059}{3}\ \text{V}$
Convert to millivolts:
$\Delta E = -\frac{59}{3}\ \text{mV} \approx -19.666\ \text{mV} \approx -19.7\ \text{mV}$
The negative sign means the potential becomes less negative (i.e., increases). In the language of the options, “increase by 19.7 mV” is exactly what we have: the potential goes from, say, $-0.74\ \text{V}$ to $-0.7203\ \text{V}$ , an increase of $+19.7\ \text{mV}$ .

Tip

A common mistake is to forget the sign in the Nernst equation for a reduction. For a metal electrode, always use $E = E^0 + \frac{0.059}{n}\log[\text{ion}]$ — dilution makes the log term more negative, so $E$ becomes less negative (increases). Also, note that the question asks for the change; the absolute potential is not needed.

Watch out

Some students incorrectly use $E = E^0 - \frac{0.059}{n}\log Q$ (the form for the full cell) and get the sign reversed. Stick to the reduction half-cell form above.

✓Final answer
$The correct option is (B) . ANSWER: B$
COMEDK 2025Set 2025-A1 markMCQ
$\begin{aligned} & 2 \mathrm{Cr}_{(\mathrm{S})}+3 \mathrm{Fe}^{2+}(0.02 \mathrm{M}) \rightarrow 2 \mathrm{Cr}^{3+}(0.2 \mathrm{M})+3 \mathrm{Fe} \\ & \mathrm{E}^0 \mathrm{Fe}^{2+} / \mathrm{Fe}=-0.42 \mathrm{~V} \quad \mathrm{E}^0 \mathrm{Cr}^{3+} / \mathrm{Cr}=-0.72 \mathrm{~V} \end{aligned}$
›Reveal solutionSolution

Use the Nernst equation. $E^\circ_{\text{cell}}=+0.30\text{ V}$ , $n=6$ , $Q=5000$ , giving $E_{\text{cell}}\approx0.2636\text{ V}$ — option (C).

The cell runs the reaction $2\mathrm{Cr}+3\mathrm{Fe}^{2+}\rightarrow 2\mathrm{Cr}^{3+}+3\mathrm{Fe}$ . Cr is oxidised (anode) and $\mathrm{Fe}^{2+}$ is reduced (cathode). Under non-standard concentrations the potential is found from the Nernst equation.

Step 1 — Standard cell potential
$E^\circ_{\text{cell}}=E^\circ_{\text{cathode}}-E^\circ_{\text{anode}}=E^\circ_{\mathrm{Fe}^{2+}/\mathrm{Fe}}-E^\circ_{\mathrm{Cr}^{3+}/\mathrm{Cr}}=(-0.42)-(-0.72)=+0.30\text{ V}$
Step 2 — Electrons transferred

Oxidation $2\mathrm{Cr}\rightarrow 2\mathrm{Cr}^{3+}+6e^-$ and reduction $3\mathrm{Fe}^{2+}+6e^-\rightarrow 3\mathrm{Fe}$ , so $n=6$ .

Step 3 — Reaction quotient (pure solids omitted)
$Q=\frac{[\mathrm{Cr}^{3+}]^{2}}{[\mathrm{Fe}^{2+}]^{3}}=\frac{(0.2)^{2}}{(0.02)^{3}}=\frac{0.04}{8\times10^{-6}}=5000$
Step 4 — Nernst equation at 298 K
$E_{\text{cell}}=E^\circ_{\text{cell}}-\frac{0.0591}{n}\log Q=0.30-\frac{0.0591}{6}\log(5000)$
With $\log(5000)=3.699$ and $\dfrac{0.0591}{6}=0.00985$ :
$E_{\text{cell}}=0.30-(0.00985)(3.699)=0.30-0.0364=0.2636\text{ V}$
✓Final answer
$E_{\text{cell}}=0.2636\text{ V}$
COMEDK 2025Set 2025-E1 markMCQ
$2 \mathrm{M}_{(\mathrm{S})}+\mathrm{O}_2(\mathrm{~g})+4 \mathrm{H}^{+} \rightarrow 2 \mathrm{M}^{2+}(\mathrm{aq})+2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})}$
›Reveal solutionSolution
The key is to apply the Nernst equation to the given cell reaction, account for the change in EMF due to non‑standard concentrations and pressure, and solve for the pH. The calculated pH is approximately 2.95, so the correct option is (B).

Concept and Intuition

The cell reaction involves H⁺ ions, so the cell potential depends on pH. Under non‑standard conditions, the Nernst equation relates the observed EMF to the concentrations (or partial pressures) of all species. Here, we know the standard potential $E^0 = +1.67\ \text{V}$ and the measured potential $E = +1.57\ \text{V}$ when $[M^{2+}] = 1.0 \times 10^{-3}\ \text{M}$ and $p(\mathrm{O}_2) = 0.1\ \text{atm}$ . The only unknown is $[H^+]$ , from which we get pH. The drop in potential from 1.67 V to 1.57 V tells us the reaction quotient $Q$ is greater than 1, meaning the system is shifted toward products relative to standard conditions — and since H⁺ is a reactant, a lower $[H^+]$ (higher pH) would cause that shift.

Step‑by‑Step Solution

1. Write the Nernst equation for the cell reaction.

The general form for a reaction
$aA + bB \rightarrow cC + dD$
is
$E = E^0 - \frac{RT}{nF} \ln Q$
where $Q$ is the reaction quotient. At 298 K, using base‑10 logs, this simplifies to
$E = E^0 - \frac{0.05916}{n} \log Q$
( $0.05916 \approx 0.059$ for most purposes).

2. Identify $n$ , the number of electrons transferred.

The half‑reactions are:

Oxidation: $\mathrm{M_{(s)}} \rightarrow \mathrm{M^{2+}} + 2e^-$ (×2 gives 4 e⁻)

Reduction: $\mathrm{O_2} + 4H^+ + 4e^- \rightarrow 2\mathrm{H_2O}$

So $n = 4$ .

3. Write the reaction quotient $Q$ .

For the overall reaction
$2\mathrm{M_{(s)}} + \mathrm{O_2(g)} + 4\mathrm{H^+} \rightarrow 2\mathrm{M^{2+}(aq)} + 2\mathrm{H_2O_{(l)}}$
solids and pure liquids are omitted. Thus
$Q = \frac{[\mathrm{M^{2+}}]^2}{p(\mathrm{O_2})\,[\mathrm{H^+}]^4}$
(Note: $p(\mathrm{O_2})$ is in atm, and concentrations are in M.)

4. Plug known values into the Nernst equation.

Given:

$E^0 = 1.67\ \text{V}$ , $E = 1.57\ \text{V}$ , $[\mathrm{M^{2+}}] = 1.0 \times 10^{-3}\ \text{M}$ , $p(\mathrm{O_2}) = 0.1\ \text{atm}$ , $n = 4$ , and using $0.05916$ at 298 K.
$1.57 = 1.67 - \frac{0.05916}{4} \log \left( \frac{(1.0 \times 10^{-3})^2}{0.1 \times [\mathrm{H^+}]^4} \right)$
5. Simplify the equation.

First, the potential difference:
$1.57 - 1.67 = -0.10\ \text{V}$
So
$-0.10 = - \frac{0.05916}{4} \log Q$
Multiply both sides by –1:
$0.10 = \frac{0.05916}{4} \log Q$ $\log Q = \frac{0.10 \times 4}{0.05916} = \frac{0.40}{0.05916} \approx 6.76$
6. Express $Q$ in terms of $[H^+]$ and solve.
$Q = \frac{(10^{-3})^2}{0.1 \times [\mathrm{H^+}]^4} = \frac{10^{-6}}{0.1 \times [\mathrm{H^+}]^4} = \frac{10^{-5}}{[\mathrm{H^+}]^4}$
Thus
$\log Q = \log(10^{-5}) - 4\log[\mathrm{H^+}] = -5 - 4\log[\mathrm{H^+}]$
Set equal to 6.76:
$-5 - 4\log[\mathrm{H^+}] = 6.76$ $-4\log[\mathrm{H^+}] = 11.76$ $\log[\mathrm{H^+}] = -2.94$
Therefore
$[\mathrm{H^+}] = 10^{-2.94}\ \text{M}$
7. Calculate pH.
$\mathrm{pH} = -\log[\mathrm{H^+}] = 2.94 \approx 2.95$
Tip

Notice that the Nernst equation gave $\log Q \approx 6.76$ . Since $Q$ is large, the reaction is product‑favoured under these conditions, which is consistent with a lower EMF. The pH turns out to be acidic, as expected from the presence of H⁺ as a reactant.

Watch out

A common mistake is forgetting to square the $[M^{2+}]$ term (coefficient 2) or to raise $[H^+]$ to the fourth power. Also, remember that $p(\mathrm{O_2})$ appears in the denominator of $Q$ , not the numerator.

✓Final answer
$The correct option is (B) . ANSWER: B$
COMEDK 2025Set 2025-M1 markMCQ
$2.2 \mathrm{M} \mathrm{Fe}^{2+}$
›Reveal solutionSolution
Use the Nernst equation for the half‑cell reduction: $E = E^0 - \frac{0.0591}{n} \log Q$ , with $n=1$ and $Q = [\mathrm{Fe}^{2+}]/[\mathrm{Fe}^{3+}]$ . Substituting gives $E \approx 0.668\ \mathrm{V}$ , so the correct option is (C).

The key idea is that the reduction potential of a half‑cell under non‑standard conditions is given by the Nernst equation. Here the reaction is

$\mathrm{Fe}^{3+} + e^- \rightarrow \mathrm{Fe}^{2+}$ , so $n = 1$ . The reaction quotient $Q$ is $[\mathrm{Fe}^{2+}]/[\mathrm{Fe}^{3+}]$ (products over reactants, but note the solid Pt electrode doesn’t appear). Because the concentrations are not 1 M, the potential shifts from the standard value.

Why this works: The Nernst equation accounts for the concentration dependence of electrochemical potential. For a reduction, a higher concentration of the reduced form (Fe²⁺) relative to the oxidized form (Fe³⁺) makes reduction less favourable, lowering the potential. Here Fe²⁺ is much more concentrated than Fe³⁺, so we expect $E < E^0$ .

Write the Nernst equation for the half‑cell reduction at 298 K:

$E = E^0 - \frac{0.0591}{n} \log Q$
where $Q = \frac{[\mathrm{Fe}^{2+}]}{[\mathrm{Fe}^{3+}]}$ for the reduction $\mathrm{Fe}^{3+} + e^- \rightarrow \mathrm{Fe}^{2+}$ .

Identify the values:

$E^0 = 0.771\ \mathrm{V}$ , $n = 1$ , $[\mathrm{Fe}^{2+}] = 2.2\ \mathrm{M}$ , $[\mathrm{Fe}^{3+}] = 0.04\ \mathrm{M}$ .

Compute the ratio:

$\frac{[\mathrm{Fe}^{2+}]}{[\mathrm{Fe}^{3+}]} = \frac{2.2}{0.04} = 55.$

Take the logarithm:

$\log(55) \approx 1.7404.$

Plug into the Nernst equation:

$E = 0.771 - 0.0591 \times 1.7404$ $E = 0.771 - 0.1028 \approx 0.6682\ \mathrm{V}.$

Round to three decimal places: $0.668\ \mathrm{V}$ .

Watch out

A common mistake is to invert the ratio in $Q$ . Remember: for the reduction $\mathrm{Ox} + ne^- \rightarrow \mathrm{Red}$ , $Q = [\mathrm{Red}]/[\mathrm{Ox}]$ . Here $\mathrm{Ox} = \mathrm{Fe}^{3+}$ , $\mathrm{Red} = \mathrm{Fe}^{2+}$ , so $Q = [\mathrm{Fe}^{2+}]/[\mathrm{Fe}^{3+}]$ , not the other way around.

Tip

Since $[\mathrm{Fe}^{2+}] \gg [\mathrm{Fe}^{3+}]$ , the log term is positive, so the potential is noticeably less than $E^0$ . Among the options, only 0.668 V and 0.598 V are below 0.771 V; the ratio 55 gives a shift of about 0.103 V, which matches 0.668 V exactly.

✓Final answer
$The correct option is (C) . ANSWER: C$
COMEDK 2023Set 2023-E1 markMCQ
$\mathrm{A}(\mathrm{s})+\mathrm{B}^{2+}\left(\mathbf{1} \mathbf{1} \mathbf{1 0} \mathbf{0}^{-\mathrm{M}} \mathbf{M}\right) \rightarrow \mathrm{B}_{(\mathrm{s})}+\mathrm{A}^{2+}(0.1 \mathrm{M}).$
›Reveal solutionSolution

Apply the Nernst equation ( $n=2$ electrons). The measured drop $E^\circ-E = 0.059$ V equals $\tfrac{0.0591}{2}\log 10^{(X-1)}$ , so $X-1=2$ and $X=3$ .

Cell reaction: $\mathrm{A(s)+B^{2+}(10^{-X}\,M)\to B(s)+A^{2+}(0.1\,M)}$ , $n=2$ .

Nernst equation:
$E = E^\circ - \frac{0.0591}{2}\log\frac{[\mathrm{A^{2+}}]}{[\mathrm{B^{2+}}]}$ $2.651 = 2.71 - \frac{0.0591}{2}\log\frac{0.1}{10^{-X}}$
Solve:
$2.71 - 2.651 = 0.02955\log\big(10^{X-1}\big)$ $0.059 = 0.02955\,(X-1)$ $X-1 = \frac{0.059}{0.02955} = 1.996 \approx 2 \;\Rightarrow\; X = 3$
✓Final answer
$The correct option is (C) — X = 3$
COMEDK 2023Set 2023-M1 markMCQ
$A(s)+B^{2+}(a q) \longrightarrow A^{2+}(a q)+B(s)$
›Reveal solutionSolution

Using $E^\circ = \frac{0.0591}{n}\log K$ with $n=2$ and $E^\circ=0.295\,\mathrm{V}$ gives $\log K \approx 10$ , hence $K = 1\times10^{10}$ .

The reaction $A(s) + B^{2+} \rightarrow A^{2+} + B(s)$ transfers 2 electrons, so $n=2$ .

At equilibrium the Nernst relation gives:
$E^\circ_{cell} = \frac{0.0591}{n}\log K.$
Solve for $\log K$ :
$\log K = \frac{n\,E^\circ}{0.0591} = \frac{2 \times 0.295}{0.0591} = \frac{0.590}{0.0591} \approx 9.98 \approx 10.$
Therefore:
$K = 10^{10} = 1\times10^{10}.$
✓Final answer
$1\times10^{10}$

KCET 2021Set B-21 markMCQ

^{2+}

›Reveal solution

Write the Nernst equation for the Zn $^{2+}$ /Zn half-cell; the electrode potential increases monotonically with $\log[\mathrm{Zn^{2+}}]$ , so just rank the four concentrations.

Step 1 — Nernst equation for the half-cell.

The reduction half-reaction is

\mathrm{Zn^{2+}} + 2e^- \rightarrow \mathrm{Zn(s)}, \qquad n = 2.

E = E^{\circ} - \frac{0.059}{n}\log \frac{[\mathrm{Zn}]}{[\mathrm{Zn^{2+}}]}

Solid Zn has unit activity, so

E = E^{\circ} - \frac{0.059}{2}\log\frac{1}{[\mathrm{Zn^{2+}}]} = E^{\circ} + \frac{0.059}{2}\log[\mathrm{Zn^{2+}}].

Step 2 — Read off the trend.

$E$ is an increasing function of $[\mathrm{Zn^{2+}}]$ . Chemically: more Zn $^{2+}$ in solution pushes the reduction equilibrium forward, making the electrode a better oxidising agent, i.e. a higher (less negative) potential.

Step 3 — Evaluate all four. With $E^\circ = -0.76\,$ V and $\frac{0.059}{2} = 0.0295$ :

Electrode	$[\mathrm{Zn^{2+}}]$	$\log[\mathrm{Zn^{2+}}]$	$E$ / V
Q	$10^{-1}$	$-1$	$-0.76 + 0.0295(-1) = -0.7895$
R	$10^{-2}$	$-2$	$-0.76 + 0.0295(-2) = -0.8190$
S	$10^{-3}$	$-3$	$-0.76 + 0.0295(-3) = -0.8485$
P	$10^{-4}$	$-4$	$-0.76 + 0.0295(-4) = -0.8780$

Step 4 — Order them. Remembering that $-0.7895$ is the largest (least negative):

Q\,(-0.790) > R\,(-0.819) > S\,(-0.849) > P\,(-0.878).

The trap in options (A) and (D) is treating the most-negative number as the biggest.

✓Final answer

The correct option is (C) — Q > R > S > P.

ANSWER: C

COMEDK 2021Set 2021-B1 markMCQ
$Q. The standard reduction potential of Pb^2+/Pb is -0.13 V. The reduction potential at pH=14 for the above electrode is :[ Given : Ksp of Pb(OH)2 is 1 * 10^15 ] (A) -0.42 V (B) -0.22 V (C) -0.31 V (D) -0.59 V$
›Reveal solutionSolution

With $[\text{Pb}^{2+}]$ suppressed to $K_{sp}=10^{-15}$ at pH 14, the Nernst equation gives $E\approx-0.57$ V, closest to $-0.59$ V.

Concentration of Pb $^{2+}$ (treating $K_{sp}=1\times10^{-15}$ ; at pH 14, $[\text{OH}^-]=1$ ):
$K_{sp} = [\text{Pb}^{2+}][\text{OH}^-]^2 \Rightarrow [\text{Pb}^{2+}] = 10^{-15}\ \text{M}$
Nernst equation for $\text{Pb}^{2+}+2e^-\to\text{Pb}$ :
$E = E^\circ + \frac{0.059}{2}\log[\text{Pb}^{2+}] = -0.13 + \frac{0.059}{2}(-15)$ $E = -0.13 - 0.443 = -0.573\ \text{V}$
Nearest listed value is $-0.59$ V.

✓Final answer
$The correct option is (D) — -0.59 V$
KCET 2019Set A-11 markMCQ
$MgCl_2$
›Reveal solutionSolution
The key idea is to use Faraday’s laws of electrolysis: the charge passed determines the moles of electrons, which relates directly to the moles of $Mg^{2+}$ reduced. The original concentration of $MgCl_2$ is found to be $5 \times 10^{-3}$ M.

The problem involves electrolysis of $MgCl_2$ solution. When we electrolyze a molten or aqueous salt, the metal ions get reduced at the cathode. Here, $Mg^{2+}$ ions are reduced to magnesium metal:
$Mg^{2+} + 2e^- \rightarrow Mg$
So each mole of $Mg$ deposited requires 2 moles of electrons. The charge passed (in coulombs) tells us the moles of electrons, and from that we can find the moles of $Mg^{2+}$ originally present in the 1 L solution. That directly gives the molarity.

Let’s work through it step by step.

Calculate total charge passed.

Current $I = 1$ A, time $t = 16$ min $5$ sec. Convert time to seconds:

$16 \times 60 = 960$ s, plus $5$ s gives $965$ s.

Charge $Q = I \times t = 1 \times 965 = 965$ C.

Find moles of electrons.

Faraday’s constant $F = 96500$ C/mol e $^-$ .

Moles of electrons $= \frac{Q}{F} = \frac{965}{96500} = 0.01$ mol.

Relate moles of electrons to moles of $Mg^{2+}$ .

From the half-reaction: $Mg^{2+} + 2e^- \rightarrow Mg$ ,

2 moles of electrons deposit 1 mole of $Mg$ .

So moles of $Mg^{2+}$ reduced $= \frac{0.01}{2} = 0.005$ mol.

Determine original concentration.

The solution volume is 1 L. The electrolysis is “complete,” meaning all $Mg^{2+}$ ions in that 1 L are reduced. So the original moles of $MgCl_2$ = moles of $Mg^{2+}$ = 0.005 mol.

Concentration $= \frac{0.005 \text{ mol}}{1 \text{ L}} = 0.005$ M $= 5 \times 10^{-3}$ M.

Watch out

A common mistake is to forget the 2-electron stoichiometry and directly equate moles of electrons to moles of $Mg^{2+}$ . That would give $0.01$ M, which is option (D) — a tempting distractor. Always check the half-reaction.

Tip

Notice that the atomic mass of Mg (24 g/mol) is not needed here because we are finding molarity, not mass deposited. If the question asked for mass of Mg deposited, you would multiply moles by 24.

✓Final answer
$MgCl_2$

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