Q.The electrode potential of a magnesium electrode varies with the concentration of Mg2+ ions according to $E_{Mg^{2+}\mid Mg} = E^{\circ}_{Mg^{2+}\mid Mg} - \frac{0.059}{2}\log\frac{1}{[Mg^{2+}]}$. Which of the following plots correctly represents $E_{Mg^{2+}\mid Mg}$ (on the y-axis) against $\log[Mg^{2+}]$ (on the x-axis)?

Showing the 12 most recent of 23 on this concept.

1. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.Observe the given cell Cr $\mid$ Cr$^{3+}$ (0.1M)$\parallel$Fe$^{2+}$ (0.01M)$\mid$Fe What is the cell potential (in V) for the above cell? (Given: $E^\circ_{\text{Cr}^{3+}/\text{Cr}} = -0.74\,\text{V} \ ;\ E^\circ_{\text{Fe}^{2+}/\text{Fe}} = -0.44\,\text{V} \ ;\ \dfrac{2.303RT}{F} = 0.06\,\text{V}$) (A) 0.52 (B) 0.26 (C) 0.13 (D) 0.39

   ›Reveal solutionSolution

   The cell potential is found using the Nernst equation for the spontaneous reaction $\ce{3Fe^{2+} + 2Cr -> 3Fe + 2Cr^{3+}}$, yielding $E_{\text{cell}} = 0.26\,\text{V}$, so the correct option is (B).

   The key idea is to treat this as a concentration cell of sorts, but with different metals. We have two half-cells: one with chromium in chromium(III) solution, the other with iron in iron(II) solution. The standard reduction potentials tell us which metal is more likely to be oxidized. Since $E^\circ_{\text{Cr}^{3+}/\text{Cr}} = -0.74\,\text{V}$ is more negative than $E^\circ_{\text{Fe}^{2+}/\text{Fe}} = -0.44\,\text{V}$, chromium is the stronger reducing agent — it will oxidize (lose electrons) and iron(II) will reduce (gain electrons). So the spontaneous cell reaction is:

   • Anode (oxidation): $\ce{Cr -> Cr^{3+} + 3e^-}$
   • Cathode (reduction): $\ce{Fe^{2+} + 2e^- -> Fe}$

   To balance electrons, multiply the anode by 2 and the cathode by 3:

   $\ce{2Cr + 3Fe^{2+} -> 2Cr^{3+} + 3Fe}$

   Now we apply the Nernst equation to find the actual cell potential under non-standard concentrations.

   1. Find the standard cell potential

      $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = (-0.44\,\text{V}) - (-0.74\,\text{V}) = +0.30\,\text{V}$.

   2. Write the Nernst equation for the cell reaction

      For the reaction $\ce{2Cr + 3Fe^{2+} -> 2Cr^{3+} + 3Fe}$, the reaction quotient is

   $Q = \frac{[\ce{Cr^{3+}}]^2}{[\ce{Fe^{2+}}]^3}$

   (solids Cr and Fe have activity 1).

   The Nernst equation at 298 K (using the given $\frac{2.303RT}{F} = 0.06\,\text{V}$) is:

   $E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.06}{n} \log Q$

   where $n$ is the total number of electrons transferred in the balanced equation. Here, from the balanced reaction, $n = 6$ (since $2 \times 3 = 6$ or $3 \times 2 = 6$).

   3. Plug in the concentrations $[\ce{Cr^{3+}}] = 0.1\,\text{M}$, $[\ce{Fe^{2+}}] = 0.01\,\text{M}$.

   $Q = \frac{(0.1)^2}{(0.01)^3} = \frac{0.01}{1 \times 10^{-6}} = 10^4$

   So $\log Q = \log(10^4) = 4$.

   4. Compute the cell potential

   $E_{\text{cell}} = 0.30 - \frac{0.06}{6} \times 4 = 0.30 - 0.01 \times 4 = 0.30 - 0.04 = 0.26\,\text{V}$

   Tip

   A common mistake is to use $n = 2$ or $n = 3$ (the individual half-reaction electron counts) instead of $n = 6$ (the least common multiple). Always balance the full redox equation first to get the correct $n$.

   Watch out

   Another pitfall: forgetting that the Nernst equation uses the reaction quotient for the balanced overall reaction, not just the ratio of concentrations from the cell diagram. Here the cell diagram $\ce{Cr | Cr^{3+} (0.1M) || Fe^{2+} (0.01M) | Fe}$ already tells you which species are on which side, but you must still write the full balanced equation.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

2. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.The pH of 0.01 M aqueous aniline solution at 298 K is 8.3. Its degree of dissociation ($\alpha$) is (K$_b$ of aniline = $4 \times 10^{-10}$; antilog (0.7) = 5.0; antilog (0.3) = 2.0; antilog (0.4) = 2.5) (A) $1 \times 10^{-4}$ (B) $4 \times 10^{-4}$ (C) $1.5 \times 10^{-4}$ (D) $2 \times 10^{-4}$

   ›Reveal solutionSolution

   The degree of dissociation of aniline is found by relating the given pH to the hydroxide concentration, then using the base dissociation expression. The result is $\alpha = 2 \times 10^{-4}$, which corresponds to option (D).

   Concept & Intuition

   Aniline is a weak base, so in water it partially accepts a proton, producing OH⁻ ions. The pH tells us the pOH, and from pOH we get [OH⁻]. For a weak base, the degree of dissociation $\alpha$ is the fraction of molecules that have reacted. Since the initial concentration is known, [OH⁻] = $C \alpha$ (if $\alpha$ is small, which we can check). We can then use the given $K_b$ to solve for $\alpha$, or directly compute it from [OH⁻] and the concentration.

   Step-by-step solution

   1. Find pOH from pH At 298 K, pH + pOH = 14. Given pH = 8.3, so

   $\text{pOH} = 14 - 8.3 = 5.7$

   2. Calculate [OH⁻] from pOH

   $[\text{OH}^-] = 10^{-\text{pOH}} = 10^{-5.7}$

   Write $10^{-5.7} = 10^{-6} \times 10^{0.3}$.

   Using antilog(0.3) = 2.0, we get

   $[\text{OH}^-] = 10^{-6} \times 2.0 = 2.0 \times 10^{-6} \ \text{M}$

   3. Relate [OH⁻] to degree of dissociation For a weak base B (here aniline) of initial concentration $C = 0.01$ M,

   $\text{B} + \text{H}_2\text{O} \rightleftharpoons \text{BH}^+ + \text{OH}^-$

   At equilibrium: $[\text{OH}^-] = C\alpha$, assuming $\alpha$ is small so $C(1-\alpha) \approx C$.

   Thus

   $C\alpha = 2.0 \times 10^{-6}$

   $\alpha = \frac{2.0 \times 10^{-6}}{0.01} = 2.0 \times 10^{-4}$

   4. Verify using $K_b$ (optional check) The base dissociation constant is

   $K_b = \frac{C\alpha^2}{1-\alpha} \approx C\alpha^2$

   Substituting: $C\alpha^2 = 0.01 \times (2\times10^{-4})^2 = 0.01 \times 4\times10^{-8} = 4\times10^{-10}$, which matches the given $K_b$. This confirms the small $\alpha$ approximation is valid.

   Watch out

   A common mistake is to forget that pH gives pOH first for a base. Using pH directly to find [H⁺] and then trying to relate that to $\alpha$ would lead to the wrong answer.

   Tip

   Since antilog(0.3) = 2.0 is provided, you can quickly compute $10^{-5.7}$ without a calculator by splitting the exponent.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

3. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.What is the approximate $E_{\text{cell}}$ (in V) for the following cell at 298 K? 298 K Sn(s)$|$Sn$^{2+}$(0.05 M)$|$H$^+$(0.02 M)$|$H$_2$(g)(1 bar)$|$Pt(s) $\left( E^\circ_{\text{Sn}^{2+}|\text{Sn}} = -0.14\,\text{V};\quad E^\circ_{\text{H}^+|\text{H}_2} = 0.0\,\text{V};\quad \log(125) = 2.097 \right)$ (A) 0.218 (B) 0.078 (C) 0.02 (D) 0.04

   ›Reveal solutionSolution

   The cell potential under non-standard conditions is calculated using the Nernst equation. First, determine the standard cell potential and the reaction quotient, then substitute these values into the Nernst equation. The approximate cell potential is $\boxed{0.078\,\text{V}}$.

   In electrochemistry, the potential of a galvanic cell (a device that converts chemical energy into electrical energy) is typically measured under standard conditions ($1\,\text{M}$ concentration for solutions, $1\,\text{atm}$ or $1\,\text{bar}$ pressure for gases, $298\,\text{K}$ temperature). However, real-world cells rarely operate under these exact conditions. The Nernst equation allows us to calculate the cell potential ($E_{\text{cell}}$) when concentrations and pressures deviate from standard values.

   The core idea is that as a reaction proceeds, reactant concentrations decrease and product concentrations increase, causing the cell potential to change. The Nernst equation quantifies this change, showing how the cell potential depends on the concentrations of the species involved in the redox reaction.

   The Nernst equation at $298\,\text{K}$ is:

   $E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0592}{n} \log Q$

   where:

   • $E_{\text{cell}}$ is the cell potential under non-standard conditions.
   • $E^\circ_{\text{cell}}$ is the standard cell potential.
   • $n$ is the number of electrons transferred in the balanced cell reaction.
   • $Q$ is the reaction quotient.

   Let's break down the calculation step-by-step.

   1. Identify Anode and Cathode, and write Half-Reactions:

      The given cell notation is Sn(s)$|$Sn$^{2+}$(0.05 M)$|$H$^+$(0.02 M)$|$H$_2$(g)(1 bar)$|$Pt(s).

      By convention, the anode (oxidation half-cell) is written on the left, and the cathode (reduction half-cell) is on the right.

      • Anode (Oxidation): Sn(s) $\rightarrow$ Sn$^{2+}$(aq) + 2e$^-$
      • Cathode (Reduction): 2H$^+$(aq) + 2e$^-$ $\rightarrow$ H$_2$(g)

      We can confirm this by comparing the standard reduction potentials:

      $E^\circ_{\text{Sn}^{2+}|\text{Sn}} = -0.14\,\text{V}$

      $E^\circ_{\text{H}^+|\text{H}_2} = 0.0\,\text{V}$

      Since Sn has a more negative standard reduction potential, it will be oxidized (act as the anode), while H$^+$ will be reduced (act as the cathode).

   2. Write the Overall Cell Reaction:

      To get the overall reaction, we sum the half-reactions, ensuring the electrons cancel out. In this case, both half-reactions involve 2 electrons, so they sum directly:

      Sn(s) + 2H$^+$(aq) $\rightarrow$ Sn$^{2+}$(aq) + H$_2$(g)

   3. Calculate the Standard Cell Potential ($E^\circ_{\text{cell}}$):

      The standard cell potential is the difference between the standard reduction potential of the cathode and the anode.

      $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$

      $E^\circ_{\text{cell}} = E^\circ_{\text{H}^+|\text{H}_2} - E^\circ_{\text{Sn}^{2+}|\text{Sn}}$

      $E^\circ_{\text{cell}} = 0.0\,\text{V} - (-0.14\,\text{V})$

      $E^\circ_{\text{cell}} = 0.14\,\text{V}$

   4. Determine the Number of Electrons Transferred ($n$):

      From the balanced overall cell reaction (or either half-reaction), we see that 2 electrons are transferred.

      So, $n = 2$.

   5. Calculate the Reaction Quotient ($Q$):

      For the reaction Sn(s) + 2H$^+$(aq) $\rightarrow$ Sn$^{2+}$(aq) + H$_2$(g), the reaction quotient $Q$ is given by:

      $Q = \frac{[\text{Sn}^{2+}] P_{\text{H}_2}}{[\text{H}^+]^2}$

      We are given:

      $[\text{Sn}^{2+}] = 0.05\,\text{M}$

      $[\text{H}^+] = 0.02\,\text{M}$

      $P_{\text{H}_2} = 1\,\text{bar}$

      Substitute these values into the expression for $Q$:

      $Q = \frac{(0.05)(1)}{(0.02)^2}$

      $Q = \frac{0.05}{0.0004}$

      $Q = \frac{5 \times 10^{-2}}{4 \times 10^{-4}}$

      $Q = \frac{500}{4}$

      $Q = 125$

   6. Apply the Nernst Equation:

      Now, substitute $E^\circ_{\text{cell}}$, $n$, and $Q$ into the Nernst equation at $298\,\text{K}$:

      $E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0592}{n} \log Q$

      $E_{\text{cell}} = 0.14\,\text{V} - \frac{0.0592}{2} \log(125)$

      We are given $\log(125) = 2.097$.

      $E_{\text{cell}} = 0.14 - 0.0296 \times 2.097$

      $E_{\text{cell}} = 0.14 - 0.0620712$

      $E_{\text{cell}} = 0.0779288\,\text{V}$

   7. Approximate the Result:

      Rounding to three decimal places, $E_{\text{cell}} \approx 0.078\,\text{V}$.

      Comparing this with the given options:

      (A) 0.218

      (B) 0.078

      (C) 0.02

      (D) 0.04

      The calculated value matches option (B).

   ✓Final answer

   The approximate $E_{\text{cell}}$ for the given cell at $298\,\text{K}$ is $\boxed{0.078\,\text{V}}$.

4. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.At 298 K, if emf of the cell corresponding to the reaction, Zn(s) + 2H$^+$(aq) $\rightarrow$ Zn$^{2+}$(0.01 M) + H$_2$(g) (1 atm) is 0.28 V, then the pH of the solution at the hydrogen electrode is $\left( \dfrac{2.303 \ RT}{F} = 0.06 \ \text{V} \right), \left( E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \ \text{V} \right)$ (A) 8 (B) 7 (C) 9 (D) 10

   ›Reveal solutionSolution

   The key idea is to use the Nernst equation for the cell, relate the measured emf to the standard cell potential and the reaction quotient, and solve for the concentration of H⁺, which gives the pH. The pH is found to be 9.

   We are given a cell reaction:

   Zn(s) + 2H⁺(aq) → Zn²⁺(0.01 M) + H₂(g) (1 atm)

   with emf = 0.28 V at 298 K.

   Given: $\frac{2.303 RT}{F} = 0.06$ V, and $E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76$ V.

   Concept and intuition:

   The cell is a galvanic cell where zinc is oxidized and H⁺ is reduced. The measured voltage depends on the concentrations (and gas pressure) via the Nernst equation. Since we know the emf and the standard potential, we can find the actual [H⁺] at the hydrogen electrode, and thus the pH. The trick is that the standard cell potential is not directly given — we must compute it from the standard reduction potentials.

   1. Determine the standard cell potential $E^\circ_{\text{cell}}$.

      The half-reactions are:

      • Oxidation: Zn(s) → Zn²⁺ + 2e⁻, $E^\circ_{\text{ox}} = +0.76$ V (reverse of given reduction).
      • Reduction: 2H⁺ + 2e⁻ → H₂(g), $E^\circ_{\text{red}} = 0$ V (by definition). So, $E^\circ_{\text{cell}} = E^\circ_{\text{red}} + E^\circ_{\text{ox}} = 0 + 0.76 = 0.76$ V.

   2. Write the Nernst equation for the cell.

      For the reaction: Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g), the reaction quotient is

   $Q = \frac{[\text{Zn}^{2+}] \cdot P_{\text{H}_2}}{[\text{H}^+]^2}$

   (solids like Zn have activity = 1).

   The Nernst equation at 298 K is:

   $E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.06}{n} \log Q$

   where $n = 2$ (electrons transferred).

   Substituting:

   $0.28 = 0.76 - \frac{0.06}{2} \log \left( \frac{(0.01) \cdot (1)}{[\text{H}^+]^2} \right)$

   3. Simplify and solve for [H⁺]. Rearranging:

   $0.28 - 0.76 = -0.03 \log \left( \frac{0.01}{[\text{H}^+]^2} \right)$

   $-0.48 = -0.03 \log \left( \frac{0.01}{[\text{H}^+]^2} \right)$

   Multiply both sides by -1:

   $0.48 = 0.03 \log \left( \frac{0.01}{[\text{H}^+]^2} \right)$

   Divide by 0.03:

   $16 = \log \left( \frac{0.01}{[\text{H}^+]^2} \right)$

   This means:

   $\frac{0.01}{[\text{H}^+]^2} = 10^{16}$

   So:

   $[\text{H}^+]^2 = \frac{0.01}{10^{16}} = 10^{-2} \times 10^{-16} = 10^{-18}$

   Thus:

   $[\text{H}^+] = 10^{-9} \ \text{M}$

   4. Find the pH.

   $\text{pH} = -\log [\text{H}^+] = -\log(10^{-9}) = 9$

   Watch out

   A common mistake is to forget that the Nernst equation uses the reaction quotient with the correct stoichiometric coefficients. Here, [H⁺] is squared because two H⁺ ions are involved. Also, note that the standard hydrogen electrode potential is 0 V, so the cell standard potential is simply the opposite of the zinc reduction potential.

   Tip

   The given value $\frac{2.303 RT}{F} = 0.06$ V is the factor that converts natural log to base-10 log at 298 K. This simplifies the Nernst equation to $E = E^\circ - \frac{0.06}{n} \log Q$.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

5. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.At 298 K the equilibrium constant for the reaction $\mathrm{M(s) + 2Ag^+(aq) \rightarrow M^{2+}(aq) + 2Ag(s)}$ is $10^{15}$. What is the $E^{\ominus}_{cell}$ (in V) for this reaction? $\left(\dfrac{2.303\,RT}{F}\right) = 0.06\ \mathrm{V}$ (A) $0.225$ (B) $0.45$ (C) $0.90$ (D) $1.10$

   ›Reveal solutionSolution

   Use $E^{\ominus}_{cell} = \dfrac{0.06}{n}\log K$ with $n=2$ and $\log K = 15$: $E^{\ominus}_{cell} = 0.03\times 15 = \mathbf{0.45\ V}$ — option (B).

   The concept first

   Why should a cell potential know anything about an equilibrium constant? Because both measure the same driving force, just in different currencies. Thermodynamics gives

   $\Delta G^{\ominus} = -nFE^{\ominus}_{cell} \qquad\text{and}\qquad \Delta G^{\ominus} = -RT\ln K.$

   Equating them,

   $nFE^{\ominus}_{cell} = RT\ln K \ \Longrightarrow\ E^{\ominus}_{cell} = \frac{RT}{nF}\ln K = \frac{2.303RT}{nF}\log K.$

   You can reach the same place from the Nernst equation, $E_{cell}=E^{\ominus}_{cell}-\frac{2.303RT}{nF}\log Q$: at equilibrium the cell is dead ($E_{cell}=0$) and $Q=K$, so $E^{\ominus}_{cell}=\frac{2.303RT}{nF}\log K$. A large $K$ therefore means a large standard cell potential — the two statements are equivalent.

   The only trap is $n$: it is the number of electrons in the balanced equation as written.

   Step-by-step

   1. Split the reaction into half-cells to find $n$:
      • Oxidation: $\mathrm{M \rightarrow M^{2+} + 2e^-}$
      • Reduction: $\mathrm{2Ag^+ + 2e^- \rightarrow 2Ag}$ Two electrons are exchanged → $\boxed{n=2}$.
   2. Write the relation with the given constant. Since $\dfrac{2.303RT}{F}=0.06\ \mathrm{V}$ at $298\ \mathrm{K}$,

   $E^{\ominus}_{cell} = \frac{0.06}{n}\log K.$

   3. Insert the data. $K = 10^{15} \Rightarrow \log K = 15$.

   $E^{\ominus}_{cell} = \frac{0.06}{2}\times 15 = 0.03 \times 15$

   4. Evaluate.

   $E^{\ominus}_{cell} = 0.45\ \mathrm{V}$

   5. Sanity check. $K \gg 1$, so the reaction is strongly product-favoured and $E^{\ominus}_{cell}$ must be comfortably positive — $+0.45\ \mathrm{V}$ is exactly the right size for a metal displacing silver. (Option (C), $0.90\ \mathrm{V}$, is what you get if you carelessly forget to divide by $n=2$.)

   ✓Final answer

   With $n=2$ and $\log K=15$, $E^{\ominus}_{cell}=0.45\ \mathrm{V}$ — option (B).

   ANSWER: B

6. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
   Q.What is the pH of 0.365% (w/V) HCl aqueous solution? (HCl = 36.5 u), (log 0.365 = -0.4377) (A) 1.301 (B) 1.0 (C) 2.699 (D) 0.4377

   ›Reveal solutionSolution

   The key is to convert the given weight/volume percentage into molarity using the molar mass of HCl, then take the negative logarithm. The pH of the solution is 1.0.

   The problem gives you a concentration as a percentage weight/volume — that means grams of solute per 100 mL of solution. For a strong acid like HCl, which dissociates completely, the pH is simply $-\log[\text{H}^+]$, and $[\text{H}^+]$ equals the molarity of the acid. So the entire task is a unit conversion: from % (w/V) to mol/L.

   Let’s walk through it.

   1. Interpret the percentage.

      $0.365\%$ (w/V) means $0.365$ grams of HCl are present in every $100$ mL of solution. That’s the definition of weight/volume percent.

   2. Find the mass in 1 litre.

      Since $100$ mL contains $0.365$ g, $1000$ mL (1 L) contains $10$ times that:

   $\text{mass per litre} = 0.365 \times 10 = 3.65 \text{ g}$

   3. Convert mass to moles using the molar mass. Molar mass of HCl is given as $36.5$ g/mol. So the number of moles in 1 L is:

   $\text{moles per litre} = \frac{3.65}{36.5} = 0.1 \text{ mol}$

   That’s exactly $0.1$ M — a very clean number.

   4. Relate to $[\text{H}^+]$.

      HCl is a strong monoprotic acid: $\text{HCl} \rightarrow \text{H}^+ + \text{Cl}^-$. So $[\text{H}^+] = 0.1$ M.

   5. Calculate pH.

   $\text{pH} = -\log_{10} [\text{H}^+] = -\log_{10} (0.1) = -(-1) = 1.0$

   The given $\log 0.365 = -0.4377$ is a distractor — you don’t need it once you see that $3.65/36.5$ simplifies to $0.1$.

   Watch out

   A common mistake is to plug the percentage value directly into the log formula. The percentage is not the molarity — you must first convert grams per 100 mL to moles per litre. Also, the given log value is for $0.365$, not for $0.1$, so using it would lead you to the wrong answer.

   Tip

   Notice that $0.365\%$ was chosen so that the molarity becomes exactly $0.1$ M. This is a typical exam trick: the numbers are designed to cancel neatly, so trust the arithmetic.

   ✓Final answer

   The pH is $\boxed{1.0}$, which corresponds to option (B).

7. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
   Q.NaCl is crystallized in the presence of small quantity of SrCl$_2$. The formula of crystallized solid is Na$_{0.9998}$Sr$_{0.0001}$Cl. The number of cationic vacancies per mole of this solid is ($N = 6 \times 10^{23}$ mol$^{-1}$) (A) $6 \times 10^{-23}$ (B) $6 \times 10^{19}$ (C) $6 \times 10^{18}$ (D) $6 \times 10^{17}$

   ›Reveal solutionSolution

   When NaCl is doped with SrCl$_2$, each Sr$^{2+}$ ion replaces two Na$^+$ ions to maintain charge neutrality, creating one cationic vacancy for every Sr$^{2+}$ ion incorporated. Given the formula Na$_{0.9998}$Sr$_{0.0001}$Cl, there are $0.0001$ moles of Sr$^{2+}$ ions, which means $0.0001$ moles of cationic vacancies, resulting in $\boxed{6 \times 10^{19}}$ cationic vacancies per mole of the solid.

   The problem describes a common type of defect in ionic crystals, specifically a vacancy defect created by doping. The core concept here is the maintenance of electrical neutrality in the crystal lattice.

   Concept and Intuition

   1. NaCl Crystal Structure: Sodium chloride (NaCl) is an ionic compound where Na$^+$ ions and Cl$^-$ ions are arranged in a crystal lattice. Each Na$^+$ ion has a charge of $+1$, and each Cl$^-$ ion has a charge of $-1$. The crystal as a whole is electrically neutral.

   2. Doping with SrCl$_2$: When a small quantity of strontium chloride (SrCl$_2$) is added during the crystallization of NaCl, the Sr$^{2+}$ ions (from SrCl$_2$) can substitute some of the Na$^+$ ions in the NaCl lattice.

   3. Charge Neutrality Principle: The crystal must remain electrically neutral.

      • A Na$^+$ ion has a charge of $+1$.
      • A Sr$^{2+}$ ion has a charge of $+2$.
      • If one Sr$^{2+}$ ion replaces one Na$^+$ ion, the site would have a charge of $+2$ instead of $+1$, leading to an excess positive charge of $+1$ at that site.
      • To compensate for this extra positive charge and maintain overall electrical neutrality, another Na$^+$ ion must be removed from the lattice. This removal of a Na$^+$ ion creates a cationic vacancy (a missing positive ion).

   4. Vacancy Formation: Therefore, for every one Sr$^{2+}$ ion that substitutes into the NaCl lattice, it replaces two Na$^+$ ions. One Na$^+$ site is occupied by the Sr$^{2+}$ ion, and the other Na$^+$ site becomes a cationic vacancy. This means the number of cationic vacancies created is equal to the number of Sr$^{2+}$ ions incorporated into the lattice.

   Step-by-Step Solution

   1. Interpret the given formula:

      The formula of the crystallized solid is Na$_{0.9998}$Sr$_{0.0001}$Cl.

      This formula indicates the relative number of moles of each cation and anion in the solid. For every 1 mole of Cl$^-$ ions, there are $0.9998$ moles of Na$^+$ ions and $0.0001$ moles of Sr$^{2+}$ ions.

   2. Determine the moles of Sr$^{2+}$ ions:

      From the formula, the subscript for Sr is $0.0001$. This means that for every mole of the solid, there are $0.0001$ moles of Sr$^{2+}$ ions incorporated into the lattice.

   3. Relate Sr$^{2+}$ incorporation to cationic vacancies:

      As explained in the concept, each Sr$^{2+}$ ion (charge $+2$) replaces two Na$^+$ ions (each charge $+1$) to maintain charge neutrality. One Na$^+$ site is occupied by Sr$^{2+}$, and the other Na$^+$ site becomes a cationic vacancy.

      Therefore, the number of cationic vacancies created is equal to the number of Sr$^{2+}$ ions incorporated.

      Moles of cationic vacancies per mole of solid = Moles of Sr$^{2+}$ ions per mole of solid.

      Important

      For every $Sr^{2+}$ ion introduced into the NaCl lattice, one cationic vacancy is created to maintain electrical neutrality.

   4. Calculate the moles of cationic vacancies:

      Since there are $0.0001$ moles of Sr$^{2+}$ ions per mole of the solid, there will be $0.0001$ moles of cationic vacancies per mole of the solid.

   5. Convert moles of vacancies to the number of vacancies:

      We are given Avogadro's number, $N = 6 \times 10^{23}$ mol$^{-1}$.

      Number of cationic vacancies = (Moles of cationic vacancies) $\times$ (Avogadro's number)

      Number of cationic vacancies = $0.0001 \text{ mol} \times 6 \times 10^{23} \text{ mol}^{-1}$

      Number of cationic vacancies = $1 \times 10^{-4} \times 6 \times 10^{23}$

      Number of cationic vacancies = $6 \times 10^{19}$

   The number of cationic vacancies per mole of this solid is $6 \times 10^{19}$.

   ✓Final answer

   The number of cationic vacancies per mole of this solid is $\boxed{6 \times 10^{19}}$.

8. TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
   Q.At 298 K, the equilibrium constant for the reaction M(s) + 2Ag$^+$(aq) $\rightarrow$ M$^{2+}$(aq) + 2Ag(s) is $10^{15}$. The entropy change for the reaction at 298 K is 10 J K$^{-1}$. What is $\Delta_r H^\Theta$ (in kJ mol$^{-1}$) for this reaction? $\left(\dfrac{RT}{F} \times 2.303 = 0.06 \text{ and } F = 96500 \text{C mol}^{-1}\right)$ (A) $-86.85$ (B) $-89.83$ (C) $-96.5$ (D) $-83.87$

   ›Reveal solutionSolution

   We use the relationship between Gibbs free energy, equilibrium constant, and entropy/enthalpy changes. First, calculate the standard Gibbs free energy change ($\Delta G^\Theta$) from the equilibrium constant ($K$), then use it along with the given entropy change ($\Delta S^\Theta$) to find the standard enthalpy change ($\Delta H^\Theta$). The reaction's standard enthalpy change is $\boxed{-83.87 \text{ kJ mol}^{-1}}$.

   The core idea here is to connect the equilibrium constant ($K$) of a reaction to its thermodynamic properties, specifically the standard Gibbs free energy change ($\Delta G^\Theta$). Once $\Delta G^\Theta$ is known, we can use the fundamental relationship between Gibbs free energy, enthalpy ($\Delta H^\Theta$), and entropy ($\Delta S^\Theta$) to find the unknown enthalpy change.

   At equilibrium, the Gibbs free energy change for a reaction is zero, but the standard Gibbs free energy change ($\Delta G^\Theta$) is related to the equilibrium constant by a specific formula. This $\Delta G^\Theta$ represents the change in Gibbs free energy when reactants in their standard states are converted to products in their standard states.

   The two key thermodynamic relationships we will use are:

   1. $\Delta G^\Theta = -RT \ln K$: This equation links the standard Gibbs free energy change to the equilibrium constant $K$ at a given temperature $T$.
   2. $\Delta G^\Theta = \Delta H^\Theta - T \Delta S^\Theta$: This is the Gibbs-Helmholtz equation, which relates the standard Gibbs free energy change to the standard enthalpy change ($\Delta H^\Theta$) and standard entropy change ($\Delta S^\Theta$) at temperature $T$.

   By combining these two equations, we can solve for the unknown $\Delta H^\Theta$.

   1. Calculate $\Delta G^\Theta$ from the equilibrium constant ($K$): The relationship between standard Gibbs free energy change and the equilibrium constant is given by:

      $\Delta G^\Theta = -RT \ln K$

      We are given $K = 10^{15}$ and $T = 298$ K.

      Recall that $\ln K = 2.303 \log K$. So, the equation becomes:

   $\Delta G^\Theta = -RT (2.303 \log K)$

   The problem provides a useful conversion factor: $\frac{RT}{F} \times 2.303 = 0.06$.
   This implies $RT \times 2.303 = 0.06 F$.
   Substitute this into the expression for $\Delta G^\Theta$:
   

   $\Delta G^\Theta = -(0.06 F) \log K$

   Now, substitute the given values: $F = 96500 \text{ C mol}^{-1}$ and $\log K = \log(10^{15}) = 15$.
   

   $\Delta G^\Theta = -(0.06 \times 96500 \text{ C mol}^{-1}) \times 15$

   $\Delta G^\Theta = -(5790 \text{ J mol}^{-1}) \times 15$

   $\Delta G^\Theta = -86850 \text{ J mol}^{-1}$

   2. Calculate $\Delta H^\Theta$ using the Gibbs-Helmholtz equation: The standard Gibbs free energy change is also related to the standard enthalpy change and standard entropy change by:

      $\Delta G^\Theta = \Delta H^\Theta - T \Delta S^\Theta$

      We need to find $\Delta H^\Theta$, so rearrange the equation:

   $\Delta H^\Theta = \Delta G^\Theta + T \Delta S^\Theta$

   We have the following values:
   *   $\Delta G^\Theta = -86850 \text{ J mol}^{-1}$ (calculated in step 1)
   *   $T = 298 \text{ K}$ (given)
   *   $\Delta S^\Theta = 10 \text{ J K}^{-1} \text{ mol}^{-1}$ (given)
   
   Substitute these values into the equation:
   

   $\Delta H^\Theta = -86850 \text{ J mol}^{-1} + (298 \text{ K} \times 10 \text{ J K}^{-1} \text{ mol}^{-1})$

   $\Delta H^\Theta = -86850 \text{ J mol}^{-1} + 2980 \text{ J mol}^{-1}$

   $\Delta H^\Theta = -83870 \text{ J mol}^{-1}$

   3. Convert $\Delta H^\Theta$ to kJ mol$^{-1}$: The question asks for $\Delta_r H^\Theta$ in kJ mol$^{-1}$.

   $\Delta H^\Theta = -83870 \text{ J mol}^{-1} \times \frac{1 \text{ kJ}}{1000 \text{ J}}$

   $\Delta H^\Theta = -83.87 \text{ kJ mol}^{-1}$

   ✓Final answer

   The standard enthalpy change for the reaction is $\boxed{-83.87 \text{ kJ mol}^{-1}}$.

9. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
   Q.At 61 K, one mole of an ideal gas of 1.0 L volume expands isothermally and reversibly to a final volume of 10.0 L. What is the work done in the expansion? (A) $-11.52$ L atm (B) $-23.04$ L atm (C) $-46.08$ L atm (D) $-5.76$ L atm

   ›Reveal solutionSolution

   For an isothermal reversible expansion of an ideal gas, work is given by $W = -nRT \ln(V_f/V_i)$. Plugging in $n=1$, $T=61\,\text{K}$, $V_f/V_i=10$, and using $R = 0.0821\,\text{L·atm/(mol·K)}$, we get $W \approx -11.52\,\text{L·atm}$, so the answer is (A).

   The key concept here is work in an isothermal reversible process. For an ideal gas, when temperature is constant, the internal energy doesn't change, so all the heat absorbed goes into doing work. The work is not simply $-P\Delta V$ because pressure changes continuously; we must integrate $P\,dV$ using the ideal gas law $P = nRT/V$. The natural logarithm appears from that integration.

   Let’s work through it step by step:

   1. Recall the formula for reversible isothermal work For an ideal gas expanding reversibly at constant temperature,

   $W = -\int_{V_i}^{V_f} P\,dV = -\int_{V_i}^{V_f} \frac{nRT}{V}\,dV = -nRT \ln\left(\frac{V_f}{V_i}\right).$

   The negative sign means work is done by the system (expansion).

   2. Identify the given values

      • $n = 1$ mole
      • $T = 61$ K
      • $V_i = 1.0$ L, $V_f = 10.0$ L, so $V_f/V_i = 10$
      • $R$ must be in consistent units: we want work in L·atm, so use $R = 0.0821\,\text{L·atm/(mol·K)}$.

   3. Plug into the formula

   $W = - (1)(0.0821)(61) \ln(10).$

   First compute $0.0821 \times 61 = 5.0081$ (approximately).

   Then $\ln(10) \approx 2.3026$.

   So $W \approx -5.0081 \times 2.3026 \approx -11.53$ L·atm.

   4. Match to the options The closest value is $-11.52$ L·atm, which is option (A).

   Tip

   A common mistake is to forget the natural log and use $W = -P\Delta V$ instead. That would give $-P(9\,\text{L})$, but pressure isn’t constant — it drops as the gas expands. Always use the integral form for reversible processes.

   Watch out

   Watch the units! If you used $R = 8.314\,\text{J/(mol·K)}$, you’d get work in joules, not L·atm. The problem asks for L·atm, so use the appropriate $R$.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

10. TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQ
    Q.What is the electrode potential (in V) of copper electrode dipped in $10^{-2}$ M Cu$^{2+}$ solution? $(E^\circ_{\text{Cu}^{2+}/\text{Cu}} = 0.34\text{V})$ (A) 0.399 (B) 0.1405 (C) 0.199 (D) 0.281

    ›Reveal solutionSolution

    Using the Nernst equation for the reduction of Cu²⁺ to Cu, the electrode potential at 298 K for a 10⁻² M Cu²⁺ solution is 0.281 V, which corresponds to option (D).

    The key idea is that the electrode potential under non‑standard conditions is given by the Nernst equation. For a metal‑ion electrode like Cu²⁺/Cu, the reduction half‑reaction is:

    $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu(s)}$

    The standard potential $E^\circ = 0.34\ \text{V}$ applies when [Cu²⁺] = 1 M. When the concentration changes, the potential shifts according to the logarithmic dependence on ion concentration.

    Why this works:

    The Nernst equation derives from the relationship between Gibbs free energy and the reaction quotient. For a reduction, a lower concentration of the oxidized form (Cu²⁺) makes reduction less favourable, so the potential becomes less positive (or more negative) than the standard value. Here, since [Cu²⁺] is lower than 1 M, we expect $E < E^\circ$.

    Step‑by‑step calculation:

    1. Write the Nernst equation for the reduction half‑cell For the reaction $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu(s)}$, the Nernst equation at 298 K is:

    $E = E^\circ - \frac{0.0591}{n} \log Q$

    where $n = 2$ (number of electrons transferred) and $Q = \frac{1}{[\text{Cu}^{2+}]}$ because the solid copper activity is 1.

    2. Substitute the known values

    $E = 0.34 - \frac{0.0591}{2} \log\left(\frac{1}{10^{-2}}\right)$

    Note: $\frac{1}{10^{-2}} = 10^2$, so $\log(10^2) = 2$.

    3. Simplify the logarithmic term

    $\frac{0.0591}{2} \times 2 = 0.0591$

    Thus:

    $E = 0.34 - 0.0591 = 0.2809\ \text{V}$

    4. Round appropriately $0.2809\ \text{V}$ rounds to $0.281\ \text{V}$.

    Tip

    A common shortcut: for a 10⁻² M solution of a divalent metal ion, the correction term is exactly $0.0591\ \text{V}$ (since $\frac{0.0591}{2} \times 2 = 0.0591$). So the potential is simply $E^\circ - 0.0591$.

    Watch out

    A frequent mistake is using $n = 1$ or forgetting that the solid copper’s activity is 1, leading to an incorrect $Q$. Always check the balanced half‑reaction for the number of electrons.

    ✓Final answer

    The correct option is (D).

    ANSWER: D

11. TG EAPCET 2023Set ap-2023-05-11-AN1 markMCQ
    Q.The rate constant of a first order reaction is $6.4 \times 10^{-5}~\mathrm{S}^{-1}$ at $47~^\circ\mathrm{C}$ and $1.6 \times 10^{-5}~\mathrm{S}^{-1}$ at $27~^\circ\mathrm{C}$. The activation energy for the reaction is around (log $4 = 0.6021$, $R = 8.314~\mathrm{JK}^{-1}~\mathrm{mol}^{-1}$) (A) $110.0~\mathrm{kJ~mol}^{-1}$ (B) $27.0~\mathrm{kJ~mol}^{-1}$ (C) $55.0~\mathrm{kJ~mol}^{-1}$ (D) $220.0~\mathrm{kJ~mol}^{-1}$

    ›Reveal solutionSolution

    Using the Arrhenius equation in its two-point logarithmic form, the activation energy is found to be approximately 55.0 kJ mol⁻¹, which corresponds to option (C).

    The heart of this problem is the Arrhenius equation, which tells us how a reaction’s rate constant $k$ depends on temperature $T$ and activation energy $E_a$:

    $k = A e^{-E_a / RT}$

    Here $A$ is the pre-exponential factor (constant for a given reaction), $R$ is the gas constant, and $T$ is the absolute temperature. When you have rate constants at two different temperatures, you can eliminate $A$ by taking a ratio — that’s the classic trick. Taking natural logs gives a linear relationship between $\ln k$ and $1/T$, so the slope directly yields $E_a$.

    Let’s work through it step by step.

    1. Convert temperatures to Kelvin.

       The Arrhenius equation uses absolute temperature.

       $T_1 = 27^\circ\mathrm{C} + 273 = 300~\mathrm{K}$

       $T_2 = 47^\circ\mathrm{C} + 273 = 320~\mathrm{K}$

    2. Write the two-point Arrhenius form.

       For two temperatures $T_1$ and $T_2$ with corresponding rate constants $k_1$ and $k_2$:

    $\ln\frac{k_2}{k_1} = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$

    A more convenient rearrangement is:

    $\log\frac{k_2}{k_1} = \frac{E_a}{2.303\,R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$

    (We switch to base-10 log because the problem gives $\log 4$.)

    3. Plug in the given values.

       $k_2 = 6.4 \times 10^{-5}~\mathrm{s}^{-1}$ (at $320~\mathrm{K}$)

       $k_1 = 1.6 \times 10^{-5}~\mathrm{s}^{-1}$ (at $300~\mathrm{K}$)

       So $\displaystyle \frac{k_2}{k_1} = \frac{6.4 \times 10^{-5}}{1.6 \times 10^{-5}} = 4$

       Hence $\log\frac{k_2}{k_1} = \log 4 = 0.6021$

    4. Compute the temperature term.

    $\frac{1}{T_1} - \frac{1}{T_2} = \frac{1}{300} - \frac{1}{320} = \frac{320 - 300}{300 \times 320} = \frac{20}{96000} = \frac{1}{4800}$

    So $\displaystyle \frac{1}{T_1} - \frac{1}{T_2} = 2.0833 \times 10^{-4}~\mathrm{K}^{-1}$

    5. Solve for $E_a$. From the equation:

    $0.6021 = \frac{E_a}{2.303 \times 8.314} \times \frac{1}{4800}$

    Rearranging:

    $E_a = 0.6021 \times 2.303 \times 8.314 \times 4800$

    Let’s calculate stepwise:

    $2.303 \times 8.314 \approx 19.147$

    $19.147 \times 4800 \approx 91905.6$

    $91905.6 \times 0.6021 \approx 55330~\mathrm{J~mol}^{-1}$

    Converting to kJ: $E_a \approx 55.33~\mathrm{kJ~mol}^{-1}$

    Watch out

    A common slip is forgetting to convert Celsius to Kelvin — using $27$ and $47$ directly gives a wildly wrong answer. Also, note that the formula uses $\log$ (base 10) when the given log value is base 10; if you use $\ln$ instead, you’d omit the $2.303$ factor.

    Tip

    The ratio $k_2/k_1 = 4$ is neat here. If you remember that $\log 4 \approx 0.6021$ and $1/300 - 1/320 = 1/4800$, the arithmetic becomes very clean — you can often do it without a calculator in an exam.

    ✓Final answer

    The activation energy is approximately 55.0 kJ mol⁻¹, which matches option (C).

12. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
    Q.The electrode potential of chlorine electrode is maximum, when the concentration of chloride ion in the solution (in mol L$^{-1}$) is X. what is the value of X? (A) $2.5 \times 10^{-3}$ (B) $7.5 \times 10^{-3}$ (C) $7.5 \times 10^{-2}$ (D) $2.5 \times 10^{-2}$

    ›Reveal solutionSolution

    The electrode potential of a chlorine electrode is maximized when the chloride ion concentration minimizes the negative term in the Nernst equation, which occurs at the lowest concentration among the options: $2.5 \times 10^{-3}$ mol L$^{-1}$.

    Understanding the Chlorine Electrode

    The chlorine electrode involves the half-reaction:

    $\text{Cl}_2(g) + 2e^- \rightleftharpoons 2\text{Cl}^-(aq)$

    The electrode potential is governed by the Nernst equation, which relates the actual potential to the standard potential and the concentrations of species involved.

    The Nernst Equation for the Chlorine Electrode

    For the chlorine electrode, the Nernst equation is:

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

    where the reaction quotient $Q$ is:

    $Q = \frac{[\text{Cl}^-]^2}{P_{\text{Cl}_2}}$

    At 25°C, converting to base-10 logarithm and assuming standard pressure for Cl₂ ($P_{\text{Cl}_2} = 1$ atm):

    $E = E^\circ - \frac{0.0591}{2}\log[\text{Cl}^-]^2$

    $E = E^\circ - 0.0591\log[\text{Cl}^-]$

    Key Relationship:

    $E = E^\circ - 0.0591\log[\text{Cl}^-]$

    Since $\log[\text{Cl}^-]$ is negative when $[\text{Cl}^-] < 1$, the term $-0.0591\log[\text{Cl}^-]$ becomes positive, adding to $E^\circ$.

    Maximizing the Electrode Potential

    Let's analyze how $E$ varies with $[\text{Cl}^-]$:

    1. When $[\text{Cl}^-]$ is small (say $10^{-3}$ M):

       • $\log(10^{-3}) = -3$
       • $E = E^\circ - 0.0591 \times (-3) = E^\circ + 0.177$ V

    2. When $[\text{Cl}^-]$ is larger (say $10^{-2}$ M):

       • $\log(10^{-2}) = -2$
       • $E = E^\circ - 0.0591 \times (-2) = E^\circ + 0.118$ V

    3. Pattern: As $[\text{Cl}^-]$ decreases, $\log[\text{Cl}^-]$ becomes more negative, making $-0.0591\log[\text{Cl}^-]$ more positive, thus increasing $E$.

    Comparing the Options

    Let's calculate the electrode potential contribution for each option:

    Option	$[\text{Cl}^-]$ (M)	$\log[\text{Cl}^-]$	$-0.0591\log[\text{Cl}^-]$ (V)
    (A)	$2.5 \times 10^{-3}$	$-2.60$	$+0.154$
    (B)	$7.5 \times 10^{-3}$	$-2.12$	$+0.125$
    (C)	$7.5 \times 10^{-2}$	$-1.12$	$+0.066$
    (D)	$2.5 \times 10^{-2}$	$-1.60$	$+0.095$

    The maximum electrode potential occurs with option (A), which has the smallest chloride concentration.

    Tip

    Physical Intuition: Lower product concentration (Cl⁻) drives the equilibrium toward the products, making the reduction of Cl₂ more favorable and increasing the electrode potential.

    ✓Final answer

    The correct option is (A).

    ANSWER: A