Questions are transcribed from NTA's official question paper. Diagrams are described in words; open the official PDF to see them. Every answer was solved twice independently and is shown only when both agree. Where they did not, we say so.
An electrochemical cell is constructed using half cells in the direction of spontaneous change
$Fe(OH)_2(s) + 2e^- \rightarrow Fe(s) + 2OH^-(aq)$ $E^0 = -0.88$ V
and $AgBr(s) + e^- \rightarrow Ag(s) + Br^-(aq)$ $E^0 = +0.07$ V
Which of the following option is correct?
Consider the following two half-cell reactions along with the standard reduction potential given:
$\text{CO}_2 + 6\text{H}^+ + 6\text{e}^- \longrightarrow \text{CH}_3\text{OH} + \text{H}_2\text{O}$ $\quad E^o_{\text{red}} = 0.02 \text{ V}$
$\frac{1}{2}\text{O}_2 + 2\text{H}^+ + 2\text{e}^- \longrightarrow \text{H}_2\text{O}$ $\quad E^o_{\text{red}} = 1.23 \text{ V}$
A fuel cell was set up using the above two reactions such that the cell operates under the standard condition of 1 bar pressure and 298 K temperature. The fuel cell works with 80% efficiency. If the work derived from the cell using 1 mol of $\text{CH}_3\text{OH}$ is used to compress an ideal gas isothermally against a constant pressure of 1 kPa, then the change in the volume of the gas,
$\Delta V = $ _________ $\text{m}^3$. (nearest integer)
Given: $F = 96500 \text{ C mol}^{-1}$
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Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
Key idea: Electrochemistry: Fuel cell efficiency, Gibbs free energy, and thermodynamic work
In order to oxidise a mixture of 1 mole each of $FeC_2O_4$, $Fe_2(C_2O_4)_3$, $FeSO_4$ and $Fe_2(SO_4)_3$ in acidic medium, the number of moles of $KMnO_4$ required is
(1) 3
(2) 2
(3) 5
(4) 7
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Answer: Option (2)
Key idea: Redox Titrations and Equivalent Concept
Identify oxidizable species: $Fe^{2+}$ can oxidize to $Fe^{3+}$ (n-factor = 1), $C_2O_4^{2-}$ can oxidize to $CO_2$ (n-factor = 2), while $Fe^{3+}$ and $SO_4^{2-}$ cannot be oxidized further
Calculate equivalents from $FeC_2O_4$: 1 mole gives 1 equiv from $Fe^{2+}$ and 2 equiv from $C_2O_4^{2-}$ = 3 equivalents
Calculate equivalents from $Fe_2(C_2O_4)_3$: 1 mole gives 0 from $Fe^{3+}$ and 6 equiv from 3 moles of $C_2O_4^{2-}$ = 6 equivalents
Calculate equivalents from $FeSO_4$: 1 mole gives 1 equiv from $Fe^{2+}$ = 1 equivalent
An electrochemical cell, consist of the following two redox couples, $M^{x+}$(aq)/$M(s)$ $[E^{\Theta}_{ox/red} = +0.15V]$ and $Fe^{3+}$(aq)/$Fe(s)$ $[E^{\Theta}_{red} = -0.036V]$. The cell EMF $(E_{cell})$ is recorded to be $0.2057$ V. If the reaction quotient of the electrochemical reaction is found to be $10^{-2}$, then the value of $x$ is __________.(Nearest integer)
[Given : M is a p-block metal and $\frac{2.303RT}{F} = 0.059$ V]
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Answer: 2
Key idea: Nernst Equation and Electrochemical Cell EMF
Interpret the given potentials: $E^\Theta_{red}(M^{x+}/M) = +0.15$ V and $E^\Theta_{red}(Fe^{3+}/Fe) = -0.036$ V
Identify cathode and anode: Since $+0.15 > -0.036$, $M^{x+}/M$ is cathode (reduction) and $Fe/Fe^{3+}$ is anode (oxidation)
One half cell in a voltaic cell is constructed by dipping silver rod in $AgNO_3$ solution of unknown concentration, other half cell is Zn rod dipped in 1 molar solution of $ZnSO_4$. A voltage of 1.60 V is measured at 298 K for this cell. What is the concentration of $Ag^+$ ions used in terms of $\log x$ ($x=[Ag^+]$)?
$E^o_{Zn^{2+}/Zn} = -0.76V$, $E^o_{Ag^+/Ag} = +0.80V$, $\frac{2.303RT}{F}=0.059V$
(1) $\frac{2}{3.9}$
(2) $\frac{4}{5.9}$
(3) $\frac{2.9}{2}$
(4) $\frac{5.9}{4}$
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Answer: Option (2)
Key idea: Nernst Equation and Cell Potential
Identify cell reaction: Zn acts as anode (oxidation) and Ag as cathode (reduction). Overall: $Zn + 2Ag^+ \rightarrow Zn^{2+} + 2Ag$
Calculate standard cell potential: $E^o_{cell} = E^o_{Ag^+/Ag} - E^o_{Zn^{2+}/Zn} = 0.80 - (-0.76) = 1.56$ V
At 298 K, the molar conductivity of $x\%$ (w/w) MX solution (aqueous) is 123.5 S cm$^2$ mol$^{-1}$. The conductance of same solution is $1.9 \times 10^{-3}$ S. The value of $x$ is __________ $\times 10^{-2}$.
(Given : cell constant = 1.3 cm$^{-1}$, molar mass of MX is 75 g mol$^{-1}$, density of aqueous solution of MX at 298 K is 1.0 g mL$^{-1}$)
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Answer: 15
Key idea: Relationship between conductivity, molar conductivity, and concentration; conversion between molarity and weight/weight percentage
Calculate conductivity: $\kappa = G \times \text{cell constant} = 1.9 \times 10^{-3} \times 1.3 = 2.47 \times 10^{-3}$ S cm$^{-1}$
Use the relation $\Lambda_m = \frac{\kappa \times 1000}{C}$ to find molarity: $C = \frac{2.47 \times 10^{-3} \times 1000}{123.5} = 0.02$ mol L$^{-1}$
For $x\%$ (w/w) solution: 100 g solution contains $x$ g of MX with volume = $\frac{100}{1.0} = 100$ mL = 0.1 L
Express molarity in terms of $x$: $C = \frac{x}{75 \times 0.1} = \frac{x}{7.5}$ mol L$^{-1}$
Solve for $x$: $0.02 = \frac{x}{7.5} \Rightarrow x = 0.15 = 15 \times 10^{-2}$
Consider the following data.
| Electrolyte | $\Lambda^\circ_m (S \text{ cm}^2 \text{ mol}^{-1})$ |
|-------------|------------------------------------------------------|
| $BaCl_2$ | $x_1$ |
| $H_2SO_4$ | $x_2$ |
| HCl | $x_3$ |
$BaSO_4$ is sparingly soluble in water. If the conductivity of the saturated $BaSO_4$ solution is $x$ S cm$^{-1}$ then the solubility product of $BaSO_4$ can be given as
(Here $\Lambda_m = \Lambda^\circ_m$)
(1) $\frac{10^6 x^2}{a^2(x_1 + x_2 - 2x_3)^2}$
(2) $\frac{x^2}{(x_1 + x_2 - 2x_3)^2}$
(3) $\frac{a^2(x_1 + x_2 - 2x_3)^2}{10^6 x^2}$
(4) $\frac{x^2}{(x_1 + x_2 + 2x_3)^2}$
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Answer: Option (1)
Key idea: Kohlrausch's Law and Solubility Product from Conductivity
Use Kohlrausch's law to find $\Lambda^\circ_m(BaSO_4)$: From the given data, $\Lambda^\circ_m(BaSO_4) = \lambda^\circ(Ba^{2+}) + \lambda^\circ(SO_4^{2-}) = (x_1 - 2\lambda^\circ(Cl^-)) + (x_2 - 2\lambda^\circ(H^+)) = x_1 + x_2 - 2x_3$
Relate conductivity to concentration: For sparingly soluble salt, $\Lambda_m = \Lambda^\circ_m = \frac{1000\kappa}{C}$ where $\kappa = x$ S cm$^{-1}$ and $C$ is concentration in mol/L
For a general redox reaction
Anode $\text{Red}_1 \to \text{Ox}_1^{n_1^+} + n_1e^-$
Cathode $\text{Ox}_2 + n_2e^- \to \text{Red}_2^{n_2^-}$
Which of the following statement is incorrect?
(1) The overall reaction can be written as $n_2\text{Red}_1 + n_1\text{Ox}_2 \rightleftharpoons n_2\text{Ox}_1^{n_1^+} + n_1\text{Red}_2^{n_2^-}$
(2) The electrons do not appear in the overall reaction because electrons produced at the anode are consumed at the cathode.
(3) [Figure: A graph showing (E-E°)/RT/F vs log₁₀Q with three lines for n=1, n=2, and n=3, all with slope ∝ 1/n. Here n is the number of electrons transferred in redox reaction.]
(4) If the reaction is carried out reversibly, the electrical work done is equal to the ratio of charge and potential difference through which charge is moved.
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Answer: Option (4)
Key idea: Nernst Equation and Electrochemical Work
Write the balanced overall reaction by multiplying anode by $n_2$ and cathode by $n_1$ to balance electrons: $n_2\text{Red}_1 + n_1\text{Ox}_2 \to n_2\text{Ox}_1^{n_1^+} + n_1\text{Red}_2^{n_2^-}$ ✓
Verify statement (2): Electrons cancel in overall reaction since electrons produced at anode equal electrons consumed at cathode ✓
Check statement (3) using Nernst equation: $E = E° - \frac{2.303RT}{nF}\log_{10}Q$, so $\frac{E-E°}{RT/F} = -\frac{2.303}{n}\log_{10}Q$ with slope $\propto \frac{1}{n}$ ✓
500 mL of 0.2 M MnO$_4^-$ solution in basic medium when mixed with 500 mL of 1.5 M KI solution, oxidises iodide ions to liberate molecular iodine. This liberated iodine is then titrated with a standard $x$ M thiosulphate solution in presence of starch till the end point. If 300 mL of thiosulphate was consumed, then the value of $x$ is __________.
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Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
Given at 298 K : $E^{\circleddash}_{Fe^{2+}/Fe} = X$ Volt
$E^{\circleddash}_{Fe^{3+}/Fe} = Y$ Volt
The $E^{\circleddash}_{Fe^{3+}/Fe^{2+}}$ in Volt at 298 K is given by :
(1) $2X - 3Y$
(2) $3Y - 2X$
(3) $3Y + 2X$
(4) $Y + X$
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Answer: Option (2)
Key idea: Relationship between standard electrode potentials using Gibbs free energy
Write the given half-reactions: $Fe^{2+} + 2e^- \rightarrow Fe$ with $E° = X$ and $Fe^{3+} + 3e^- \rightarrow Fe$ with $E° = Y$
Use the relation $\Delta G° = -nFE°$ for each reaction: $\Delta G°_1 = -2FX$ and $\Delta G°_2 = -3FY$
The target reaction $Fe^{3+} + e^- \rightarrow Fe^{2+}$ can be obtained by: Reaction 2 - Reaction 1
