Electrochemistry JEE Advanced previous year questions with solutions

4 solved JEE Advanced questions on Electrochemistry, free to read — no sign-in needed. The full chapter has 35 questions; sign in to attempt the remaining 31 in the exam simulator.

1. Q1JEE Advanced Adv 2025 (Paper 2)
   An electrochemical cell is fueled by the combustion of butane at 1 bar and 298 K . Its cell potential is $\frac{\boldsymbol{X}}{\boldsymbol{F}} \times 10^3$ volts, where $\boldsymbol{F}$ is the Faraday constant. The value of $\boldsymbol{X}$ is $\qquad$ . Use : Standard Gibbs energies of formation at 298 K are : $\Delta_f G_{\mathrm{CO}_2}^{\circ}=-394 \mathrm{~kJ} \mathrm{~mol}^{-1}$; $\Delta_f G_{\text {water }}^{\circ}=-237 \mathrm{~kJ} \mathrm{~mol}^{-1} ; \Delta_f G_{\text {butane }}^{\circ}=-18 \mathrm{~kJ} \mathrm{~mol}^{-1}$
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   Answer: 105.5

   $\begin{aligned} & \mathrm{C}_4 \mathrm{H}_{10}(\mathrm{~g})+\frac{13}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow 4 \mathrm{CO}_2(\mathrm{~g})+5 \mathrm{H}_2 \mathrm{O}(l) \\ & \Delta_{\mathrm{r}} \mathrm{G}^{\circ}=4 \Delta_{\mathrm{f}} \mathrm{G}_{\mathrm{CO}_2}^{\circ}+5 \Delta_{\mathrm{f}} \mathrm{G}_{\mathrm{H}_2 \mathrm{O}}^{\circ}-\Delta_{\mathrm{f}} \mathrm{G}_{\mathrm{C}_4 \mathrm{H}_{10}}^{\circ} \\ & =4 \times(-394)+5(-237)+18 \\ & =-2743 \mathrm{~kJ} / \mathrm{mol} \\ & \Delta_{\mathrm{r}} \mathrm{G}^{\circ}=-\mathrm{nFE}^{\circ} \\ & -2743 \times 1000=-26 \times \mathrm{FE}^{\circ} \\ & \mathrm{E}^{\circ}=\frac{105.5}{\mathrm{~F}} \times 10^3=105.50\end{aligned}$
2. Q2JEE Advanced Adv 2025 (Paper 1)
   In an electrochemicalcell, dichromate ions in aqueous acidic medium are reduced to $\mathrm{Cr}^{3+}$. The current (in amperes) that flows through the cell for 48.25 minutes to produce 1 mole of $\mathrm{Cr}^{3+}$ is _____ . Use: 1 Faraday $=96500 \mathrm{C} \mathrm{mol}^{-1}$
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   Answer: 100

   For reduction of dichromate, balanced reaction is : $\mathrm{Cr}_2 \mathrm{O}_7^{-2}(\mathrm{aq})+6 \mathrm{e}^{-}+14 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow 2 \mathrm{Cr}^{3+}(\mathrm{aq})+7 \mathrm{H}_2 \mathrm{O}(\mathrm{l})$ $\qquad \qquad \qquad3 Mole$ $\qquad \qquad \qquad \qquad 1 Mole$ Number of Farads required $=3 \mathrm{~mol}$ Let current $=\mathrm{I}$ amperes $\begin{aligned} & \Rightarrow \frac{I \times 48.25 \times 60}{96500}=3 \\ & I=100 \mathrm{~A} \end{aligned}$
3. Q3JEE Advanced Adv 2024 (Paper 2)
   An aqueous solution of hydrazine $\left(\mathrm{N}_2 \mathrm{H}_4\right)$ is electrochemically oxidized by $\mathrm{O}_2$, thereby releasing chemical energy in the form of electrical energy. One of the products generated from the electrochemical reaction is $\mathrm{N}_2(\mathrm{~g})$. Choose the correct statement(s) about the above process :
   1. A.$\mathrm{OH}^{-}$ions react with $\mathrm{N}_2 \mathrm{H}_4$ at the anode to form $\mathrm{N}_2(\mathrm{~g})$ and water, releasing 4 electrons to the anode.
   2. B.At the cathode, $\mathrm{N}_2 \mathrm{H}_4$ breaks to $\mathrm{N}_2(\mathrm{~g})$ and nascent hydrogen released at the electrode reacts with oxygen to form water.
   3. C.At the cathode, molecular oxygen gets converted to $\mathrm{OH}^{-}$.
   4. D.Oxides of nitrogen are major by-products of the electrochemical process.
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   Answer: A,C

   $\underset{\substack{\text { Oxidation } \\ \text { (Anode) }}}{\stackrel{-2\qquad}{N_2H_4}}+\underset{\substack{\text { Reduction } \\ \text { (Cathode) }}}{\stackrel{0}{O_2}} \longrightarrow \stackrel{0}{N_2}+\mathrm{H}_2 \mathrm{O}^{-2}$ At anode: $\mathrm{N}_2 \mathrm{H}_4+4 \mathrm{OH}^{-} \longrightarrow \mathrm{N}_2+4 \mathrm{H}_2 \mathrm{O}+4 e^{-}$ At cathode: $\mathrm{O}_2+2 \mathrm{H}_2 \mathrm{O}+4 e^{-} \longrightarrow 4 \mathrm{OH}^{-}$ Complete reaction: $\mathrm{N}_2 \mathrm{H}_4+\mathrm{O}_2 \longrightarrow \mathrm{N}_2+2 \mathrm{H}_2 \mathrm{O}$ Statements (A) and (C) are correct.
4. Q4JEE Advanced Adv 2023 (Paper 1)
   Plotting $1/{\Lambda }_{m}$ against ${c\Lambda }_{m}$ for aqueous solutions of a monobasic weak acid $(HX)$ resulted in a straight line with $y-$axis intercept of $P$ and slope of $S$. The ratio $P/S$ is $\left[{\Lambda }_{m}=\right.\text{molar conductivity}$ ${\Lambda }_{m}^{o}=limitingmolarconductivity c=molarconcentration$ $\left.{K}_{a}=\text{dissociation constant of}HX\right]$
   1. A.${K}_{a}{\Lambda }_{m}^{o}$
   2. B.${K}_{a}{\Lambda }_{m}^{o}/2$
   3. C.$2{K}_{a}{\Lambda }_{m}^{o}$
   4. D.$1/\left({K}_{a}{\Lambda }_{m}^{o}\right)$
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   Answer: (A)

   The degree of dissociation is the ratio of molar conductivity to the limiting molar conductivity. $\alpha =\dfrac{{\Lambda }_{m}}{{\Lambda }_{m}^{o}}$ The equilibrium reaction is given below, $HX\rightleftharpoons ^{}{H}^{+}+{X}^{-} initialc Atequilibriumc-c\alpha c\alpha c\alpha$ The acid dissociation constant, ${K}_{a}=\dfrac{{c\alpha }^{2}}{1-\alpha }$ $\Rightarrow {K}_{a}=\dfrac{c{\left({\Lambda }_{m}/{\Lambda }_{m}^{o}\right)}^{2}}{1-\left({\Lambda }_{m}/{\Lambda }_{m}^{o}\right)}$ $\Rightarrow {K}_{a}=\dfrac{{c\Lambda }_{m}^{2}}{{\Lambda }_{m}^{o}\left({\Lambda }_{m}^{o}-{\Lambda }_{m}\right)}$ ${K}_{a}{{\Lambda }_{m}^{o}}^{2}-{K}_{a}{\Lambda }_{m}^{o}{\Lambda }_{m}={c\Lambda }_{m}^{2}$ $\dfrac{{K}_{a}{{\Lambda }_{m}^{o}}^{2}}{{\Lambda }_{m}}-{K}_{a}{\Lambda }_{m}^{o}={C\Lambda }_{m}$ $\dfrac{{K}_{a}{{\Lambda }_{m}^{o}}^{2}}{{\Lambda }_{m}}={c\Lambda }_{m}+{K}_{a}{\Lambda }_{m}^{o}$ $\dfrac{1}{{\Lambda }_{m}}=\left(\dfrac{{c\Lambda }_{m}}{{K}_{a}{{\Lambda }_{m}^{o}}^{2}}\right)+\dfrac{1}{{\Lambda }_{m}^{o}}$ From the above line equation, slope and y-intercept can be written as follows, $P=\dfrac{1}{{\Lambda }_{m}^{o}}$ $S=\dfrac{1}{{K}_{a}{{\Lambda }_{m}^{o}}^{2}}$ $\dfrac{P}{S}=\left(\dfrac{\dfrac{1}{{\Lambda }_{m}^{o}}}{\dfrac{1}{{K}_{a}{{\Lambda }_{m}^{o}}^{2}}}\right)={K}_{a}{\Lambda }_{m}^{o}$