Top 50 Most Repeated ELECTROCHEMISTRY PYQs | JEE MAINS
A curated collection of the most important questions from ELECTROCHEMISTRY, fully solved with step-by-step concepts to prepare for JEE MAINS.
A curated collection of the most important questions from ELECTROCHEMISTRY, fully solved with step-by-step concepts to prepare for JEE MAINS.
Use Nernst eq: $E = 1.1 - \frac{0.059}{2}\log(10) = 1.1 - 0.0295 = 1.07$ V....
Read Full Step-by-Step Solution →The standard EMF of the cell is calculated as $ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} $. Here, Cu is the cathode a...
Read Full Step-by-Step Solution →In standard notation, left electrode is anode. Zn is on left, so it undergoes oxidation: $ \text{Zn} \to \text{Zn}^{2+} + 2e^- $. Cu is reduced at cat...
Read Full Step-by-Step Solution →Use $\Delta G = -nFE$. For water formation, $n=4$. $E = \frac{240 \times 10^3}{4 \times 96500} \approx 1.24$ V....
Read Full Step-by-Step Solution →Standard EMF: E°cell = 0.34 − (‑0.76) = 1.10 V. Reaction quotient Q = [Zn²⁺]/[Cu²⁺] = 1/0.01 = 100. Nernst correction: (RT/nF) ln Q = (0.025693 / 2) ×...
Read Full Step-by-Step Solution →$\text{Na}^+ + e^- \to \text{Na}$. 23 g = 1 mol Na → requires 1 mol $e^-$....
Read Full Step-by-Step Solution →The EMF of a concentration cell is given by \(E = \frac{RT}{nF}\ln\frac{c_{high}}{c_{low}}\). At 298 K, \(\frac{RT}{F}=0.025693\,\text{V}\); for Cu²⁺,...
Read Full Step-by-Step Solution →Molar conductance = (specific conductance × 1000) / concentration. Substitute values directly....
Read Full Step-by-Step Solution →$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.34 - (-0.76)$...
Read Full Step-by-Step Solution →Using Nernst: $E = 1.10 - \frac{0.059}{2} \log\frac{0.1}{0.01} = 1.10 - 0.0295 = 1.07\,\text{V}$...
Read Full Step-by-Step Solution →EMF = E$_{\text{cathode}}$ - E$_{\text{anode}}$ = 0.34 - (-0.76) = 1.10 V....
Read Full Step-by-Step Solution →$Cu^{2+}$ requires 2 moles of electrons. 1 Faraday deposits 1 Eq. 2F deposits 2 Eq = 63.5g....
Read Full Step-by-Step Solution →$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.34 - (-0.76) = 1.10\,\text{V}$....
Read Full Step-by-Step Solution →$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.34 - (-0.76) = 1.10\,V$...
Read Full Step-by-Step Solution →Using Nernst equation: $E = E^\circ - \frac{0.059}{2}\log\frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}$...
Read Full Step-by-Step Solution →For a 1:1 electrolyte, the equivalent conductance equals the molar conductivity because one mole provides one equivalent. Hence $\Lambda_{eq}=\Lambda_...
Read Full Step-by-Step Solution →Foundational check for Electrochemistry. Study the core principles carefully for competitive exams....
Read Full Step-by-Step Solution →This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Electrochemistry....
Read Full Step-by-Step Solution →Use Nernst eq: $E = 1.1 - \frac{0.059}{2}\log(10) = 1.1 - 0.0295 = 1.07$ V....
Read Full Step-by-Step Solution →The standard EMF of the cell is calculated as $ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} $. Here, Cu is the cathode a...
Read Full Step-by-Step Solution →In standard notation, left electrode is anode. Zn is on left, so it undergoes oxidation: $ \text{Zn} \to \text{Zn}^{2+} + 2e^- $. Cu is reduced at cat...
Read Full Step-by-Step Solution →Use $\Delta G = -nFE$. For water formation, $n=4$. $E = \frac{240 \times 10^3}{4 \times 96500} \approx 1.24$ V....
Read Full Step-by-Step Solution →Standard EMF: E°cell = 0.34 − (‑0.76) = 1.10 V. Reaction quotient Q = [Zn²⁺]/[Cu²⁺] = 1/0.01 = 100. Nernst correction: (RT/nF) ln Q = (0.025693 / 2) ×...
Read Full Step-by-Step Solution →$\text{Na}^+ + e^- \to \text{Na}$. 23 g = 1 mol Na → requires 1 mol $e^-$....
Read Full Step-by-Step Solution →The EMF of a concentration cell is given by \(E = \frac{RT}{nF}\ln\frac{c_{high}}{c_{low}}\). At 298 K, \(\frac{RT}{F}=0.025693\,\text{V}\); for Cu²⁺,...
Read Full Step-by-Step Solution →Molar conductance = (specific conductance × 1000) / concentration. Substitute values directly....
Read Full Step-by-Step Solution →$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.34 - (-0.76)$...
Read Full Step-by-Step Solution →Using Nernst: $E = 1.10 - \frac{0.059}{2} \log\frac{0.1}{0.01} = 1.10 - 0.0295 = 1.07\,\text{V}$...
Read Full Step-by-Step Solution →EMF = E$_{\text{cathode}}$ - E$_{\text{anode}}$ = 0.34 - (-0.76) = 1.10 V....
Read Full Step-by-Step Solution →$Cu^{2+}$ requires 2 moles of electrons. 1 Faraday deposits 1 Eq. 2F deposits 2 Eq = 63.5g....
Read Full Step-by-Step Solution →$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.34 - (-0.76) = 1.10\,\text{V}$....
Read Full Step-by-Step Solution →$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.34 - (-0.76) = 1.10\,V$...
Read Full Step-by-Step Solution →Using Nernst equation: $E = E^\circ - \frac{0.059}{2}\log\frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}$...
Read Full Step-by-Step Solution →For a 1:1 electrolyte, the equivalent conductance equals the molar conductivity because one mole provides one equivalent. Hence $\Lambda_{eq}=\Lambda_...
Read Full Step-by-Step Solution →Foundational check for Electrochemistry. Study the core principles carefully for competitive exams....
Read Full Step-by-Step Solution →This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Electrochemistry....
Read Full Step-by-Step Solution →Use Nernst eq: $E = 1.1 - \frac{0.059}{2}\log(10) = 1.1 - 0.0295 = 1.07$ V....
Read Full Step-by-Step Solution →The standard EMF of the cell is calculated as $ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} $. Here, Cu is the cathode a...
Read Full Step-by-Step Solution →In standard notation, left electrode is anode. Zn is on left, so it undergoes oxidation: $ \text{Zn} \to \text{Zn}^{2+} + 2e^- $. Cu is reduced at cat...
Read Full Step-by-Step Solution →Use $\Delta G = -nFE$. For water formation, $n=4$. $E = \frac{240 \times 10^3}{4 \times 96500} \approx 1.24$ V....
Read Full Step-by-Step Solution →Standard EMF: E°cell = 0.34 − (‑0.76) = 1.10 V. Reaction quotient Q = [Zn²⁺]/[Cu²⁺] = 1/0.01 = 100. Nernst correction: (RT/nF) ln Q = (0.025693 / 2) ×...
Read Full Step-by-Step Solution →$\text{Na}^+ + e^- \to \text{Na}$. 23 g = 1 mol Na → requires 1 mol $e^-$....
Read Full Step-by-Step Solution →The EMF of a concentration cell is given by \(E = \frac{RT}{nF}\ln\frac{c_{high}}{c_{low}}\). At 298 K, \(\frac{RT}{F}=0.025693\,\text{V}\); for Cu²⁺,...
Read Full Step-by-Step Solution →Molar conductance = (specific conductance × 1000) / concentration. Substitute values directly....
Read Full Step-by-Step Solution →$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.34 - (-0.76)$...
Read Full Step-by-Step Solution →Using Nernst: $E = 1.10 - \frac{0.059}{2} \log\frac{0.1}{0.01} = 1.10 - 0.0295 = 1.07\,\text{V}$...
Read Full Step-by-Step Solution →EMF = E$_{\text{cathode}}$ - E$_{\text{anode}}$ = 0.34 - (-0.76) = 1.10 V....
Read Full Step-by-Step Solution →$Cu^{2+}$ requires 2 moles of electrons. 1 Faraday deposits 1 Eq. 2F deposits 2 Eq = 63.5g....
Read Full Step-by-Step Solution →$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.34 - (-0.76) = 1.10\,\text{V}$....
Read Full Step-by-Step Solution →$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.34 - (-0.76) = 1.10\,V$...
Read Full Step-by-Step Solution →