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Given below are two statements: 1text M aqueous solution of each of textCu(NO_3)_2, textAgNO_3, textHg_2(textNO_3)_2; textMg(NO_3)_2 are electrolysed using inert electrodes, Given: E_textAg^+/textAg^theta = 0.80textV , E_textHg_2^2+/textHg^theta = 0.79textV, E_textCu^2+/textCu^theta = 0.24textV and E_textMg^2+/textMg^theta = -2.37textV Statement (I): With increasing voltage, the sequence of deposition of metals on the cathode will be textAg, textHg and textCu Statement (II): Magnesium will not be deposited at cathode instead oxygen gas will be evolved at the cathode. In the light of the above statement, choose the most appropriate answer from the options given below [cite: 426, 427]

Solution & Explanation

### Related Formula textEase of discharge at Cathode propto textStandard Reduction Potential (E^0) ### Core Logic - At the cathode, the metal ion with the highest standard reduction potential (E^0) gets reduced and deposited first. Arranging the given potentials: E^0_textAg^+/textAg (0.80textV) > E^0_textHg_2^2+/textHg (0.79textV) > E^0_textCu^2+/textCu (0.24textV) Thus, deposition follows the order textAg ightarrow textHg ightarrow textCu as voltage is steadily increased, confirming Statement I. - For textMg^2+, its reduction potential is highly negative (-2.37text V), much lower than that of water (-0.83text V). Consequently, water undergoes reduction at the cathode instead of magnesium: 2textH_2textO + 2e^- ightarrow textH_2(g) + 2textOH^- This results in the evolution of **Hydrogen gas** at the cathode, not oxygen gas. Oxygen gas is evolved at the *anode* via water oxidation. Thus, Statement II is incorrect. [cite: 1042, 1044] ### Step 1: Conclusion Match Since Statement I is correct and Statement II is incorrect, we select option (2). ### Pattern Recognition Cathode vs Anode Gas Trap: During the aqueous electrolysis of highly reactive metals (Groups 1, 2, and textAl), textH_2 gas is always discharged at the cathode due to water's easier reduction profile. Oxygen gas (textO_2) is an anodic product generated by water oxidation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry

Reference Study Guides

Electrochemistry

More Electrochemistry Previous-Year Questions — Page 4

Q48 2025 Faraday's Laws of Electrolysis
Electrolysis of 600mathrm~mL aqueous solution of NaCl for 5mathrm\ min changes the mathrmpH of the solution to 12. The current in Amperes used for the given electrolysis is ______ (Nearest integer).
Numerical Answer. Answer: 2 to 2

Solution

### Related Formula Faraday's law of electrolysis equation: textMoles of electrons (equivalents) = fracI cdot tF Water ion product relation: textpH + textpOH = 14 ### Core Logic During the electrolysis of brine (NaCl(aq)), hydroxide ions (OH^-) are generated at the cathode: 2H_2O + 2e^- rightarrow H_2 + 2OH^- Given metrics: - Final textpH = 12 implies textpOH = 14 - 12 = 2 - [OH^-] = 10^-2mathrm\ M - textVolume = 600mathrm\ mL = 0.6mathrm\ L - textTime = 5mathrm\ min = 300mathrm\ s ### Step 1: Calculate Moles of Hydroxide Produced Find the absolute moles of OH^- ions generated: textMoles = textMolarity times textVolume (L) = 10^-2 times 0.6 = 6 times 10^-3text moles ### Step 2: Relate to Electrical Current Since 1 mole of electrons produces 1 mole of OH^-, the moles of charge equals 6 times 10^-3. Applying Faraday's equation: 6 times 10^-3 = fracI times 30096500 I = frac6 times 10^-3 times 96500300 = 1.93mathrm\ A Rounding to the nearest integer gives 2. ### Pattern Recognition Always convert a given textpH value into [OH^-] concentration when dealing with cathodic water reduction. Tracking the relationship where 1\ e^- equiv 1\ OH^- provides a direct shortcut to link textpH changes to current flow. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry
Q27 2025 Standard Reduction Potential and Oxidising Power
The standard reduction potential values of some of the p-block ions are given below. Predict the one with the strongest oxidising capacity.
  • A. mathrmE_mathrmSn^4+/mathrmSn^2+^ominus = +1.15mathrmV
  • B. mathrmE_mathrmTl^3+/mathrmTl^ominus = +1.26mathrmV
  • C. mathrmE_mathrmAl^3+/mathrmAl^ominus = -1.66mathrmV
  • D. mathrmE_mathrmPb^4+/mathrmPb^2+^ominus = +1.67mathrmV

Solution

### Related Formula textOxidising Capacity propto textStandard Reduction Potential (E^ominus) ### Core Logic A higher positive value of standard reduction potential (E^ominus) indicates a stronger tendency to undergo reduction, hence behaving as a stronger oxidising agent. Comparing the given values: * mathrmE_mathrmSn^4+/mathrmSn^2+^ominus = +1.15mathrmV * mathrmE_mathrmTl^3+/mathrmTl^ominus = +1.26mathrmV * mathrmE_mathrmAl^3+/mathrmAl^ominus = -1.66mathrmV * mathrmE_mathrmPb^4+/mathrmPb^2+^ominus = +1.67mathrmV Since +1.67mathrmV is the highest value, mathrmPb^4+ possesses the strongest oxidising capacity[cite: 745, 746]. ### Pattern Recognition Strongest oxidising agent = Most positive reduction potential. Weakest oxidising agent = Most negative reduction potential. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry Class 12 Chemistry: p-Block Elements
Q28 2025 Variation of Molar Conductivity with Concentration
The molar conductivity of a weak electrolyte when plotted against the square root of its concentration, which of the following is expected to be observed?
  • A. A small decrease in molar conductivity is observed at infinite dilution.
  • B. A small increase in molar conductivity is observed at infinite dilution.
  • C. Molar conductivity increases sharply with increase in concentration.
  • D. Molar conductivity decreases sharply with increase in concentration.

Solution

### Related Formula For weak electrolytes, the degree of dissociation alpha increases sharply near infinite dilution according to Ostwald's Dilution Law: alpha = sqrtfracK_aC ### Core Logic When a weak electrolyte is diluted (concentration C ightarrow 0), its molar conductivity increases steeply. Conversely, when plotted against sqrtC, as concentration increases, the degree of dissociation drops rapidly, causing a sharp decrease in molar conductivity[cite: 188, 204]. This matches the curve given below:
Variation of Molar Conductivity with Concentration diagram for Q28 - JEE Main 2025 Morning
Variation of Molar Conductivity with Concentration diagram for Q28 - JEE Main 2025 Morning
### Pattern Recognition Weak electrolyte plots feature a steep asymptotic exponential-like rise towards the y-axis as C ightarrow 0, meaning a sharp decrease occurs with increasing concentration[cite: 204, 752]. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry
Q35 2025 Nernst Equation
For a mathrmMg mid mathrmMg^2+ (aq) parallel mathrmAg^+(mathrmaq) mid mathrmAg the correct Nernst Equation is :
  • A. mathrmE_cell = E_cell^o - fracRT2Flnfrac[Ag^+][Mg^2 + ]
  • B. mathrmE_mathrmcell = mathrmE_mathrmcell^circ + fracmathrmRT2 mathrm~F ln frac[mathrmAg^+]^2[mathrmMg^2+]
  • C. mathrmE_cell = E_cell^o - fracRT2Flnfrac[Mg^2 + ][Ag^+]
  • D. mathrmE_cell = E_cell^o - fracRT2Flnfrac[Ag^+]^2[Mg^2 + ]

Solution

### Related Formula E_textcell = E_textcell^circ - fracRTnF ln Q ### Core Logic Let us explicitly formulate the complete chemical oxidation-reduction equations : Anode oxidation: mathrmMg_(s) ightarrow mathrmMg^2+_(aq) + 2e^- Cathode reduction: 2mathrmAg^+_(aq) + 2e^- ightarrow 2mathrmAg_(s) Net total equation : mathrmMg(s) + 2mathrmAg^+(aq) ightleftharpoons mathrmMg^2+(aq) + 2mathrmAg(s) Total transferred moles of electrons n = 2 [cite: 838, 841]. Reaction quotient : Q = frac[mathrmMg^2+][mathrmAg^+]^2 Substituting into Nernst form : E_textcell = E_textcell^circ - fracRT2F lnleft( frac[mathrmMg^2+][mathrmAg^+]^2 ight) Inverting the inside quotient changes the sign of the logarithm term from negative to positive: E_textcell = E_textcell^circ + fracRT2F lnleft( frac[mathrmAg^+]^2[mathrmMg^2+] ight) ### Pattern Recognition A standard negative logarithmic quotient can always toggle into an addition configuration by inverting the products/reactants variables concentration ratio. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry

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