Exams › JEE Main › Chemistry

Using the data E°(MnO4⁻ / Mn²+) = 1.51 V and E°(MnO2 / Mn²+) = 1.23 V, determine the standard electrode potential E° for the couple MnO4⁻ / MnO2.

1. 1.70 volt
2. 1.51 volt
3. 1.23 volt
4. 1.00 volt

Correct answer: 1.70 volt

Solution

The MnO4⁻/MnO2 couple is a 3-electron change obtained by subtracting the 2-electron MnO2/Mn2+ reduction from the 5-electron MnO4⁻/Mn2+ reduction. Because potentials cannot be added directly, the free-energy combination gives E° = (5*1.51 - 2*1.23)/3 = 1.70 V.

Related JEE Main Chemistry questions

• During electrolysis of an aqueous solution of dipotassium succinate, which gas is evolved at the electrodes?
• Which expression gives the equivalent conductance at infinite dilution of Al2(SO4)3, if Λ°Al3+ and Λ°SO4²− denote the equivalent conductances at infinite dilution of the corresponding ions?
• The standard emf of a Daniell cell is 1.10 V. What is the greatest amount of electrical work that can be extracted from this cell?
• During the recharging of a lead storage battery, which reaction takes place at the cathode?
• The limiting molar ionic conductivities of the ions X2+ and Y2− are 57 and 73, respectively. What will be the molar conductivity of the electrolyte formed by these ions?
• According to Kohlrausch’s law, at what condition does each ion contribute a fixed amount to the conductance of an electrolyte, irrespective of the other ion present?

⚔️ Practice JEE Main Chemistry free + battle 1v1 →