+91-8076753736[email protected]
logo
logo
-->

Electrochemistry

Question
CBSEENCH12011094
Wired Faculty App

When 0.1 mol MnO42- is oxidised, the quantity of electricity required to completely oxidise MnO42- to MnO42- is

  • 95600C

  • 2 x 96500C

  • 9650 C

  • 96.50 C

Solution

C.

9650 C

stack MnO subscript 7 superscript 2 minus end superscript with plus 7 below space rightwards harpoon over leftwards harpoon with straight i space straight e to the power of minus space left square bracket space 1 equals space straight F right square bracket on top space stack MnO subscript 4 superscript minus with plus 6 below
As per the equation, for 1 mole of MnO42-, 1F of electricity is required. Thus, for 0.1 mole of MnO4-, 0.1 F of electricity is required.
Since, 1 F = 96500 C
0.1 F = 0.1 x  96500 C
= 9650 C
Hence, 9650 C of electricity is required to completely oxidise MnO42- to MnO4-

Some More Questions From Electrochemistry Chapter

Why does the conductivity of a solution decrease with dilution?

Suggest a way to determine the Λ°m value of water.

The molar conductivity of 0.025 mol L–1 methanoic acid is 46.1 S cm2 mol–1. Calculate its degree of dissociation and dissociation constant. Give λ°(H+) = 349.6 S cm2 mol–1 and λ° (HCOO– ) = 54.6 s cm2 mol–1.

If the current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons flow through the wire?

Suggest a list of metals that are extracted electrolytically.

Consider the reaction: Cr2O72– + 14H+ + 6e- → 2Cr3+ + 8H2O
What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr2O72-?

Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging.

Suggest two materials other than hydrogen that can be used as fuels in fuel cells.

Mock Test Series

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation