Moderate

8H++ MnO4- → Mn2+ + 4H2O , which one is correct about given equation?

Correct answer: B. 5e- on L.H.S

  • A. 5e- on R.H.S
  • B. 5e- on L.H.S
  • C. 3e- on R.H.S
  • D. 3e- on L.H.S

Explanation

Option B is correct.The balanced equation for the reaction is: MnO4- + 8H+ + 5e- → Mn2+ + 4H2O The oxidation state of manganese changes from +7 in MnO4- to +2 in Mn2+. This means that manganese has lost 5 electrons in the reaction. The number of electrons transferred can be calculated by subtracting the oxidation state of manganese in the reactant from the oxidation state of manganese in the product. In this case, the number of electrons transferred is 7 - 2 = 5. The addition of 5 electrons on the left-hand side of the equation represents the reduction of manganese. Therefore, the answer is 5e- in L.H.S.

Last updated

About Electrochemistry

Electrochemistry describes electron transfer through oxidation and reduction, identifies oxidizing and reducing agents, and applies oxidation numbers to balance redox equations. It also covers electrode potential and the standard hydrogen electrode, which provides the reference for comparing the reduction tendencies of other electrodes.

Practise Electrochemistry

696 free Electrochemistry MCQs from Chemistry, each with the correct answer and an explanation. Unlimited attempts, no account needed.

Exams that ask Chemistry questions like this

Chemistry is on 12 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.

Related questions