Rebecca S.

asked • 07/22/21

How to balance the following redox reaction using half-reaction method

How to balance the following redox reaction using half-reaction method under basic conditions


Sn(OH)42- + MnO4- ---> MnO2 + Sn(OH)62-



thank u for taking the time to help me with this question

2 Answers By Expert Tutors

By:

Kristy L. answered • 07/22/21

Tutor
5 (4)

Undergraduate Student for Math and Science Tutoring

Kristy L.

Additionally, we can check our work by adding up the charges for both sides and seeing if they are equal. If they are equal, that means that we have balanced the redox reaction correctly. If not, this means that we have made an error in one of the steps above. For the reactant side, 3Sn(OH)42- has a charge of -6, 2MnO4- has a charge of -2, and 4H2O has a charge of 0. Adding those charges up, we get -6 + (-2) + 0 = -8 for the reactant (left) side. Now, for the product side, 3Sn(OH)62- has a charge of -6, 2MnO2 has a charge of 0, and 2OH- has a charge of -2. Adding the charges up, we have -6 + 0 + (-2) = -8 for the product (right) side. As we have a charge of -8 on the reactant (left) side and a charge of -8 on the product (right) side, we know that we have balanced the redox reaction under basic conditions correctly as -8 = -8.
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07/22/21

Martin P. answered • 07/22/21

Tutor
5 (5)

BA Biology, further advanced professional degrees-50 years teaching

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