ew?assignmentProblemID-79380684 Chem 101 Spring 2017 1of4 nacide the ion can be

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ew?assignmentProblemID-79380684 Chem 101 Spring 2017 1of4 nacide the ion can be used react anumber of era ore suchreacson i Balancing Redox Equations: Harreaction Method since reacton takes place in solution H100) and H (aq) wil be involved the reaction Places forthese species are indicated by the blanks inthe in the equation electrons lost inthe following restatement be balanced such that the number of mehed arte oxidato number metod The metho balances electrons lostinte oidator hasreacion whee ederos gained in te reduction hafeeaction. Part A balanced equation above? Remember noude Hao() and H (aq)inthe appropriate blanks and H (aq) may be added to What are the coeffoients muss balance. Which bstances are used depends on madton conditions Enter the equation coefficients in order separated by commas (eg.2,2,1 6.4.3 Include coefficients of 1, as required, for grading purposes. 316313 MLAnssers Give up Review Part Incorrect Try Again: 3 attempts remaining Potassium pemangarale. KMno is a powerful oxiduring agent The products of a given redox reaction with the permanganate in bascsok with a sodum sumo ion depend on the ruaction sinceris reacton takes place in basic soluson, H400) and OH (aq) wil beshown in reaction places for these species are indicated by the blanks inee following restatement of the equation Mao (s) +so (aq) Part B in the blanks where appropriate Your answer should have six terms. in order separated by commas 2,2,1.....». Include coefficients of 1 as required, for grading purposes Enter the equation coemeients My Answers up Review part

Explanation / Answer

We have to balance the redox reaction here:-

IO2- + Sn2+ ---> I- + Sn4+

First we will split the unbalanced half reactions:-

IO2- ----> I-

Sn2+ ------> Sn4+

we will balance the charges in each half reaction:

Iodine goes from oxidation no. +3 (x+2(-2) = -1) to -1.So to balance,

IO2- + 4e- -----> I-

Similarly to balance Sn

Sn2+ ----> Sn4+  + 2e-

Now we will add H2O to balance the oxygen in iodine reaction.it needs to be balanced by adding 2 H2O molecules

IO2- + 4e- ----> I-  + 2H2O

Balance hydrogen by adding 4protons(4H+)

IO2- + 4e- + 4H+ -----> I- + 4H2O

Now the iodine reaction has 4 electrons and Sn reaction has 2 electrons so we will multiply the Sn reaction by 2 to scale the electrons.so,

2Sn2+ ---> 2Sn4+  + 4e-

Adding the reaction and cancelling the common term,we get

IO2-(aq) + 2Sn2+(aq) + 4H+(aq)

----> I-  + 2Sn4+  + 2H2O(l)

SO the coefficients will be,

1,2,4,1,2,2

Part B:

MnO2- + SO32------> MnO2 + SO42-

First,splitting the unbalanced half equations:

MnO2-  -----> MnO2

SO32- ------> SO42-

We will balance their charges:

Mn goes from O.N +7 (x +4(-2) = -1) to O.N +4 ( x +2(-2)=0)

So,

MnO4-  + 3e- ------> MnO2

S goes from O.N +4(x+3(-2)=-2) to O.N. +6(x+4(-2)=-2)

So,

SO32- -----> SO42- + 2e-

Now for reactions in a basic solution balance the charges so that both sides have the same total charge by adding OH- ion to the side deficient in negative charge:

MnO4- + 3e- ------> MnO2 + 4OH-

SO32- + 2OH- -------> SO42- + 2e-

Now balance oxygen by adding water molecule:

MnO4- + 3e- + 2H2O --------> MnO2 + 4OH-

SO32- + 2OH- -----> SO42- + 2e- + H2O

Now to scale the electrons we will multiply Mn equation by 2 and SO3 equation by 3 so we will get:

2MnO4- + 6e- + 4H2O  ----> 2MnO2 + 8OH-

3SO32- + 6OH- -----> 3SO42- + 6e- + 3H2O

Adding the reactions and cancelling the common part we get:

2MnO4-(aq) + 3SO32-(aq) + H2O(l)    ----->2MnO2(s) + 3SO42-(aq) + 2OH-(aq)

So coefficients are:

2,3,1,2,3,2

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