# 2mno4(-)+10fe(2+)+16H(+)=... +5fe(3+)+8h20?

he molecular equation for it is 2kmn4+8h2so4+10fes04=>k2so4+2m...

which is obtained by adding these two

2kmno4+3h2so4=>k2so4+2mnso4+3h...

10fes04+5h2so4+5(o)=>5fe3(so4)...

so here it is shown that first 3hso and in second equation 5 h2o r obtained but in ionic equation it is clear that all the 16 H+ ions join with all the 04 of 2mno4 but in mol.equation 5h2o r obtained by rection of nascent equation with h2so4

thats a tough one

Your query is justified.

This ionic equation represents a redox reaction.

Reduction: MnO4- ------> Mn2+

Since, the formal charge of Mn changes from 7 to 2. MnO4- must gain 5 electrons to balance out. In other words, reduction takes place.

MnO4- + 5e- -----> Mn2+

Now, we need to balance the no. of oxygen atoms;

MnO4- + 5e- -----> Mn2+ + 4H20

Now, oxygen is balanced, we need to balance hydrogen.

Since, we're using H2SO4, we'll consider acidic medium(H+).

MnO4- + 5e- + 8H+ -----> Mn2+ + 4H20..(a)

Oxidation: Fe2+ -----> Fe3+

Since, the formal charge changes from +2 to +3, it means in order to balance, Fe3+ must lose 1 electron. In other words, oxidation takes place.

Fe2+------> Fe3+ + e-

Now, we just need to balance the no. of e- in (a). So, we multiply both sides by 5 and we get;

5Fe2+------>5Fe3+ + 5e-...(b)

Now, add (a) and (b). Cancel out that is equal and common just like you do in linear equations.

5Fe2+ + MnO4- + 5e- + 8H+--->Mn2+ + 4H20 + 5Fe3+ + 5e-

5Fe2+ + MnO4- + 8H+-----> Mn2+ + 4H2O + 5Fe3+

This is the actual equation. If, you multiply both sides by 2, you'll get the equation that has put you in doubt.

10Fe2+ + 2MnO4- + 16H+ ---> 2Mn2+ + 8H2O + 10Fe3+

So, you see H+ and MnO4- have given rise to H2O as in equation (a).

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which is obtained by adding these two

2kmno4+3h2so4=>k2so4+2mnso4+3h...

10fes04+5h2so4+5(o)=>5fe3(so4)...

so here it is shown that first 3hso and in second equation 5 h2o r obtained but in ionic equation it is clear that all the 16 H+ ions join with all the 04 of 2mno4 but in mol.equation 5h2o r obtained by rection of nascent equation with h2so4

**Answer:**thats a tough one

Your query is justified.

This ionic equation represents a redox reaction.

Reduction: MnO4- ------> Mn2+

Since, the formal charge of Mn changes from 7 to 2. MnO4- must gain 5 electrons to balance out. In other words, reduction takes place.

MnO4- + 5e- -----> Mn2+

Now, we need to balance the no. of oxygen atoms;

MnO4- + 5e- -----> Mn2+ + 4H20

Now, oxygen is balanced, we need to balance hydrogen.

Since, we're using H2SO4, we'll consider acidic medium(H+).

MnO4- + 5e- + 8H+ -----> Mn2+ + 4H20..(a)

Oxidation: Fe2+ -----> Fe3+

Since, the formal charge changes from +2 to +3, it means in order to balance, Fe3+ must lose 1 electron. In other words, oxidation takes place.

Fe2+------> Fe3+ + e-

Now, we just need to balance the no. of e- in (a). So, we multiply both sides by 5 and we get;

5Fe2+------>5Fe3+ + 5e-...(b)

Now, add (a) and (b). Cancel out that is equal and common just like you do in linear equations.

5Fe2+ + MnO4- + 5e- + 8H+--->Mn2+ + 4H20 + 5Fe3+ + 5e-

5Fe2+ + MnO4- + 8H+-----> Mn2+ + 4H2O + 5Fe3+

This is the actual equation. If, you multiply both sides by 2, you'll get the equation that has put you in doubt.

10Fe2+ + 2MnO4- + 16H+ ---> 2Mn2+ + 8H2O + 10Fe3+

So, you see H+ and MnO4- have given rise to H2O as in equation (a).

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