Question

What is the number of electrons transferred from the reducing agent to the oxidizing agent for Bi(OH)3 + SnO22- --> Bi + SnO32- and Sn(OH)62- + Si ---> HSnO2- + SiO32

Answer 1

Just check the change in oxidation number for each:
For the first equation: Sn2+ becomes Sn4+ which requires 2 electrons
And Bi3+ becomes Bi+ which again requires 2 electrons .
Hence in the first equation 2 electrons are transferred.

Answer 2

So would the second equation also be 2 because Si went from 0 to 2?

Answer 3

Is that SiO3 2-?

Answer 4

Yes , but a lone element always has the oxidation number of 0 , and to balance out oxygen's -2 m it would have to be +2 , right?

Answer 5

If it is then Si is going from O to 4+ not 2+.
So 4e- are needed.
One more thing:
If you have like Mn7+ becoming Mn2+ and simultaneously fe2+ becoming Fe3+ then you may think that since Fe2+ became Fe3+ only one electron is transferred but it is not so. 5 electrons are transfrerred. You have to write both oxidation and reduction equations and then take LCM.
Like in the second question ,
Si -> Si4+ and Sn4+ becomes Sn3+.
In silicon reaction 4 electrons are needed while in Tin reaction only 1 is needed So you take LCM of 1 and 4 which is 4 itself so 4 electrons would be transferred.

Answer 6

Si in elemental state is 0. But in SiO3 2- you have 3 oxygen so if Si has x oxidation state the x -3*2=-2 you get x=4.