# Write a balanced net ionic equation for the following reaction in a basic solution?

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Bi(OH)3 (s) + Sn(OH)3^- (aq) —> Bi (s) + Sn(OH)6^2-(aq)

I’m pretty confused by this one. Any help/an explanation would be appreciated.

I think I need to eliminate the spectator ions but I’m not sure which ones those are. 🙁

• if you have a compound that does not dissociate completely, you cannot consider that to be spectator ions.

So.. none of the Bi(OH)3(s), because it’s a solid, ends up as spectator ions

and NONE of the Sn(OH)3- is spectator ions either.. It’s not dissociated is it? and 1 Sn(OH)3 –> 1 Sn(OH)6

so there are NO spectator ions to consider

*********

now since 1 Bi +3 —> 1 Bi(0).. it’s easy to see we have a redox reaction.. so let’s balance this like a redox reaction…

1) identify the oxidation states

2) identify which species was oxidized and which was reduced

3) write half reactions including electrons

4) balance electrons in half reactions..

5) combine half reactions and cancel electrons

like this…

*** 1 ***

Bi in Bi(OH)3(s) is +3

Sn in Sn(OH)3(-1) is +2

Bi in Bi(s) is 0

Sn in Sn(OH)6(-2) is +4

All the O is -2 and all the H is +1

*** 2 ***

Bi has gone from +3 to 0 and was reduced

Sn has gone from +2 to +4 and was oxidized

*** 3 ***

Bi(+3) + 3 e’s —> Bi(s)

Sn(+2) —> Sn(+4) + 2 e’s

*** 4 ***

multiply the first by 2 and the 2nd by 3 to balance e’s

2 Bi(+3) + 6 e’s —> 2 Bi(s)

3 Sn(+2) —> 3 Sn(+4) + 6 e’s

*** 5 ***

2 Bi(+3) + 6 e’s + 3 Sn(+2) —> 2 Bi(s) + 3 Sn(+4) + 6 e’s

canceling e’s

2 Bi(+3) + 3 Sn(+2) —> 2 Bi(s) + 3 Sn(+4)

*** 6 ***

2 Bi(OH)3 + 3 Sn(OH)3(-1) —> 2 Bi(s) + 3 Sn(OH)6(-2)

*** 7 ***

2 Bi(OH)3 + 3 Sn(OH)3(-1) + __ OH- —> 2 Bi(s) + 3 Sn(OH)6(-2)

balancing the OH’s.. 18 on the right.. needs 18 on the left…

2 Bi(OH)3 + 3 Sn(OH)3(-1) + 3 OH- —> 2 Bi(s) + 3 Sn(OH)6(-2)

************

and again.. no net ions to drop.. and you don’t write Bi+3 and OH- seperate because it’s a solid.. so this is your final equation

2 Bi(OH)3(s) + 3 Sn(OH)3(-1)(aq) + 3 OH-(aq) —> 2 Bi(s) + 3 Sn(OH)6(-2)(aq)

• What complicates things are the complex ions and the fact this is an oxidation/reduction reaction. I suppose a good start would be ignoring the fact Bi(OH)3 is solid and assuming it forms ions:

Bi(OH)3 -> Bi(+3) + 3OH(-)

this makes the equation:

BI(+3) + 3OH(-) + Sn(OH)3(-) -> Bi + Sn(OH)6(-2)

Oxidation/reduction reactions can be written as a pair of half reactions:

Bi(+3) + 3e(-) -> Bi (reduction reaction)

Sn(OH)3(-) + 3OH(-) -> Sn(OH)6(-2) + 2e(-) (oxidation reaction)

However, the two half reactions don’t balance in terms of electrons, so multiply them by a constant:

2Bi(+3) + 6e(-) -> 2Bi

3Sn(OH)3(-) + 9OH(-) -> 3Sn(OH)6(-2) + 6e(-)

This gives:

2Bi(+3) + 3Sn(OH)3(-) + 9OH(-) -> 2Bi + 3Sn(OH)6(-2)

Frankly, the only spectators are the electrons themselves, and these are not ions. This is another example of poorly taught chemistry because:

1.) It did not include a complete equation (the number of hydroxides was omitted)

2.) it asked for spectator ions. There weren’t any!

Source(s): I’m a chemist!

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