Is it AuCl4^- mixed with Sn^2+?
As you have written it I don't know whether to start with AuCl4^- or AuCl^-.
A solution of AuCl- is mixed with a solution of Sn2+ under standard conditions. What is the equation? the answer is
2AuCl4- + 3Sn2+ -> 3Sn4+ + 2Au + 8Cl-
how do i get this? ITS desperate... plz help me! THANK_YOU SO MUCH!
4 answers
yes
Au has changed from +3 in AuCl4^- top zero in Au.
Sn has changed from +2 to +4.
I think the easy way is this.
Au^3+ + 3e ==> Au
Sn^2+ ==> Sn^4+ + 2e
----------------------
Multiply eqn 1 by 2 and eqn 2 by 3 and add.
2Au^3+ + 3Sn^2+ ==> 2Au + 3Sn^4+
Then add 8Cl^- to each side.
2AuCl4^- + 3Sn^2+ ==> 3Sn^4+ + 2Au + 8 Cl^-
Sn has changed from +2 to +4.
I think the easy way is this.
Au^3+ + 3e ==> Au
Sn^2+ ==> Sn^4+ + 2e
----------------------
Multiply eqn 1 by 2 and eqn 2 by 3 and add.
2Au^3+ + 3Sn^2+ ==> 2Au + 3Sn^4+
Then add 8Cl^- to each side.
2AuCl4^- + 3Sn^2+ ==> 3Sn^4+ + 2Au + 8 Cl^-
THX SO MUCH!