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!

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
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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^-

Is it AuCl4^- mixed with Sn^2+?

As you have written it I don't know whether to start with AuCl4^- or AuCl^-.

yes

THX SO MUCH!

To determine the equation for the reaction between AuCl- and Sn2+, you need to consider the oxidation states and balance the reaction using the method of balancing redox reactions.

1. Write the oxidation half-reaction:
Start by identifying which element is being oxidized and which is being reduced. In this case, Sn2+ is being oxidized to Sn4+. The oxidation half-reaction is:
Sn2+ -> Sn4+

2. Balance the number of atoms:
Since Sn2+ is being oxidized, it increases in oxidation state from +2 to +4. To balance the number of atoms, you need to multiply the half-reaction by 2:
2Sn2+ -> 2Sn4+

3. Write the reduction half-reaction:
Next, identify which element is being reduced. In this case, AuCl- is being reduced to Au. The reduction half-reaction is:
AuCl- -> Au

4. Balance the number of atoms:
To balance the number of atoms, you need to add enough chlorine ions (Cl-) to the reactant side:
AuCl- + 3Cl- -> Au

5. Balance the charges:
Now, both half-reactions have balanced atoms, but the charges are not balanced. The oxidation half-reaction has a charge of +2 on the left, while the reduction half-reaction has a charge of -1 on the left.
To balance the charges, multiply the reduction half-reaction by 2:
2AuCl- + 6Cl- -> 2Au

6. Combine the half-reactions:
Now that both half-reactions are balanced, you can combine them to form the overall balanced equation:
2Sn2+ + 2AuCl- + 6Cl- -> 2Sn4+ + 2Au + 6Cl-

As there are 3 moles of Sn2+ present in the reaction, you can multiply both sides of the equation by 3:
2Sn2+ + 2AuCl- + 6Cl- -> 2Sn4+ + 2Au + 6Cl-
3(2Sn2+ + 2AuCl- + 6Cl-) -> 3(2Sn4+ + 2Au + 6Cl-)
6Sn2+ + 6AuCl- + 18Cl- -> 6Sn4+ + 6Au + 18Cl-
Simplifying this equation gives:
2AuCl4- + 3Sn2+ -> 3Sn4+ + 2Au + 8Cl-

Therefore, the balanced equation is:
2AuCl4- + 3Sn2+ -> 3Sn4+ + 2Au + 8Cl-