Write a balanced overall reaction from these unbalanced half-reactions.

Sn->Sn2+
Ag+->Ag
Balanced overall equation

I got Sn+2Ag+->Sn+2Ag, but its wrong.

To write a balanced overall reaction from the given unbalanced half-reactions, we need to ensure that the number of electrons gained in one half-reaction is equal to the number of electrons lost in the other half-reaction.

Given half-reactions:
Sn -> Sn2+
Ag+ -> Ag

To balance the number of electrons in the half-reactions, we can multiply the first half-reaction by two:

2Sn -> Sn2+ + 2e-

Now, we have two electrons gained in the first half-reaction and one electron lost in the second half-reaction. To make the number of electrons equal, we need to multiply the second half-reaction by two:

2Ag+ -> 2Ag + 2e-

Now, the number of electrons gained and lost are balanced. The overall balanced equation is:

2Sn + 2Ag+ -> Sn2+ + 2Ag

To balance the overall reaction, we need to ensure that the number of atoms and charges are equal on both sides of the equation. Let's balance the given half-reactions step by step:

Unbalanced Half-Reactions:
1) Sn -> Sn2+
2) Ag+ -> Ag

To balance the oxidation half-reaction (Sn -> Sn2+), we need to add two electrons to the left side:

1) Sn -> Sn2+ + 2e-

Now, let's balance the reduction half-reaction (Ag+ -> Ag). Since Ag does not change its charge, there is no need to add or subtract electrons. Thus, the half-reaction is already balanced.

Now, we need to manipulate the half-reactions to ensure that the number of electrons involved is the same in both reactions. Multiplying the oxidation half-reaction by two would give us an even number of electrons on both sides:

2Sn -> 2Sn2+ + 4e-
2Ag+ -> 2Ag

Finally, combining the half-reactions will give us the balanced overall equation:

2Sn + 4Ag+ -> 2Sn2+ + 4Ag

I think you just made a a couple of typos. You left no space between Sn+2Ag and you omitted the 2Ag on the right.

Sn + 2Ag ==> An^2+ + 2Ag