balance the following ionic equation:

Question

Sn(2+)(aq)+ Mn(2+)(aq)----- Mn(5+)(aq)+ Sn (s)

Answer 1

First separate the reaction into reduction and oxidation reactions. Then add the right number of electrons to balance the charges.
Reduction:
Mn(2+) -> Mn(5+)
Mn(2+) -> Mn(5+) + 3e-
Oxidation:
Sn(2+) -> Sn
Sn(2+) + 2e- -> Sn
And then balance both reactions such that the number of electrons will be equal. Then add:
2 * ( Mn(2+) -> Mn(5+) + 3e- )
3 * ( Sn(2+) + 2e- -> Sn )
----------------------------------------
2 Mn(2+) + 3 Sn(2+) -> 2 Mn(5+) + 3 Sn

Hope this helps :3