Question

the redox reaction given below occurs in basic solution.\\(\ce{tco^-_{4} + sn^{+2} -> tc^{+4} + sn^{+4}}\\)\
balance the half-reaction:\\(\ce{sn^{+2} -> sn^{+4}}\\)\
how many electrons are transferred?\\(\ce{?e^-}\\)

Explanation:

Step1: Determine oxidation state change

The oxidation state of Sn in \( \text{Sn}^{+2} \) is +2, and in \( \text{Sn}^{+4} \) is +4. The change in oxidation state is \( 4 - 2 = +2 \).

Step2: Relate to electron transfer

In oxidation (since Sn is losing electrons, going from +2 to +4), the number of electrons lost is equal to the change in oxidation state. For each Sn atom, the number of electrons transferred is \( 2 \) (because the oxidation state increases by 2, meaning 2 electrons are lost).

Answer:

2