Li^+(aq) + Zn(s) → Li(s) + Zn^2+(aq)
How many electrons are transferred in the balanced equation?
I came up with an answer of 2, is this correct?
yes
Yes, that is correct. In the balanced equation:
Li^+(aq) + Zn(s) → Li(s) + Zn^2+(aq)
One lithium ion (Li^+) is reduced to lithium (Li), which involves gaining one electron. On the other hand, one zinc atom (Zn) is oxidized to a zinc ion (Zn^2+), which involves losing two electrons.
Since two electrons are transferred, your answer of 2 is correct.
To determine the number of electrons transferred in the balanced equation, we need to look at the oxidation state of each element involved in the reaction.
In the given equation:
Li^+(aq) + Zn(s) → Li(s) + Zn^2+(aq)
Let's assign the oxidation states to each element:
Li^+: +1
Zn: 0
Li: 0
Zn^2+: +2
From the oxidation states, we can see that Li is being reduced (going from an oxidation state of +1 to 0), and Zn is being oxidized (going from an oxidation state of 0 to +2).
The number of electrons involved in the redox reaction is equal to the difference in oxidation numbers. In this case, we have:
(+2) - (+1) = +1
Therefore, 1 electron is transferred in this balanced equation, not 2.
So, the correct answer is 1 electron.