Balance the redox equation. It occurs in acid solution.
Sn(s) + H+(aq) = Sn2+(aq) + H2(g)
I understand that first the equation needs to be put into halves.
Sn(s) = Sn2+(aq) and H+(aq) = H2(g)
Then,
Sn(s) = Sn2+(aq) + 2e- and ?
From here I am lost. Please help.
Sn ==> Sn^2+ + 2e
2H^+ + 2e ==> H2
--------------------add 1 to 2.
Sn + 2H^+ + 2e ==> Sn^2+ + 2e + H2
Then cancel the two e on each side and you're left with the balanced equation.
Oh, let's get this redox equation balanced and have some fun while we're at it! First, we need to balance the electrons. We have Sn(s) on the left side, which means it's reduced, gaining electrons to become Sn2+(aq). So, we add 2e- to the right side to balance the charge. Now, let's focus on the hydrogen. We have H+(aq) on the left side, which means it's being consumed. And where does it go? It becomes H2(g) on the right side! So, we need to add the same number of hydrogen ions on the right side to make them equal. That means adding 2H+ to the right side. And voila, our balanced equation is:
Sn(s) + 2H+(aq) = Sn2+(aq) + H2(g)
Easy peasy, lemon squeezy! Keep in mind, my friend, that this balancing act is all about conserving charge and making sure each element is accounted for.
To balance a redox equation, you need to ensure that the number of electrons lost in the oxidation half-reaction is equal to the number of electrons gained in the reduction half-reaction.
Let's start with the oxidation half-reaction:
Sn(s) → Sn2+(aq) + 2e-
Now, let's balance the reduction half-reaction:
H+(aq) + 2e- → H2(g)
To balance the hydrogen atoms, we add two H+ ions to the right side of the equation:
H+(aq) + 2e- → H2(g) + 2H+(aq)
Now, to balance the charge on both sides, we need to multiply the oxidation half-reaction by 2, and the reduction half-reaction by 1:
2Sn(s) → 2Sn2+(aq) + 4e-
H+(aq) + 2e- → H2(g) + 2H+(aq)
Finally, combine both half-reactions:
2Sn(s) + 2H+(aq) → 2Sn2+(aq) + H2(g)
Now, the redox equation is balanced in acid solution.
To balance the redox equation, you need to ensure that the number of electrons lost (oxidation) is equal to the number of electrons gained (reduction).
In the given equation:
Sn(s) + H+(aq) = Sn2+(aq) + H2(g)
Start by writing the oxidation half-reaction and the reduction half-reaction:
Oxidation half-reaction: Sn(s) → Sn2+(aq) + 2e-
Now, balance the number of atoms on each side of the equation. In this case, since tin (Sn) is already balanced, you only need to balance the charge. To balance the charge on the left-hand side (LHS) of the equation, add 2H+ ions to both sides of the equation:
Sn(s) + 2H+(aq) → Sn2+(aq) + 2e-
Now, let's write the reduction half-reaction:
Reduction half-reaction: 2H+(aq) + 2e- → H2(g)
To balance the reduction half-reaction, start by balancing the hydrogen (H) atoms. In this case, there are two hydrogen atoms on the left-hand side (LHS) and only two hydrogen atoms on the right-hand side (RHS) of the equation, so hydrogen is already balanced.
Finally, balance the charge by adding two electrons (2e-) to the left-hand side (LHS) of the reduction half-reaction:
2H+(aq) + 2e- → H2(g)
Now that both half-reactions are balanced, multiply each half-reaction by a factor such that the total number of electrons lost in the oxidation half-reaction is equal to the number of electrons gained in the reduction half-reaction.
Multiply the oxidation half-reaction by 2:
2Sn(s) + 4H+(aq) → 2Sn2+(aq) + 4e-
Multiply the reduction half-reaction by 4:
4H+(aq) + 4e- → 2H2(g)
Now that both half-reactions have the same number of electrons, add the two half-reactions together:
2Sn(s) + 4H+(aq) + 4H+(aq) + 4e- → 2Sn2+(aq) + 4e- + 2H2(g)
Simplifying the equation:
2Sn(s) + 8H+(aq) → 2Sn2+(aq) + 2H2(g)
Therefore, the balanced redox equation in acid solution for the oxidation of Sn(s) to Sn2+(aq) and the reduction of H+(aq) to H2(g) is:
Sn(s) + 4H+(aq) → Sn2+(aq) + 2H2(g)