The following reaction occurs spontaneously. 2H+(aq)+ca(s) Ca2+(aq)+H2(g)

write the balanced reduction half-reaction

Don't you realize that we can only guess when you don't include an arrow? We can only guess as to the reactants versus the products. I'm guessing you made two mistakes. First ca(s) PROBABLY is Ca(s) and the arrow PROBABLY goes like this.

2H^+(aq) + Ca(s) ==> Ca^+2(aq) + H2(g).
Do you know the definitions of oxidation and reduction?
Reduction is the gain of electrons. Oxidation is the loss of electrons. First we pick out what is what.
H goes from an oxidation state of +1 on the left to 0 on the right. That's a gain of electrons.
Ca goes from an oxidation state of 0 on the left to +2 on the right. That's a loss of electrons.
So Ca is oxidized and H^+ is reduced.
Reduction half.
2H^+(aq) + 2e ==> H2(g)
Oxidation half.
Ca(s) ==> Ca^+2(aq) + 2e

Hey man just wondering is the reductant in the equation considered 2H^+ or simply H^+

simply H+

Well, this seems like a pretty electrifying reaction! Let's break it down and figure out the reduction half-reaction.

In the reaction, the calcium (Ca) metal is being oxidized (losing electrons) to form calcium ions (Ca2+), while the hydrogen ions (H+) are being reduced (gaining electrons) to form hydrogen gas (H2).

To write the reduction half-reaction, we need to show the species involved and the electrons gained. In this case, the reduction half-reaction can be written as:

2H+(aq) + 2e- → H2(g)

Now, remember to always keep an electrolyte balance, kids!

To write the balanced reduction half-reaction, we need to identify the species that is being reduced and write the associated half-reaction.

In this reaction, the species being reduced is H+(aq). To determine the reduction half-reaction, we need to consider the change in oxidation state of hydrogen.

Since H+ has an oxidation state of +1, and the product H2(g) has an oxidation state of 0, hydrogen is going from an oxidation state of +1 to 0. Therefore, it undergoes a reduction.

The reduction half-reaction can be written as follows:
2H+(aq) + 2e- → H2(g)

The number of electrons in the reduction half-reaction is determined by balancing the charges on both sides of the equation. In this case, each hydrogen ion (H+) gains two electrons (2e-) to form one molecule of hydrogen gas (H2).