# Redox reaction with tin and sulfuric acid solution

I need to write the redox reaction from tin and a sulfuric acid solution. Here is the solution given:

The first half-reaction: $\ce{Sn->Sn^2+\+2e^-}$
The second half-reaction: $\ce{2H+\+2e^-->H2}$
Adding both reactions will give you: $\ce{Sn\+2H+->Sn^2+\+H2}$

Why is the sulfuric acid solution replaced with $\ce{H^+}$?

The sulfuric acid dissociates in water: $$\ce{H2SO4 -> H+ + HSO4-}$$$$\ce{HSO4- -> H+ + SO4^{2-}}$$
In this case it is the hydrogen ions which are oxidising the tin and the sulfate and hydrogensulfate ions play no part in the reaction so they are left out. Including them in the hydrogen half equation would give (with sulfate ions): $$\ce{2H+ + SO4^{2-} +2e- -> H2 + SO4^{2-}}$$
I doubt the answer because I'm sure you'll get $\ce{Sn_{(s)} + 2H2SO4_{(aq)} → SnSO4_{(aq)} + 2H2O_{(l)} + SO2_{(g)}}$ as your Products
anyways , The Arrhenius concept of acids and bases is as follows: an acid is a substance that when dissolved in water increases the concentration of the hydrogen ion, $$\ce{H+}$$ When Sulphuric acid is dissolved in water it gives you $$\ce{H+}$$and $$\ce{SO4^{2-}}$$ ions. Sulphuric acid is a VERY Strong Oxidising agent, that means that it will oxidise your Tin therefore we get $\text{Sn}\rightarrow \text{Sn}^{2+}+2\text{ e}^-$ in the second part I think you have reacted your 2 electrons formed in the previous reaction with Hydrogen ion, But I doubt this part