Just wanted to check this. Are the two half reactions for 2K+CaSO4 yields K2SO4+Ca
1K yields K^+1 + e^-
and Ca^+2 + e^-2 yields Ca?
Very good.
You can write --> or ==> to show the arrow and its faster than writing yields.
Yes, you have correctly written the two half-reactions for the reaction 2K + CaSO4 → K2SO4 + Ca. Here's the breakdown of each half-reaction:
1. K → K+ + e-
This half-reaction represents the oxidation of potassium (K) atom. The potassium atom loses one electron (e-) and is oxidized to form a potassium ion (K+).
2. Ca2+ + 2e- → Ca
This half-reaction represents the reduction of calcium ion (Ca2+). Two electrons (2e-) are added to the calcium ion, reducing it to neutral calcium (Ca) atom.
When you combine these two half-reactions, you'll notice that the electrons on each side cancel out, ensuring that the total number of electrons is conserved. This allows the potassium ions (K+) and calcium (Ca) to form the products potassium sulfate (K2SO4) and calcium (Ca), respectively.
Remember, when balancing redox reactions, it is important to ensure that both charge and mass are conserved. Balancing half-reactions and combining them allows you to balance the overall redox reaction.