For the cell diagram
Pt(s)|H2(g)|H+(aq)||Cu2+(aq)|Cu(s)
Write a balanced equation for the half-reaction at the cathode.
The cathode is the site of reduction so copper gets reduced.
So wouldn't the answer be:
Cu2+(aq) -> Cu(s) + 2e-?
the cathode is the site of reduction..but you should check your equation: Note that reduction can be characterized by the gaining of electrons.
Yes, you are correct. The copper ions (Cu2+) in the solution are reduced to copper metal (Cu) at the cathode. The balanced equation for this half-reaction is:
Cu2+(aq) + 2e- -> Cu(s)
To determine the half-reaction at the cathode, you need to identify the species being reduced. In this case, the copper ions are being reduced to copper metal as you correctly stated. The coefficient of 2 in front of the electrons (e-) balances the charge on both sides of the equation, ensuring conservation of charge.
Remember that in electrode notation, the reactants are on the left side of the arrow, and the products are on the right side.
Yes, you are correct. The cathode is the site of reduction, and in this cell diagram, copper (Cu) is being reduced from Cu2+ to Cu.
The balanced equation for the half-reaction at the cathode is:
Cu2+(aq) + 2e- -> Cu(s)