A galvanic cell has the following overall reaction: Zn(s) + 2Eu(NO3)3(aq) → Zn(NO3)2(aq) + 2Eu(NO3)2(aq) What is the half reaction that occurs at the CATHODE?

Answer 1

I don't know what 2Eu(NO₃)₃ is. Is it europium nitrate? If so, then we have the following:

Zn(s) ==> Zn²⁺(aq) ... oxidation half reaction

Eu²⁺ + 2e- ==> Eu(s) ... reduction half reaction

The Eº_red for Zn²⁺ + 2e- ==> Zn(s) is -0.76 V

The Eº_red for Eu²⁺ + 2e- ==> Eu(s) is -0.43 V (had a little trouble finding this value, so yours may differ)

This means that Eu will be the cathode and this is where reduction takes place.

So, the half reaction that occurs at the cathode is...

Eu²⁺ + 2e- ==> Eu(s)