Pb(aq)2+ + 2e- → Pb(s) Ero = - 0.13 V
Sc(aq)3+ + 3e- → Sc(s) Ero = - 2.02 V
a) Write a balanced chemical equation for a spontaneous reaction involving these half reactions.
b) Explain your rationale for the reaction going in this direction.
I don't know what Ero stands for. These are reductions that you've written. To make a spontaneous reaction you want Ecell to be + so you want to reverse the most negative reaction and add the other to it. See if you can do that. Remember you ant to have the electrons to balance so you will need 2*3e and 3*2e.
a) To write a balanced chemical equation for a spontaneous reaction involving these half reactions, we need to combine them in a way that the electrons cancel out. Firstly, let's write the oxidation and reduction half reactions:
Oxidation half-reaction:
Pb(aq)2+ + 2e- → Pb(s)
Reduction half-reaction:
2Sc(aq)3+ + 6e- → 2Sc(s)
To balance the electrons, we multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2:
3Pb(aq)2+ + 6e- → 3Pb(s)
4Sc(aq)3+ + 12e- → 4Sc(s)
Now, we can add the two half-reactions together:
3Pb(aq)2+ + 6e- + 4Sc(aq)3+ + 12e- → 3Pb(s) + 4Sc(s)
Simplifying the equation, we get the balanced chemical equation for the spontaneous reaction:
3Pb(aq)2+ + 4Sc(aq)3+ → 3Pb(s) + 4Sc(s)
b) The rationale for the reaction going in this direction can be determined by comparing the standard reduction potentials (Ero) of the half-reactions. The more positive the Ero value, the greater the tendency for a substance to be reduced. In this case, since the Ero value for the reduction of Sc(aq)3+ is more negative (-2.02 V) compared to the reduction of Pb(aq)2+ (-0.13 V), Sc(aq)3+ has a higher tendency to be reduced.
Therefore, to achieve a spontaneous reaction, where the electrons flow from anode to cathode, we need to reverse the reduction half-reaction of Sc(aq)3+ and multiply it by 3 and the oxidation half-reaction of Pb(aq)2+ and multiply it by 4, resulting in the balanced chemical equation mentioned above.