Answer:The standard cell potential, E˚cell =2.234 V, Pb metal is the cathode.
Explanation:
The Half cell reactions are
Pb2+(aq) + 2e– > Pb(s) ------E˚ = –0.136 V
Mg2+(aq) + 2e–>Mg(s)------- E˚ = –2.37 V
In a voltaic cell, reduction occurs at the cathode and oxidation occurs at the anode
Mg(s)---->Mg2+(aq) + 2e– ( anodic oxidation)
Pb2+(aq) + 2e----> Pb(s) ( Cathodic reduction)
We can see that Mg has a more negative reduction potential value leading to having a low reduction potential and therefore will occur at the anode since it will be oxidized. On the other hand Pb2+ has a less negative reduction potential and therefore will have a high reduction making it to occur at the cathode and be reduced
Using The standard cell potential, E˚cell = E˚cathode – E˚anode
E˚cell = –0.136 - (–2.37)
E˚cell=2.234 V
