Does Pt(s)+NaI(aq) react

Question

aaronq · Answer

no, take a look at their reduction potentials (on a table). Na is unable to oxidize Pt.

abb0t · Answer

@aaronq could you explain this a bit. I am actually unsure how you used reduction potential to explain why platinum and sodium iodide will not react. I just have never used reduction potential values to explain reactions and I am curious to know more. I'm really awful with electrochemistry.

aaronq · Answer

It's an electron-transfer rxn, so you would determine if the reaction would be spontaneous, i.e. $E_{cell}=+ve$

$E_{cell}=E^{cathode}_{red}-E^{anode}_{red}$

Cathode gains e, anode loses e. So, as, it's written

anode: $ Pt_{(s)} \rightarrow Pt^{2+}+2e $;  $E_{red}=+1.2~V$
cathode: $Na^+ +e \rightarrow Na_{(s)}$;  $E_{red}=-2.71~V$

$E_{cell}=E^{cathode}_{red}-E^{anode}_{red}=-2.71-(+1.2)=-3.91~V$

Again, a negative value means that it's not spontaneous, so the reaction would not proceed. (I assumed values at standard conditions, btw)

aaronq · Answer

btw, it's spontaneous because of $\Delta G=-nFE_{cell}$

where $E_{cell}=+ve$ gives $\Delta G=-ve$ and $E_{cell}=-ve$ gives $\Delta G=+ve.$

abb0t · Answer

Oh! I see. I completly forgot about all of this. Thanks!

aaronq · Answer

no problem, dude !