Question

What will happen to Sn metal when placed into CuSO₄ solution? Was this what you expected? Explain your answer. Did a reaction occur? How do you know?

Answer 1

Sn (s) + CuSO4(aq)-------------> SnSO4(aq) + Cu(s)

Tin displace the the copper from its salt solution

Sn (s) + Cu^2+(aq) +SO4^2-(aq)-------------> Sn^2+(aq) +SO4^2-(aq) + Cu(s)

removal of spectator ions to get net ionic equation

Sn (s) + Cu^2+(aq) -------------> Sn^2+(aq) + Cu(s)

at anode

     Sn(s) ---------------------> Sn^2+ + 2e^-      E0 = 0.14v    (oxidation reaction)

at cathode

    Cu^2+ (aq) + 2e^- ----------> Cu(s)              E0 = 0.34v      ( reduction reaction)

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Sn (s) + Cu^2+(aq) -------------> Sn^2+(aq) + Cu(s)   E0 cell = 0.48v

      n= 2

G⁰   = -nE0cell * F

             = -2*0.48*96500

             = -92640J

G⁰ <0 it is spontaneous reaction