This is what I have so far. Im sorry for the bad picture. If it\'s hard to see i

Question

This is what I have so far. Im sorry for the bad picture.  If it's hard to see ill post another one. but im unsure of the other ones. any help will be appreciated.

anodes are  
cu-zn ---> Zn  
Cu-Fe--->Fe
zn-Fe---> Zn

Anode reactions are:

oxidation    Zn^0 - 2 e -->Zn2+

   Fe^0 - 2 e -->Fe2+

  Zn^0 -2e -->  Zn^2+
Cathode:    

reduction  Cu2+  +2 e  -->Cu^0

Fe2+  +2e -->  Fe^0

Write balanced equation for the six cell reaction. What is the oxidization agent in the Zn-Mg cell? Compare the sum of the Cu-Zu and Zn-Mg cell potentials with the Cu-Mg cell potential. Explain. Compare the sum of the Zn-Fe and Zn-Mg cell potentials with the Fe-Mg cell potential. Explain.

Explanation / Answer

1) Cu2+(aq) + Zn(s) <--------> Cu(s) + Zn2+ (aq)

Cu2+(aq) + mg(s) <-----------> Cu(s) + Mg2+(aq)

Cu2+(aq) + Fe(s) <---------> Cu(s) + Fe2+(aq)

Zn2+(aq) + Mg (s) <---------> Zn(s) + Mg2+(aq)

Fe2+(aq) + Mg(s) <--------> Mg2+(aq) + Fe(s)

Fe2+(aq) + Zn (s) <--------> Zn2+(aq) + Fe(s)

2) Zn2+(aq) is oxidizing agent in Zn-Mg cell

3) sum of Zn-Fe and Zn-Mg cells potentials = 0.302 + 0.799 = 1.101 volts

equations are Zn(S) + Fe2+(aq) <-----> Zn2+(aq) + Fe(S)

                 Zn 2+(aq) + Mg (s) <--------> Mg2+(aq) + Zn

net eq is Mg(S) + Fe2+(aq) < ------> Mg2+(aq) + Fe(s) which has emf = 1.1 volts

Fe-Mg cell potential = 1.053

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