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4. Use the information above to write net ionic equations for the two reactions that you have observed.
5. For each reaction,
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Those are different from the original post, I am trying to work on it!
Sn2+ (aq) + 2Hg2+ (aq) + 2Cl (aq)--> Sn4+ (aq) + Hg2Cl2 (s) Hg2Cl2 (s) + Sn2+ (aq) --> 2Hg (I) + Sn4+ (aq) + 2Cl- (aq) 5. Sn(2+) --> Sn(4+) + 2e- Eo = -0.15 V 2e+ + 2Hg(2+) + 2Cl --> Hg2Cl2 Eo = 0.85 V Total: 0.75 V Hg2Cl2 + 2e- --> 2Hg(2+) + 2Cl- Eo= 0.27
