Monday
April 21, 2014

Homework Help: chemistry

Posted by someone on Tuesday, March 31, 2009 at 5:31pm.

For
Hg2+(aq) + 2e -> Hg(l) E=0.851 V

This means that:
A) Hg(l) is more readily oxidized than H_2(g);
B) Hg^2+(aq) is more readily reduced than H^+(aq)
C) Hg(l) will dissolve in 1 M HCl
D) Hg(l) will displace Zn(s) from an aqueous solution of Zn^2+ ion

~~~~~~
Sooo...this half rxn shows a reduction.
The E value for 2H^+ + 2e -> H_2(g) is 0 V. Since Hg has a E, it will be more readily to reduced?
I THINK...the answer is B.

I don't understand how to interperate C and D.
The E value for Zn2+ + 2e -> Zn(s) is -0.763.

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