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.
Mastering Chemistry Question:
What is produced at each electrode in the electrolysis of an aqueous solution of both NaBr and AgF?
Answer is:
Anode=Br
Cathode=Ag
Not Produced= all the other junk
is what came back as correct answer when entered into mastering Chem.
C and D. Look at the activity series and that is easier to see than looking at reduction potentials.
C. Hg is below H in the activity series; therefore, Hg will NOT displace H^+ from an acid which means it will not dissolve in HCl.
D. D is a crazy answer (because Zn(s) is a solid so it can't be in solution). Zn is ABOVE Hg in the activity series; therefore, Zn will displace Hg^+2 from solution but the reverse is not true. It is true that Zn(s) will react with Hg(l) and form an amalgam.
Your answer of B is correct.
Thank you.
Also I have another question:
What is produced at each electrode in the electrolysis of an aqueous solution of both NaBr and AgF?
My answers are...
anode:
- Na(s)
- Br2(l)
cathode:
- Ag(s)
- F2(g) ?
not produced:
- H2(g)
- O2(g)
oops I meant
anode:
- Na(s)
- F2(g)
cathode:
- Br2(l)
- Ag(s)
Something for you to read.
http://en.wikipedia.org/wiki/Electrolysis
For NaBr, Na^+ is more difficult to reduce at the negative electrode (the cathode); therefore, H2 gas is evolved at the cathode. Br^- can be oxidized at the anode (the positive electrode) and Br2 will be produced.
For AgF, Ag^+ can be reduced at the cathode and Ag will be deposited. F^- has the highest reduction potential listed (-2.87) so oxygen will be evolved at the anode(the positive electrode).
You may, if you wish, think Na will be deposited at the cathode BUT remember from the activity series that Na metal is extremely reactive so even if a deposition of Na appeared, it would immediately react with water to produce H2 gas and NaOH.
only br and h are produced, br in the anode and h in the cathode
The oxidation or reduction of water will not occur if there is another available half-reaction with a higher potential. The answer is Ag only at the cathode and Br only at the anode.
Happy Mastering Chemistrying :)
sarah is wrong.
wtf are you talkin about? sarah is right.
anode - br
cathode - ag
not produced- everything else