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Na+ Br-
Na can get reduced. na at the cathode competes with reduction of h2o that produces H2 for the least extreme value. -2.71 vs. -0.83. H2 at the cathode?
br can get oxidized. br at the anion competes with h2o for least extreme.
0.54 vs. 1.23. Br wins and will be produced at the anode.
Na+ + e- --> Na -2.71
Br2 (l) + 2e- --> 2br +0.54
Aqueous AgF
Ag +1, F is always -1
Ag can get reduced (cathode): --> +0.80
it competes with the reduction of water H2 -0.83
for cations, the highest reduction potential will be reduced. Ag wins.
F can get oxidized (anode) <-- +2.87
competes with oxidation of water which produces O2 +1.23
for anions, the lowest reduction potential will be oxidized, O2 wins.
3. metal plating. Cathode: Au3+ + 3e- --> Au (s)
reduction here.
givens: 24.5 A, 14 mins. what is mass of Au in g.
charge: A x time (s) = 24.5 A x (14 min x 60 sec/min) = 20,580 C
3 mol e to produce 1 mol of Au (s).
840 s x
strongest ox agent:
substance is being reduced. we want the left side and the most positive.
strongest reducing agent:
substance is being oxidized, we want the right side and the most negative.
which can be reduced by sn2+?
sn itself is a reducing agent, so it is being oxidized. look for a substance that is being reduced, so it is on the left side and more negative Ered.
options are between br2 at +1.06 and Ca2+ -2.87
at the cathode is ?? at the anode is sn2+ +0.15
E cell = 1.06 - 0.15 = + positive; Br2?
E cell with ca2+ = -2.87 - 0.15 = - very negative;
which can be oxidized by I2 (s)?
I2 is an oxidizing agent, so it is being reduced. look for a substance on the right side and find a combo that produces the most positive Ecell.
br- or H2
Br- = +1.06 and H2 0.
E cell = 0.54 - 0= +, so H2 wins.
voltaic cell with
Fe is being oxidied, anode. loses mass and loses e
Cu is being reduced, cathode. gains mass and gains e
standard cell potential:
I is being oxidixed by losing 2 electrons. anode. <-- 0.54
cu is being reduced by gaining 2 electrons. cathode -->
E cell = 0.34 - 0.54
voltaic cell with zn and h
E cell = 0 - -.76 = +.76
Q = [H2] [Zn2+] / [H+]^2
fe reduced by 1, cathode
ca oxidized by 2, anode.
e cell = .77 - -2.87 = +3.64 spont
salt bridge notation:
sn is oxidized by 2 anode
cu is reduced by 2 cathode,
e cell = 0.34 - -0.14 = 0.48 spont
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