Hi Ultima, hope u could examine this
qn : 2004 A level chemistry
https://www.dropbox.com/s/e5rkrmko3uhrac0/20160720_163024-1.jpg?dl=0
, part 2cii
my answer and teacher's answer
https://www.dropbox.com/s/0jqju117dpcdesm/20160720_163753-1.jpg?dl=0
my doubt : E°cell = 0.00V , can reaction still proceed/
thermodynamically favourable?
qn say 'two reagents mixed together'
can VO2+ form?
is it VO2+ can form because its aqueous medium, where H2O and H+
are present uh?
In this case, you're wrong, but your teacher isn't fully correct
either.
Vanadium's maximum OS is +5, not +3.
Based on standard oxidation potentials of V 3+, compared to the
standard reduction potential of Cu+2 to Cu, you need only consider
oxidation of V 3+ to VO 2+ (ie. OS of V = +4), and to nothing
further (eg. to VO3 - or VO2 +) under standard conditions. But of
course, the exam question can be more sadistic and replace V 3+
with V, and Cu2+ with a stronger oxidizing agent, then you need to
consider each and of the possible oxidation products for vanadium
all the way to +5, including memorizing (if asked by the question)
all the color changes.
Minor point : Your teacher used the 'minus' version of the cell
potential formula, but the 'plus' version is actually more
meaningful, chemistry-wise (ie. for students & teachers who
understand their chemistry more deeply).
Cell potential = Reduction potential @ Cathode + Oxidation
potential @ Anode = (+0.34 V) + (-0.34V) = 0.00 V
If the calculated (standard or otherwise) cell potential is 0 V,
it means the system with standard molarities on both sides of the
equation would be at equilibrium (ie. Qc = Kc; it just so happens
that in rare cases like this case, also that Kc = 1).
If the system is already at equilibrium, no net visible change
occurs (since rate of forward reaction = rate of backward reaction,
but be careful : k(forward) = k(backward) only if Kc = 1, a rare
case). But note that at initial, you only mixed standard molarities
(ie. 1M) of V 3+ and Cu 2+ together, which means Qc is not yet =
Kc. Hence a net visible change will indeed occur, until Qc = Kc
(which happens to be Kc = 1 in this reaction). But (your teacher
didn't mention this) the reaction will be significantly slower
(exactly how much slower, depends on kinetics and activation energy
for the forward reaction), compared to if the calculated (standard
or otherwise) cell potential was a lot more positive. This is
because of the intimate relationship between electrochemistry and
equilibria and kinetics.
(Flying Grenade, do you think your school JC students, or even
your school JC teachers, already understand all the above?)
