Which of the following statements is correct if
Standard reduction potentials of Cu and Sn are 0.337V and -0.136V resp. (both from +2 charge)
a) Cu2+ ions can be reduced by H2(g)
b) Cu can be oxidised by H+
c) Sn2+ ions can be oxidised by H2(g)
d) Cu can reduce Sn2+
I am confused between reduction/oxidation of ions/metals by H2 or H+. I don't have any idea of correctly employing the given reduction potential
values in each case. I tried and deduced that c) and d) must be incorrect. No idea about a) and b)
Please provide me clear explanation of this.
Thanks in advance. blogfast25 - 27-12-2010 at 09:57
Since as I feel I’m doing your homework, I’ll only do one part to get you going. Here’s a good table with reduction potentials in alphabetical
order (convenient):
The REDUCTION potential for Cu2+ === > Cu is + 0.3419 V, so the OXIDATION potential of Cu === > Cu2+ is – 0.3419 V.
The REDUCTION potential for Sn2+ === > Sn is – 0.1375 V
Add up: Cu + Sn2+ === > Cu2+ + Sn: - 0.3419 V + (- 0.1375 V) < 0, this reaction cannot proceed: the cell potential (REDUCTION potential +
OXIDATION potential) must be POSITIVE for it to proceed.
This means of course that Cu2+ + Sn === > Cu + Sn2+ DOES proceed, with a cell potential of + 0.4794 V. Stick some tin in a CuSO4 solution and
copper plates out and tin enters the solution as Sn2+…
Now work out the validity/invalidity of your other assertions by means of the same principle and report back. Extra marks for clear reporting…
[Edited on 27-12-2010 by blogfast25]Claisen - 28-12-2010 at 03:37
Thank you very much for showing me a different method. I slightly modified it for generality as I found it more convenient
a) Cu2+ -----> Cu Reduction P = 0.337 V (cathode)
H2----->2H+ Reduction P = 0 v (anode)
For the cell reaction to be spontaneous, Er.p.(cathode) - Er.p.(anode) >0
0.337-0 > 0
So this reaction is spontaneous
b) Cu---> Cu2+ Reduction P = 0.337 V (anode)
2H+----> H2 Reduction P = 0 V (cathode)
Here Er.p.(cathode) - Er.p.(anode) <0 so the reaction is not spontaneous.
c) I am stuck up here.
To oxidise Sn2+, H2 should reduce itself. Does it become a hydride ion for this?
[Edited on 28-12-2010 by Claisen]blogfast25 - 28-12-2010 at 09:04
c) I am stuck up here.
To oxidise Sn2+, H2 should reduce itself. Does it become a hydride ion for this?
[Edited on 28-12-2010 by Claisen]
Reduction potential series shows that H2 reduces Sn4+ to Sn2+ (used in practice for quant. tin determinations) but not Sn2+ to Sn
(0):
Sn2+ === > Sn, RP = -0.1375 V
H2 === > 2 H+, OP = 0 V
Sum is negative: no reaction!
In case you’re wondering about the 0 V value of hydrogen, it’s just a convention really. All reduction/oxidation potentials are half-reactions,
the potential of which is measured relative to H2/2H+ which conventionally has been set to 0 V. A Cu/Cu2+ H2/2H+ galvanic cell (battery) yields an
electromotive force (voltage) of + 0.3419 V, so we say that for the half reaction Cu2+ === > Cu the reduction potential is + 0.3419 V, relative to
H2/H+…
