is is confusing:
What is the theoretical voltage required to electrolyze molten NaCl that forms Cl2(g) at anode and Na(l) at cathode, given Eo for Na+/Na is 2.71 V and Eo for Cl2/Cl- is +1.36 V?
A. 0.00 V B. 1.35 V C. 1.36 V D. 2.71 V E. 4.07 V
answer is E: why?
here is one rule if the reaction or salt bridge reaction is not given.
a couple of things to remember:
1) Reduction will occur which have higher REDUCTION POTENTIAL
2) reduction occurs at CATHODE
3) oxidation occurs at ANODE.
once you figure out your cathode and anode reaction use this formula:
(a) E* = E(reduction) + E(oxidation)
OR
(b) E*= E(reduction potential at cathode) - E(reduction potential at anode)
In your question:
Cl2/Cl has higher reduction potential so it will undergo reduction and at CATHODE.
Na+/Na has lower reduction potential, so it will undergo oxidation at ANODE.
now u can use 3 steps from here:
1) so E(Cl2/Cl)= 1.36 V at CATHODE and
E(Na+/Na) = -2.71 V at ANODE.
Using formula (b)
E* = E(reduction potential at cathode) - E(reduction potential at anode)
E* = 1.36 - [-2.71] = 4.07 V
NOTE that both should use their reduction potentials
2) Reduction at CATHODE : E(Cl2/Cl) = +1.36 V
Oxidation at ANODE : E(Na/Na+) = +2.71 V......(reverse the reaction and reverse the sign or the E)
with the formula (a)
E* = E(reduction) + E(oxidation)
E* = 1.36 + 2.71 = 4.07 V
3) A third way would be, after deciding the cathodic and anodic reaction write out the reaction:
Cl2 + 2e ----> 2Cl- ........E*= 1.36 V (Reduction at CATHODE)
[Na -----> Na+ + 1e]x2.......E* = 2.71 V (Oxidation at ANODE)
Add these two reactions:
2Na + Cl2 ------> 2Cl- + 2Na+ .....E* = 1.36 + 2.71 = 4.07 V
so ur salt bridge notation would be
Na/Na+||Cl2/Cl-
remember LOAN
Left Oxidation Anode Negative
LEFT side of notation has OXIDATION at ANODE and its NEGATIVE.
