Question

Calculate ECell , deltaG°, and K for the following reactions at 25 °C:

(2a): Br2 (l) + 2 I– (aq) → I2 (s) + 2 Br– (aq)

2 H2O (l) + 2 Cl2 (g) → 4 H+ (aq) + 4 Cl– (aq) + O2 (g)

Answer 1

for Br2 + 2 I-. = I2 +2Br-

its a net cell reaction, we have to write oxidation and reduction half cell reactions

oxidation 2I- = I2 +2e- .[anode

reduction Br2 +2e- = 2Br- [cathode

we know that

Ecell = Ecathode - E anode = E (Br2/Br-) - E (I2/I-)

now putting standard reduction potential for B2/Br- = +1.07 and I2/I- = +0.53

Ecell = 1.07 - 0.53 = 0.54 volt ---------[1]

Delta G = - n .Ecell. F = - 2x 0.54x96500 =

=- 104220 = - 104.22 kilo joules ------_----[2]

here n is number of electrons lost or gained =2 , F is Faraday constant 96500 quolumb.

also ∆G = -RT ln K . here K is equillibrium constant, R is gas constant ,

-104220 = - 8.314x (273+25) x ln K

ln K = 104220 / 2477.57 = 42 .065

log K = 18.27

K = 1.86 x 10 ^18 .-------------[3]

for reaction >........>........................>......>:"

oxidation 2 H2O = 4H+ + O2 + 4 e- .[ anode

reduction. 2 Cl2 + 4e- = 4 Cl- [ cathode

Ecell = E Cl2/cl- - E O2/O-2 = [ +1.36 - 0.68] = 0.68 volt

Ecell = 0.68 volt ---------[1]

delta G = - n Ecell F = - 4x 0.68 x 96500 = -189040 joules = -189.04 kilo joules ------[2]

ln K = 189040 / 8.314x298 = 76.30

log K = 33.13

K = 1.35 x 10 ^33 .----------[3]