IIT JEE , AIEEE Forum - Chemistry , Physical Chemistry

rajat

plz. answer this

CLO3--->ClO- E.= -0.54 V

CLO- ---->1/2 CL2 E.= -0.45 V

1/2 CL2-----> CL- E.=1.07 V

CLO- ----> CL- E.= 0.17 V

THEN

CLO3 -------> CL- E.= ?

rajat

dont just see my post also try to answer it plzz.

arvind_413

In ur question ,

CLO3--->ClO- E.= -0.54 V --------------- 1

CLO- ---->1/2 CL2 E.= -0.45 V ---------------- 2

1/2 CL2-----> CL- E.=1.07 V -------------------3

CLO- ----> CL- E.= 0.17 V ----------------- 4

First , eq 1+eq 3

2CLO- ----> 1/2 CL2 + CL- E = -0.28 V -------------------------- 5

Now , 2 * eq 1 + eq 5

2CLO3- ----> 1/2 CL2 + CL- E= -1.36 V ------------------------ 6

Now eq 6 +eq 3

We get

2CLO3- ----> 2CL- E = -0.29V

CLO3 -------> CL- E= -0.145 V

plz reply if u have any doubt..........

rajat

sorry arvind ur answer does not match da rite answer plus i think ur reasoning is wrong. u cant multiply the E. values wen u multiply an equation like u multiply H,G.

nrki99

see G= -nFE , u know n of electrons their in rxn

calculate G of all rxns separately

G = G1+G2+----

Aatish

hey rajat just have alook at it:

ClO₃ ---------> ClO^- E= -0.54V **************** 1

ClO^- ----------> 1/2 Cl₂ E= - 0.45V ***************2

1/2 Cl₂ ----------> Cl^- E=0.17 V ***************3

ClO₃ ------------->Cl^-- E=?

eqn 1 + eqn 2 implies

ClO₃ ---------> 1/2 Cl₂ E= - 0.99V ***************4

1/2 Cl₂ ----------> Cl^- E=0.17 V ***************3

eqn 4 +eqn 3 implies

ClO₃ ---------> Cl^- E= -0.82 V

is this your answer ?

if any mistakes then plz let me know

dfgdf

is the ans 0.08V

v_gurucharan

please tell me the way in which i can distinguish between an acid and a base.

manish_banga

do this with respect to change in Gibbs free energy as the electrons involved in diff. rxn. are not the same

dGfinal = dG1 +dG2 + dG3+ dG4

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