1
Question
Why in the question- class 12 chemistry- chapter electrochemistry - Pg 92 Q.3.5 (iv)
have we taken Br- to Br2 rather than the other way round?
Why in the question- class 12 chemistry- chapter electrochemistry - Pg 92 Q.3.5 (iv)
have we taken Br- to Br2 rather than the other way round?
Open in App
Solution
Ecell is given by using following formula :
Ecell = E0cell - 0.0591/n log Kc
In given equation:
Pt(s) | Br2(l) | Br−(0.010 M) || H+(0.030 M) | H2(g) (1 bar) | Pt(s).
Pt is used as electrode. Br- is oxidising to Br2 and H+ is reducing to H2. Br2 can not be oxidised as Br- so the equation can be written like this only.
2Br–(aq) +2H+(aq)↔ Br2(l)+ H2(g)
For writing Kc in Ecell we consider ionic species only.
Kc = 1/ [Br-]2[H+]2
As both are the product term will not be included in writing Kc the numerator will be considered as 1. That is why Ecell equation will be :
Ecell = E0cell - 0.0591/n log 1/ [Br-]2[H+]2
Ecell = E0cell - 0.0591/n log Kc
In given equation:
Pt(s) | Br2(l) | Br−(0.010 M) || H+(0.030 M) | H2(g) (1 bar) | Pt(s).
Pt is used as electrode. Br- is oxidising to Br2 and H+ is reducing to H2. Br2 can not be oxidised as Br- so the equation can be written like this only.
2Br–(aq) +2H+(aq)↔ Br2(l)+ H2(g)
For writing Kc in Ecell we consider ionic species only.
Kc = 1/ [Br-]2[H+]2
As both are the product term will not be included in writing Kc the numerator will be considered as 1. That is why Ecell equation will be :
Ecell = E0cell - 0.0591/n log 1/ [Br-]2[H+]2
Suggest Corrections
0
Similar questions
View More
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
