reduction potential is given by:
From the given data PH =14 and KsP (Cu(OH2) =1.0 x 10-19
we get
H+ =10-14 M thus,
[OH]- =
ans -0.22V
KKCl = 0.0129 s cm–1
RKCl = 58 Ω
cell constant= k x R
cell constant= 0.0129 x 58
Cell constant = 0.7482 cm–1
As AgNO3 is also in the same conductivity cell, the cell constant remains same. Therefore,
Conductivity of
We ave given that
R = 1500 Ω
K = 1.46 x 10–4 Ω–1 cm–1
Cell constant = K x R
= 1.46 x 10–4 x 1500 = 0.219
Consider the cell Zn/Zn2+ (aq) (1.0 M) || Cu2+ (aq) (0.1 M) | Cu The standard reaction potentials are + 0.35 V for 2e– + Cu2+ (aq) → Cu and – 0.763 V for 2e– + Zn2+ (aq) → Zn
(i) Write down the cell reaction.
(ii) Calculate the emf of the cell.
(iii) Is the cell reaction spontaneous or not?
(I)Cell Reaction:
Oxidation: Zn----> Zn2+ +2e-
Reduction: Cu2+ +2e------> Cu
Therefore overall reaction
Zn +Cu2+ ------> Zn2+ +Cu
(ii) =+0.35-(-0.763)
=1.113 volts
(iii) Since EMF of cell is positive, it is a spontaneous reaction.
The standard reduction potential of the reaction at 25°C
2H2O + 2e– H2(g) + 2OH– is – 0.8277 V
Calculate equilibrium constant for the reaction
at
Consider the given reaction as the net cell reaction
two half reaction :
Oxidation:
H2O +1/2H2 ----> H3O+ +e- Ecell0 =0
reduction:
H2O +e- -----> 1/2H2(g) + OH- Ecell0 = -0.8277V
It is evident from the cell reaction that it involves the transfer of one electron so that n= 1