Q. 265.0( 1 Vote )
(a) A cell is pre
(a) The Ecell decreases.
As we know the Ecell of a cell is calculated by the Nerst equation:
E(Zn2+/Zn) = E0(Zn2+/Zn) -
This is the equation for the cell given above:
Therefore ecell here is inversely proportional to the concentration of Zn2 + ions as we see.
Clearly, if the concentration is increased, Ecell will decrease.
(b) At the Anode: Chlorine gas is evolved.
At the Cathode: Hydrogen gas is evolved
in the electrolysis of aqueous NaCl,
the ions produced will be –
Na+,Cl- ,H+ and OH- released by the dissociation of NaCl and H2O respectively.
At the Cathode:
Na+ (aq) + e-→ Na (s), E0(cell) = - 2.71V
H+ (aq) + e-→ 1/2 H2 (s), E0(cell) = 0.00 V
As the reaction with high Ecell is preferred,His released at cathode
At the anode:
Cl- (aq) → 1/2 Cl2 (g), E0(cell) = - 2.71V
Chlorine gets oxidized at anode to give chlorine at the anode.
(c) The overall cell - reaction is –
Cu2 + (aq) + Ni (s) → Ni2 + (aq) + Cu(s)
FORMULA TO FIND E0CELL
E0cell = E0cathode - E0anode
E0cell = 0. 34 - ( - 0. 25)
E0 = 0. 59V
FORMULA TO FIND ECELL
Ecell = E0cell - log OR Ecell = E0cell - log
n = the no. of exchanged electrons
r = gas constant
T = absolute temperature 298K
F = Faradays’s constant
Ecell = 0. 59 - log [n = 2]
Ecell = 0. 6195V
Ecell of the given cell is = 0. 6195V
(a) If an electrolyte on dissociation gives cations and anions,then its limiting molar conductivity is given by,
For calcium chloride,
(b) The conductivity of the NaCl solution decreases on dilution as the number of ions per unit volume decreases.
The molar conductivity of NaCl increases on dilution as on dilution the inter - ionic interactions are overcome and ions are free to move about. This is also clearly seen in the data given above.
Resistance = 100 ohms
C = 0. 1M
Cell constant G* = 1. 29 cm - 1
G* = K . R
K = = 0. 0129 S cm - 1
Formula to calculate limiting molar conductivity is =
= 129 S cm2 mol – 1
The molar conductivity of the KCl solution is 129 S cm2 mol - 1
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