zigya tab
The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500Ω. What is the cell constant if conductivity of 0.001 M KCl solution at 298 K is 0.146 x 10–3 s cm–1

Given that ,
Conductivity of the cell , κ = 0.146 × 10−3 S cm−1
Resistance of the cell , R = 1500 Ω
Formula of cell constant
Cell constant = κ × R
Plug the values we get
= 0.146 × 10−3 × 1500
= 0.219 cm−1
219 Views

Advertisement

Write the Nernst equation and emf of the following cells at 298 k.
Sn(s) | Sn2+ (0.050 M) || H+(0.020 M) H2(g) (1bar) Pt (s)


For the given cell Anode reaction:
Sn(s) → Sn2+ (aq) + 2e 

Cathode reaction:
2H+(aq) + 2e → H2(g) 

Overall cell reaction:
Sn(s) + 2H(aq) → Sn2+(aq) + H2(g)

Here, n = 2, E°cell = E°cathode – E°anode = 0 – (– 0.14 V) = + 0.14 V.
The Nernst equation for Ecell and 298 k can be written as:
Ecell = E0cell - 0.059nlog Sn2+H+2        = 0.14 - 0.059n log 0.050.022        = 0.14 V - 0.0295 (log 1.25 × 102)         = 0.14 V -0.0295 × 2.0969         = 0.14 - 0.0619          = 0.0781 V

324 Views

Advertisement
Write the Nernst equation and emf of the following cells at 298 k.
Fe(s) | Fe2+ (0.001 M) || H+ (1M) | H2 (g) (1 bar) | Pt(s)

Anode reaction:
Fe(s)  Fe2+ (aq) + 2e-

Cathode reaction:
2H++2e- H2(g)

Overall cell reaction:
              Fe(s) + 2H+(aq)  Fe2+(aq) + H2(g)Here, n = 2,  E0cell = E0cathode - E0anode = 0-(-0.44 V) = + 0.44 V.

The Nernst equation for E
cell at 298 k can be written as:
Ecell = E0cell - 0.059n log Fe2+H+        = 0.44 - 0.059n log 10-31        =0.44-0.05912(-3)

= 0.44 – 0.0295 (–3) = 0.44 – 0.0885 = 0.5285 V

174 Views

In the button cell widely used in watches and other devices the following reaction takes place:
Zn(s) + Ag2O(s) + H2O(l) → Zn2+(aq) + 2Ag(s) + 2OH (aq)
Determine ΔrG° and E° for the reaction.


The formula of standard cell potential is
Eocell = Eo right  – Eoleft
Eocell = 0.344 – ( – 0.76)
Eocell = 0.344+0.076 V
Eocell = +1.104 V
 
In balanced reaction there are 2 electron are transferring so that n = 2
Faraday constant, F = 96500 C mol−1
Eocell = + 1.104 V
Use formula
rGθ = – nFEocell
Plug the value we get   
Then, = −2 × 96500 C mol−1 × 1.104 V
= −212304 CV mol−1
= −212304J mol−1
= −212.304kJ mol−1
= −213.04 kJ
550 Views

Write the Nernst equation and emf of the following cells at 298 k.
Mg(s) | Mg2+ (0.001 M) || Cu2+ (0.0001 M) | Cu(s)

For an electrochemical cell reaction
aA + bB → cC + dD
The Nernst’s equation for cell reaction is
Ecell = E0cell-0.059n log Cc DdAa Bb
The values of a, b, c, d and n can be obtained from the balanced cell reaction. (i) Anode reaction:
Mg(s)  Mg2+(aq) + 2e-
Cathode reaction:
Cu2+(aq) + 2e-  Cu(s)
Overall cell reaction:
          Mg(s) + Cu2+(aq)  Mg2+(aq) + Cu(s)
Here,  n= 2,  E0cell = E0cathode - E0anode = 0.34 V - (-2.37) V = 2.71 V
The Nernst equation for Ecell at 298 can be written as
Ecell =E0cell         = -0.059nlog10-310-4          = 2.71- 0.0295 = 2.68 V

174 Views

Advertisement