﻿ When a certain conductivity cell was filled with 0.1 M KCl, it has a resistance of 85 Q at 25°C. When the same cell was filled with an aqueous solution of 0.052 M unknown electrolyte, the resistance was 96 Ω. Calculate the molar conductivity of the unknown electrolyte at this concentration. (Specific conductivity of 0.1 M KCl = 1.29 x 10–2 ohm–1cm–1). from Chemistry Electrochemistry Class 12 Nagaland Board

## Book Store

Currently only available for.
CBSE Gujarat Board Haryana Board

## Previous Year Papers

Download the PDF Question Papers Free for off line practice and view the Solutions online.
Currently only available for.
Class 10 Class 12
When a certain conductivity cell was filled with 0.1 M KCl, it has a resistance of 85 Q at 25°C. When the same cell was filled with an aqueous solution of 0.052 M unknown electrolyte, the resistance was 96 Ω. Calculate the molar conductivity of the unknown electrolyte at this concentration. (Specific conductivity of 0.1 M KCl = 1.29 x 10–2 ohm–1cm–1).

Resistance of KCl solution,
R = 85 $\mathrm{\Omega }$
Cell constant  = K x R

Resistance of unknown electrolyte solution,

Specific conductance

Concentration,

Molar conductance,

931 Views

How would you determine the standard electrode potential of the system Mg2+/Mg?

Use standard hydrogen electrode as anode and Mg2+ | Mg as a cathode we can measure the standard electrodepotential of systemMg2+ | Mg. Standard hydrogen electrode, represented by Pt(s), H2(g) (1 atm) | H+ (aq) and dip the electrode of Magnesium wire in a 1M MgSO4 solution .The standard hydrogen electrode is always zero.
Use formula
Eocell = Eo right  – Eoleft
The standard hydrogen electrode is always zero.
So that the value of
Eoleft =0
Hence
Eocell = Eo Mg|Mg2+
Or
Eo Mg|Mg2+= Eocell
2203 Views

Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.

For hydrogen electrode, ,
given that      pH = 10
use formula [H+] = 10–pH
so that  [H+] = 10−10 M
Electrode reaction will
H+  + e –  →1/2 H2
Use the formula

1279 Views

Calculate the emf of the cell in which the following reaction takes place:
Ni(s) + 2Ag+ (0.002 M) $\to$ Ni2+ (0.160 M) + 2Ag(s)
Given that

$\mathrm{Ni}\left(\mathrm{s}\right)+2{\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)\to {\mathrm{Ni}}^{2+}\left(\mathrm{aq}\right)+2\mathrm{As}\left(\mathrm{s}\right)$

or

The equation is also written as

or

= 1.05 V – 0.0295 x log 80

= 1.05 V – 0.0295 x 1.9031

= 1.05 V – 0.056 = 0.99 V.

1585 Views

Can you store copper sulphate solutions in a Zinc pot?

No. Because zinc is more reactive than copper and thus holes will be developed in zinc pot.

Cu2+(aq) + Zn(s) → Zn2+ (aq) + Cu(s)
1384 Views

Consult the table of standard electrode potentials and suggest three substance that can oxidize ferrous ions under suitable conditions.