﻿ The conductivity of 0.1 M KCl solution at 298 K is 0.0129 s cm–1. The resistance of this solution in a conductivity cell is found to be 58 ohms. What is the cell constant of the cell? The 0.1 M AgNO3 solution at 298 K in the same conductivity cell offered a resistance of 60.5 ohms. What is the conductivity of 0.1 M AgNO3 solution? from Chemistry Electrochemistry Class 12 Nagaland Board

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The conductivity of 0.1 M KCl solution at 298 K is 0.0129 s cm–1. The resistance of this solution in a conductivity cell is found to be 58 ohms. What is the cell constant of the cell? The 0.1 M AgNO3 solution at 298 K in the same conductivity cell offered a resistance of 60.5 ohms. What is the conductivity of 0.1 M AgNO3 solution?

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

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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

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

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.

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