﻿ The specific conductance of a saturated solution of AgCl in water is 1.826 x 10–6ohm–1 cm–1 at 25°C. Calculate its solubility in water at 25°C. [Given Λ∞m (Ag+) = 61.92 ohm–1 cm2 mol–1 and Λ∞m (CI– ) = 76.34 ohm–1 cm2 mol–1] from Chemistry Electrochemistry Class 12 Nagaland Board

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The specific conductance of a saturated solution of AgCl in water is 1.826 x 10–6ohm–1 cm–1 at 25°C. Calculate its solubility in water at 25°C. [Given Λm (Ag+) = 61.92 ohm–1 cm2 mol–1 and Λm (CI ) = 76.34 ohm–1 cm2 mol–1]

Λm (AgCl) = λmm (Ag+) + λm (Cl)

Solubility (in mol L-1) =

1356 Views

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

oxidation of ferrous ion means :

Fe2+--> Fe3+ +e-
Any substance which standard electrode potential is more than that of Fe+3 /F+2 can oxidise ferrous ions.
(refer to the table given in book)

The EMF of the substance whose reduction potentials greater than 0.77V will oxidised ferrous ion.
for example Br2, Cl2,and F2 .

879 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 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

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