﻿ The molar conductivity of 0.025 mol L–1 methanoic acid is 46.1 S cm2 mol–1. Calculate its degree of dissociation and dissociation constant. Give λ°(H+) = 349.6 S cm2 mol–1 and λ° (HCOO– ) = 54.6 s cm2 mol–1. from Chemistry Electrochemistry Class 12 Nagaland Board

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The molar conductivity of 0.025 mol L–1 methanoic acid is 46.1 S cm2 mol–1. Calculate its degree of dissociation and dissociation constant. Give λ°(H+) = 349.6 S cmmol–1 and λ° (HCOO ) = 54.6 s cm2 mol–1.

Given that
λ0(H+)= 349.6 S cm2 mol–1
λ0(HCOO) = 54.6 S cm2 mol–1
Concentration ,C = 0.025 mol L−1
λ(HCOOH) = 46.1 S cm2 mol−1
use formula
λo(HCOOH)   = λ0(H+)  +   λ0(HCOO)
plug the values we get
λo(HCOOH)   = 0.349.6 + 54.6
=404.2 S cm2 mol−1
Formula of degree of dissociation:
ά = λo(HCOOH)/ λo(HCOOH)
ά = 46.1 / 404.2
ά = 0.114
Formula of dissociation constant:
K = (c ά2)/(1 – ά)
Plug the values we get

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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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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
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Fe2+--> Fe3+ +e-
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(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 .

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Cu2+(aq) + Zn(s) → Zn2+ (aq) + Cu(s)
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