Question
Explain relationship between $E _{\text {cell }}^{\circ}$ and equilibrium constant with the help of Nernst equation.

Answer

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Match the items of Column I and Column II.
  Column I   Column II
(i) Mathematical expression for rate of reaction. (a) Rate constant.
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Match the items of Column I and Column II on the basis of data given below:
$\text{E}^{\ominus}_{\text{F}_2/\text{F}^-}=2.87\text{V},\text{ E}^{\ominus}_{\text{Li}^+/\text{Li}}=-3.5\text{V},\text {E}^{\ominus}_{\text{Au}^{3+}/\text{Au}}=1.4\text{V},\text{ E}^{\ominus}_{ \text{Br}_2/\text{Br}^-}=1.09\text{V}$
 
Column I
 
Column II
i.
F2
a.
Metal is the strongest reducing agent.
ii.
Li
b.
Metal ion which is the weakest oxidising agent.
iii.
Au3+
c.
Non metal which is the best oxidising agent.
iv.
Br-
d.
Unreactive metal.
v.
Au
e.
Anion that can be oxidised by Au3+
vi.
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f.
Anion which is the weakest reducing agent.
vii.
F-
g.
Metal ion which is an oxidising agent.
The experimental data for decomposition of N2O5
[2N2O5 → 4NO2 + O2]
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t/s 0 400 800 1200 1600 2000 2400 2800 3200
102 × [N2O5]/mol L-1 1.63 1.36 1.14 0.93 0.78 0.64 0.53 0.43 0.35
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