Selective Discharge of Ions

The standard reduction potential values of three metallic cations, X, Y and Z are $0.52,-3.03 \mathrm{and}-1.18\mathrm{V}$ respectively. The order of reducing power of the corresponding metals is:

${\mathrm{H}}_2$ cannot be displaced by

The standard electrode potential of ${\mathrm{M}}^{2+}/\mathrm{M}$ is positive for

Lower the position of an ion in an electrochemical series more easily it will get discharged at the respective electrode.

How does nature of electrode affect selective discharge of ions.

If a solution contains more than two cations and anions, the selective discharge of ions depends on electrochemical series.

Define selective discharge of ions.

Identify the easily oxidisable halide.

Select the correct option, for the given standard ${\mathrm{E}}^{\mathrm{o}}$ values of few redox couples:
$\mathrm{Z}{\mathrm{n}}^{2+}/\mathrm{Z}\mathrm{n}=-0.76\mathrm{V}, \mathrm{A}{\mathrm{g}}^{+}/\mathrm{A}\mathrm{g}=+0.80\mathrm{V}, {\mathrm{Cu}}^{2+}/\mathrm{Cu}=0.34\mathrm{V}$
 

Mark the correct statement based on the ${\mathrm{E}}^{\mathrm{o}}$ values of redox couple of halogens given below
$\mathrm{ }{{\mathrm{E}}^{\mathrm{o}}}_{{\mathrm{I}}_2/{\mathrm{I}}^{-}}=+0.54\mathrm{V},{{\mathrm{E}}^{\mathrm{o}}}_{{\mathrm{Br}}_2/{\mathrm{Br}}^{-}}=+1.08\mathrm{V};{{\mathrm{E}}^{\mathrm{o}}}_{{\mathrm{Cl}}_2/{\mathrm{Cl}}^{-}}=+1.36\mathrm{V}$.

One metal $\mathrm{X}$ displaces nickel from nickel sulphate solution but does not displace manganese from manganese sulphate solution. The correct order of their reducing powers is:

Which element can reduce Aluminium?

Match the column :

Out of which of the following statements is true- 

Given the reduction potential data as-  ${\mathrm{E}}_{{\mathrm{Ag}}^{+}/\mathrm{Ag}}^{\mathrm{o}}=+0.8\mathrm{ }\mathrm{V},\mathrm{ }{\mathrm{E}}_{{\mathrm{Ni}}^{+2}/\mathrm{Ni}}^{\mathrm{o}}=-\mathrm{ }0.25\mathrm{ }\mathrm{V}\mathrm{ }$ .

Select the correct statements based on standard electrode potential of some half cell reactions are given below:
(A)    ${\mathrm{Sn}}^{+4}+2{\mathrm{e}}^{-}⟶{\mathrm{Sn}}^{2+}, \mathrm{E}°=+0.15\mathrm{V}$
(B)    $2{\mathrm{Hg}}^{+2}+2{\mathrm{e}}^{-}⟶{\mathrm{Hg}}_2^{+2}, \mathrm{E}°=+0.92\mathrm{V}$
(C)    ${\mathrm{PbO}}_2+4{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}⟶{\mathrm{Pb}}^{+2}+2{\mathrm{H}}_2\mathrm{O}, \mathrm{E}°=+1.45\mathrm{V}$

Among the following, which is the most reactive metal which displaces other metals from their salts in solution?

${\mathrm{E}}_{\mathrm{Z}\mathrm{n}/\mathrm{Z}{\mathrm{n}}^{2+}}^0=+0.76\mathrm{V},\mathrm{ }{\mathrm{E}}_{\mathrm{A}\mathrm{g}/\mathrm{A}{\mathrm{g}}^{+}}^0=-0.80\mathrm{V},\mathrm{ }$

${\mathrm{E}}_{\mathrm{C}\mathrm{u}/\mathrm{C}{\mathrm{u}}^{2+}}^0=-0.34\mathrm{V},\mathrm{ }{\mathrm{E}}_{\mathrm{C}\mathrm{o}/\mathrm{C}{\mathrm{o}}^{+2}}^0=0.28\mathrm{V}$

We have taken two electrode, with potential values: ${\mathrm{E}}_{{\mathrm{Mg}}^{2+}/\mathrm{Mg}}^{\mathrm{a}}=- 2.37 \mathrm{V} \mathrm{and} {{\mathrm{E}}^{\mathrm{b}}}_{{\mathrm{Fe}}^{3+}/\mathrm{Fe} }=-0.04 \mathrm{V}$. Which of the following will act as the best reducing agent: