The reduction potential of a hydrogen half-cell will be negative if:

If the concentration of a $Zn^{2+}$ solution is diluted $10$ times,what is the change in the potential of the $Zn/Zn^{2+}$ electrode?

Consider the following $4$ electrodes:

$A$. $Ag^{+}(0.0001 \ M) / Ag_{(s)}$	$B$. $Ag^{+}(0.1 \ M) / Ag_{(s)}$
$C$. $Ag^{+}(0.01 \ M) / Ag_{(s)}$	$D$. $Ag^{+}(0.001 \ M) / Ag_{(s)}$

$E^{\circ}_{Ag^{+} / Ag} = +0.80 \ V$
Arrange the reduction potential of these electrodes in decreasing order.

The Nernst equation is given by $E = E^o - \frac{RT}{nF} \ln Q$. If $Q = K_{eq}$,then which of the following is not correct?

$EMF$ of the following concentration cell will be ............. $V$
$Cl_2 \ (1 \ atm) | Cl^{-} \ (1 \ M) || Cl^{-} \ (0.1 \ M) | Cl_2 \ (2 \ atm)$

At $298 \ K$,some standard electrode potentials are given below:

$Pb^{2+} / Pb$	$-0.13 \ V$
$Ni^{2+} / Ni$	$-0.24 \ V$
$Cd^{2+} / Cd$	$-0.40 \ V$
$Fe^{2+} / Fe$	$-0.44 \ V$

Metal rods $X$ and $Y$ are inserted into a solution containing $0.001 \ M$ $X^{2+}$ and $0.1 \ M$ $Y^{2+}$ at $298 \ K$ and connected by a conducting wire. This results in the dissolution of $X$. The correct combination$(s)$ of $X$ and $Y$ are,respectively:
(Given: Gas constant,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$,Faraday constant,$F = 96500 \ C \ mol^{-1}$)
$(A) \ Cd$ and $Ni \ \ (B) \ Cd$ and $Fe \ \ (C) \ Ni$ and $Pb \ \ (D) \ Ni$ and $Fe$