Write the Nernst equation for the $E_{cell}$ reaction in the Daniell cell. How will the $E_{cell}$ be affected when the concentration of $Zn^{2+}$ ions is increased?

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 $Fe$

Which of the following is correct as a Nernst equation for the given electrochemical cell ?
$Mg_{(s)}|Mg_{(aq)}^{2+}(0.1 \ M)||Cl_{(aq)}^{-}(0.1 \ M)|Cl_{2_{(g)}}(1 \ bar)|Pt_{(s)}$

For a certain redox reaction in a galvanic cell $X(s) + Y^{2+}(aq) \rightarrow X^{2+}(aq) + Y(s)$, $E^0_{cell} = 0.0296 \text{ V}$ at $298 \text{ K}$. What is the equilibrium constant $(K_c)$ of the reaction?

At $298 \ K$,the following reaction takes place for a hydrogen electrode:
$H^{+}_{(aq)} + e^{-} \longrightarrow \frac{1}{2} H_2(1 \ bar)$
The solution $pH$ is $10.0$. What is the hydrogen electrode potential in volts?
$\left(\frac{2.303 RT}{F} = 0.06 \ V\right)$

The pressure of hydrogen gas is increased from $1 \, atm$ to $100 \, atm$ keeping the $[H^+]$ constant at $1 \, M$. The potential of the hydrogen half-cell at $25 \, ^\circ C$ will be .......... $V$.