The standard $e.m.f.$ of a cell involving a one-electron change is found to be $0.591 \ V$ at $25^{\circ} C$. What is the equilibrium constant for the reaction? $(F = 96500 \ C \ mol^{-1}; R = 8.314 \ J \ K^{-1} \ mol^{-1})$

At $298 \, K$,find out the $emf$ for the cell: $Al_{(s)} | Al^{+3} (0.1 \, M) || Fe^{+2} (0.001 \, M) | Fe_{(s)}$. Given $E^o_{Al^{+3}/Al} = -1.66 \, V$ and $E^o_{Fe^{+2}/Fe} = -0.44 \, V$. (in $, V$)

If $E^{\circ}_{\text{cell}}$ is $1.049 \ V$ and the reaction involves the transfer of $2$ electrons,calculate the equilibrium constant $(K)$ of the cell.

In the given electrochemical cell, $Ag_{(s)} | AgCl_{(s)} | Cl^-_{(aq)}, Fe^{2+}_{(aq)}, Fe^{3+}_{(aq)} | Pt_{(s)}$ at $298 \ K$, the cell potential $(E_{cell})$ will increase when :
$(A)$ Concentration of $Fe^{2+}$ is increased.
$(B)$ Concentration of $Fe^{3+}$ is decreased.
$(C)$ Concentration of $Fe^{2+}$ is decreased.
$(D)$ Concentration of $Fe^{3+}$ is increased.
$(E)$ Concentration of $Cl^-$ is increased.
Choose the correct answer from the options given below :

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?
$\left(\frac{2.303 RT}{F} = 0.06 \ V\right)$

Which of the following is the correct expression for the electrode potential of a cell?