For a cell involving one electron $E_{cell}^{\ominus} = 0.59 \; V$ at $298 \; K$,the equilibrium constant for the cell reaction is.
Given that $\frac{2.303 \; RT}{F} = 0.059 \; V$ at $T = 298 \; K$.

If $Zn/Zn^{+2}$ electrode is diluted $100$ times,then the change in electromotive force will be:

For a cell,$Cu_{(s)} \mid Cu^{2+}(0.001\,M) \mid\mid Ag^{+}(0.01\,M) \mid Ag_{(s)}$,the cell potential is found to be $0.43\,V$ at $298\,K$. The magnitude of standard electrode potential for $Cu^{2+}/Cu$ is $......... \times 10^{-2}\,V$. $[\text{Given}: E^{\Theta}_{Ag^{+}/Ag} = 0.80\,V \text{ and } \frac{2.303RT}{F} = 0.06\,V]$

The reduction potential of a hydrogen electrode at $pH = 1$ is ........ $V$.

What is the change in potential of the following cell $Zn_{(s)}|Zn^{2+} (1 \ M)||Pb^{2+} (1 \ M)|Pb_{(s)}$ if the concentration of ions at the anode is increased $10$ times?

Find the $emf$ of the cell in which the following reaction takes place at $298 \ K$ (in $V$):
$Ni_{(s)} + 2Ag^{+}(0.001 \ M) \rightarrow Ni^{2+}(0.001 \ M) + 2Ag_{(s)}$
(Given that $E_{cell}^{\circ} = 10.5 \ V$,$\frac{2.303 RT}{F} = 0.059$ at $298 \ K$)