Calculate the equilibrium constant of the following reaction:
$\text{Cu(s)} + 2\text{Ag}^+_{\text{(aq)}} \rightarrow \text{Cu}^{2+}_{\text{(aq)}} + 2\text{Ag(s)}$,given $E^\circ_{\text{cell}} = 0.46 \text{ V}$.

If a solution of $Cu^{+2}/Cu$ at $298 \, K$ is diluted $100$ times,how will the electrode potential change?

The photoelectric current from $Na$ (work function,$w_{0}=2.3 \ eV$) is stopped by the output voltage of the cell
$Pt_{(s)} | H_{2}(g, 1 \ bar) | HCl(aq, pH=1) | AgCl_{(s)} | Ag_{(s)}$
The $pH$ of aqueous $HCl$ required to stop the photoelectric current from $K$ $(w_{0}=2.25 \ eV)$,all other conditions remaining the same,is..........$\times 10^{-2}$ (to the nearest integer).
Given,$2.303 \frac{RT}{F}=0.06 \ V; E_{AgCl|Ag|Cl^{-}}^{0}=0.22 \ V$

The standard emf for the cell $Cd_{(s)}|Cd^{2+}_{(aq)}(1 \ M)||Cu^{2+}_{(aq)}(1 \ M)|Cu_{(s)}$ is $0.74 \ V$. If the concentration of $Cd^{2+}_{(aq)}$ and $Cu^{2+}_{(aq)}$ both decrease by $10$ times at $298 \ K$,calculate the emf of the cell.

The relationship between standard reduction potential of a cell and the equilibrium constant is shown by:

For the cell reaction,$Cu | Cu^{2+}(0.1 \ M) || Cu^{2+}(1.0 \ M) | Cu$,the emf of the cell at $25^{\circ}C$ is given that $E^{\circ}_{Cu^{2+}/Cu} = 0.34 \ V$. (in $V$)