The initial $EMF$ of the concentration cell shown in the figure at $298 \ K$ is .............. $V$.

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,$2.303 \frac{RT}{F}=0.06 \ V; E_{AgCl|Ag|Cl^{-}}^{0}=0.22 \ V$

When a copper plate is kept in a $0.1 \ M$ solution of $CuSO_4$ at $298 \ K$ temperature and if $70 \ \%$ dissociation has occurred,then calculate the potential of the copper electrode.

For the electrochemical cell shown below:
$Pt \mid H_{2}(p=1 \, atm) \mid H^{+}(aq., x \, M) \mid\mid Cu^{2+}(aq., 1.0 \, M) \mid Cu_{(s)}$
The potential is $0.49 \, V$ at $298 \, K$. The $pH$ of the solution is closest to:
[Given: Standard reduction potential,$E^{\circ}$ for $Cu^{2+}/Cu$ is $0.34 \, V$; Gas constant,$R = 8.31 \, J \, K^{-1} \, mol^{-1}$; Faraday constant,$F = 9.65 \times 10^{4} \, J \, V^{-1} \, mol^{-1}$]

Calculate $pH$ of $HCl$ solution at $298\,K$ temperature for the following cell: $Pt_{(s)} \mid H_2 \,(1\,bar) \mid HCl\,(xM) \parallel Cu^{2+}\,(0.02\,M) \mid Cu_{(s)}$. Given that the standard cell potential $E^{\circ}_{cell} = 0.34\,V$ and the measured cell potential $E_{cell} = 0.45\,V$.

At what $pH$,given half cell $MnO_4^{-} (0.1 \ M) \mid Mn^{2+} (0.001 \ M)$ will have electrode potential of $1.282 \ V$? (Nearest Integer) Given $E_{MnO_4^{-} / Mn^{2+}}^{o} = 1.54 \ V, \frac{2.303 RT}{F} = 0.059 \ V$