For a reaction,$A + B^{2+} \to B + A^{2+}; E^{\circ} = 0.2955 \ V$. Hence,the equilibrium constant of the reaction at $25 \ ^oC$ is:

The $EMF$ of the cell $M | M^{n+} (0.02 \, M) || H^{+} (1 \, M) | H_{2(g)} (1 \, atm), Pt$ at $25 \, ^\circ C$ is $0.81 \, V$. Calculate the valency of the metal $(n)$ if the standard oxidation potential of the metal is $0.76 \, V$. (Use $\frac{2.303 \, RT}{F} = 0.06, \log \, 2 = 0.3$)

For the cell,$Mn_{(s)}|Mn_{(aq)}^{2+}(0.4\,M)||Sn_{(aq)}^{2+}(0.04\,M)|Sn_{(s)}$,calculate the free energy change $(\Delta G)$ at $298\,K$ in $kJ$.
Given: $E_{Mn^{2+}|Mn}^o = -1.18\,V$; $E_{Sn^{2+}|Sn}^o = -0.14\,V$; $\frac{2.303\,RT}{F} = 0.06$

For the cell $Zn | Zn^{2+}(0.01 \, M) || Fe^{2+}(0.001 \, M) | Fe$ at $25^o C$,the $E_{cell} = 0.2905 \, V$. The equilibrium constant $K_c$ is:

The e.m.f. of the cell $Zn | Zn^{2+} (0.01 \ M) || Fe^{2+} (0.001 \ M) | Fe$ at $298 \ K$ is $0.2905 \ V$. The value of the equilibrium constant for the cell reaction is:

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