What will be the electromotive force of the following cell $Fe|Fe^{+2}\,(0.2\, M)\,||\, Au^{+3}\,(0.02\, M)\,|Au$ (in $,V$)?
Given: $E^o_{Fe^{+2}/Fe} = -0.44\,V$ and $E^o_{Au^{+3}/Au} = +1.50\,V$.

$A$ solution of $Fe^{2+}$ is titrated potentiometrically using $Ce^{4+}$ solution. When $80 \%$ of $Fe^{2+}$ is titrated,the $EMF$ of the system in $V$ is (Given,$E^{\circ}_{Fe^{3+}/Fe^{2+}} = 0.77 \ V$ and $Fe^{2+} + Ce^{4+} \longrightarrow Fe^{3+} + Ce^{3+}$)
$(\log 2 = 0.3, \log 3 = 0.5, \log 4 = 0.6)$

The standard Gibbs energy change in $kJ \ mol^{-1}$ for a galvanic cell $A_{(s)} + B_{(aq)}^{3+} \longrightarrow A_{(aq)}^{3+} + B_{(s)}$ that has a standard emf of $0.5 \ V$ is: $\left(F = 96500 \ C \ mol^{-1}\right)$

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

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 $E^\circ_{\text{cell}} = 0.46 \text{ V}$.

Calculate the e.m.f. of the half-cell given below: $Fe | FeSO_4$ $(a = 0.1 \ M)$ where $E^o_{OP} = 0.44 \ V$. (in $V$)