Considering the reaction $Cl_{2(g)} + 2Br^{-}_{(aq)} \rightarrow 2Cl^{-}_{(aq)} + Br_{2(l)}$,calculate the cell $emf$ in $V$ when $[Cl^{-}] = [Br_2] = [Br^{-}] = 0.01 \ M$ and $Cl_2$ gas is at $1 \ atm$ pressure. (Given $E^o = 0.29 \ V$ for the reaction)

For the disproportionation reaction $2 Cu ^{+}( aq ) \rightleftharpoons Cu ( s ) + Cu ^{2+}( aq )$ at $298 \ K$,$\ln K$ (where $K$ is the equilibrium constant) is....... $\times 10^{-1}$.
Given: $(E _{ Cu ^{2+} / Cu ^{+}}^{0} = 0.16 \ V, E _{ Cu ^{+} / Cu }^{0} = 0.52 \ V, \frac{ RT }{ F } = 0.025 \ V)$

In the cell $Pt_{(s)} | H_2(g, 1 \, bar) | HCl_{(aq)} | AgCl_{(s)} | Ag_{(s)} | Pt_{(s)}$,the cell potential is $0.92 \, V$ when a $10^{-6} \, m$ $HCl$ solution is used. The standard electrode potential of the $(AgCl/Ag, Cl^-)$ electrode is ............. $V$ $\{ \text{Given, } \frac{2.303RT}{F} = 0.06 \, V \text{ at } 298 \, K \}$

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

Calculate the cell potential for $Cr_{(s)} | Cr^{3+} (0.1 \, M) || Fe^{2+} (0.01 \, M) | Fe_{(s)}$ at $298 \, K$. Given $E^{\circ}_{Cr^{3+}/Cr} = -0.74 \, V$ and $E^{\circ}_{Fe^{2+}/Fe} = -0.44 \, V$. (in $, V$)

The $emf$ of a $Daniell$ cell at $298 \, K$ is $E_1$:
$Zn | ZnSO_4 \,(0.01 \, M) || CuSO_4 \,(1.0 \, M) | Cu$
When the concentration of $ZnSO_4$ is $1.0 \, M$ and that of $CuSO_4$ is $0.01 \, M$,the $emf$ changes to $E_2$. What is the relation between $E_1$ and $E_2$?