The cell reaction of the given cell is spontaneous if:
$Pt | Cl_2 (P_1 \, atm) | Cl^{-} (1 \, M) || Cl^{-} (1 \, M) | Cl_2 (P_2 \, atm) | Pt$

The Gibbs energy change of the reaction (in $kJ \ mol^{-1}$) corresponding to the following cell $Cr | Cr^{3+} (0.1 \ M) || Fe^{2+} (0.01 \ M) | Fe$ is: (Given: $E^{\circ}_{Cr^{3+}/Cr} = -0.74 \ V$,$E^{\circ}_{Fe^{2+}/Fe} = -0.44 \ V$)

Calculate $E_{cell}^{\circ}$ if the equilibrium constant for the following reaction is $1.2 \times 10^6$.
$2 Cu_{(aq)}^{+} \longrightarrow Cu_{(aq)}^{2+} + Cu_{(s)}$ (in $V$)

The magnitude of the change in oxidising power of the $MnO_4^- / Mn^{2+}$ couple is $x \times 10^{-4} \, V$,if the $H^{+}$ concentration is decreased from $1 \, M$ to $10^{-4} \, M$ at $25^{\circ} C$. (Assume concentration of $MnO_4^-$ and $Mn^{2+}$ to be same on change in $H^{+}$ concentration). The value of $x$ is ....... .
(Rounded off to the nearest integer)
$[\text{Given} : \frac{2.303 RT}{F} = 0.059]$

In which of the following conditions will the reduction potential of a hydrogen half-cell be negative?

In the electrochemical cell $:$
$Zn \,|\,ZnSO_4\,(0.01\,M)\,||\,CuSO_4\,(1.0\,M)\,|\,Cu$
the $emf$ of this Daniell cell is $E_1.$ When the concentration of $ZnSO_4$ is changed to $1.0\,M$ and that of $CuSO_4$ changed to $0.01\,M,$ the $emf$ changes to $E_2.$ From the followings,which one is the relationship between $E_1$ and $E_2$ $?$ (Given,$RT/F = 0.059$)