For a cell reaction involving two electron changes,$E_{\text{cell}}^{\circ} = 0.3 \text{ V}$ at $25^{\circ}\text{C}$. The equilibrium constant of the reaction is:

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

Consider the cell whose $emf$ is $1.01 \ V$.
$Pt, H_2(1 \ atm) | H^{+}(pH = 4) || Ag^{+}(xM) | Ag$
What is the value of $x$? (Given: $E^o_{Ag^{+}|Ag} = +0.8 \ V$,$\frac{2.303 \ RT}{F} = 0.06$)

$A$ $Daniel$ cell is made at $25\,^oC$ by connecting $Zn/Zn^{2+} (0.1\ M, 1\ L)$ and $Cu/Cu^{2+} (0.9\ M, 1\ L)$ electrodes. The cell is discharged until its $emf$ reaches $1.10\ V$,then it is charged (reversing the discharge process) by passing $0.6\ F$ charge.
[Given: ${E^0}_{Zn^{2+}/Zn} = -0.76\ V, {E^0}_{Cu^{2+}/Cu} = 0.34\ V, \frac{2.303RT}{F} = 0.06, \log 2 = 0.30]$
Select the incorrect option.

What will be the $EMF$ for the given cell?
$Pt | H_2(g, P_1) | H^{+}(aq.) (1 \ M) || H^{+}(aq.) (1 \ M) | H_2(g, P_2) | Pt$

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)
$[\text{Given} : \frac{2.303 RT}{F} = 0.059]$