Consider the cell at $25^{\circ} C$:
$Zn | Zn^{2+}_{(aq)} (1 \ M) || Fe^{3+}_{(aq)}, Fe^{2+}_{(aq)} | Pt_{(s)}$
The fraction of total iron present as $Fe^{3+}$ ion at the cell potential of $1.500 \ V$ is $X \times 10^{-2}$. The value of $X$ is $.....$ (Nearest integer).
(Given $E^{0}_{Fe^{3+} / Fe^{2+}} = 0.77 \ V, E^{0}_{Zn^{2+} / Zn} = -0.76 \ V$)

In acidic medium,$MnO_4^-$ acts as an oxidising agent: $MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O$. If the $H^+$ ion concentration is doubled,the electrode potential of the half-cell will:

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.

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)

Give the equation to calculate the equilibrium constant $K_C$ of a Daniell cell.