The value of the reaction quotient $(Q)$ for the cell $Zn_{(s)} | Zn^{2+}(0.01 \ M) || Cu^{2+}(1.25 \ M) | Cu_{(s)}$ is:

Calculate the equilibrium constant $(K_C)$ for the cell obtained by connecting two electrodes with standard electrode potentials $E^o_{(Sn^{2+}|Sn)} = -0.14 \ V$ and $E^o_{(Ni^{2+}|Ni)} = -0.23 \ V$ at $298 \ K$.

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

For the cell $Cu_{(s)}|Cu^{2+}_{(aq)}(0.1 \ M) || Ag^{+}_{(aq)}(0.01 \ M)| Ag_{(s)}$,the cell potential $E_{1} = 0.3095 \ V$. For the cell $Cu_{(s)}|Cu^{2+}_{(aq)}(0.01 \ M) || Ag^{+}_{(aq)}(0.001 \ M)| Ag_{(s)}$,the cell potential $= ..... \times 10^{-2} \ V$. (Round off to the Nearest Integer). [Use: $\frac{2.303 \ RT}{F} = 0.059$]

The standard $emf$ of a galvanic cell involving $2$ moles of electrons in its redox reaction is $0.59 \ V$. The equilibrium constant for the redox reaction of the cell is

For the cell $Fe | Fe^{+2} (x \, M) || Cu^{+2} (0.01 \, M) | Cu$,if $E_{cell} = 0.78 \, V$,$E^{\circ}_{Fe^{+2} | Fe} = -0.44 \, V$,and $E^{\circ}_{Cu^{+2} | Cu} = +0.34 \, V$,determine the value of $x$.