Consider the following electrode processes of a cell: $Cl^{-} \rightarrow \frac{1}{2} Cl_2 + e^{-}$ and $MCl + e^{-} \rightarrow M + Cl^{-}$. If the $EMF$ of this cell is $-1.140 \ V$ and the $E^{\circ}$ value of the cell is $-0.55 \ V$ at $298 \ K$,the value of the equilibrium constant $(K_{sp})$ of the sparingly soluble salt $MCl$ is in the order of:

An electrochemical cell consists of the following two redox couples, $M^{x+}(aq)/M(s)$ $[E^{\ominus}_{red} = +0.15 \text{ V}]$ and $Fe^{3+}(aq)/Fe(s)$ $[E^{\ominus}_{red} = -0.036 \text{ V}]$. The cell $EMF$ is recorded to be $0.2057 \text{ V}$. If the reaction quotient of the electrochemical reaction is found to be $10^{-2}$, then the value of $x$ is . . . . . . . (Nearest integer) [Given: $M$ is a $p$-block metal and $\frac{2.303RT}{F} = 0.059 \text{ V}$]

$E_{1}, E_{2}, E_{3}$ are the $EMF$ values of three galvanic cells with the reaction $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$ at $298 \ K$ with different concentrations: $(I) [Zn^{2+}] = 1 \ M, [Cu^{2+}] = 0.1 \ M$; $(II) [Zn^{2+}] = 1 \ M, [Cu^{2+}] = 1 \ M$; $(III) [Zn^{2+}] = 0.1 \ M, [Cu^{2+}] = 1 \ M$. Compare the $EMF$ values.

Calculate the cell potential for a $Cu$ plate kept in $0.2 \ M$ $CuSO_4$ solution. Given: $E_{Cu^{2+} \mid Cu}^o = 0.34 \ V$. (in $V$)

The cell potential for the given cell at $298 \, K$ is $Pt \mid H_2 (g, 1 \, bar) \mid H^{+}_{(aq)} \parallel Cu^{2+}_{(aq)} \mid Cu_{(s)}$. The cell potential is $0.31 \, V$. The $pH$ of the acidic solution is $3$,and the concentration of $Cu^{2+}$ is $10^{-x} \, M$. The value of $x$ is $.....$ (Given: $E^{\ominus}_{Cu^{2+}/Cu} = 0.34 \, V$ and $\frac{2.303 RT}{F} = 0.06 \, V$)

The e.m.f. of the cell $Ag | Ag^{+}(0.1 \ M) || Ag^{+}(1 \ M) | Ag$ at $298 \ K$ is ........... $V$.