For the redox reaction occurring in a cell: $Zn_{(s)} + Cu^{2+}(0.1 \ M) \to Zn^{2+}(1 \ M) + Cu_{(s)}$,if $E^o_{cell} = 1.10 \ V$,calculate the value of $E_{cell}$ in $V$. (Given: $2.303 \frac{RT}{F} = 0.0591$)

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 $emf$ of cell $Tl | Tl^+_{(0.001M)} || Cu^{2+}_{(0.01M)} | Cu$ is $0.83 \ V$ at $298 \ K$. It could be increased by:

The hydrogen electrode is dipped in a solution of $pH=3$ at $25^{\circ} C$. The potential of the electrode will be . . . . . . $\times 10^{-2} \ V$. $\left(\frac{2.303 RT}{F}=0.059 \ V\right)$

For the cell $Co|Co^{2+}(C_2)||Co^{2+}(C_1)|Co$,$\Delta G$ is negative if

In an electrochemical reaction of lead,at standard temperature,if $E^0_{(Pb^{2+}/Pb)} = m \ V$ and $E^0_{(Pb^{4+}/Pb)} = n \ V$,then the value of $E^0_{(Pb^{2+}/Pb^{4+})}$ is given by $m - x n$. The value of $x$ is $...........$. (Nearest integer)