The cell potential for the following reaction is $0.3305 \ V$ at $298 \ K$. Find the value of $x$ for the reaction: $Zn | Zn^{2+} (0.1 \ M) || Cd^{2+} (x \ M) | Cd$. (Given: $E^{\circ}_{Zn^{2+}/Zn} = -0.76 \ V$, $E^{\circ}_{Cd^{2+}/Cd} = -0.40 \ V$) (in $M$)

Which will increase the voltage of the cell $Sn_{(s)} + 2Ag^{+}_{(aq)} \to Sn^{2+}_{(aq)} + 2Ag_{(s)}$?

At $298 \ K$,the standard reduction potentials are $1.51 \ V$ for $MnO_4^- \ | \ Mn^{2+}$,$1.36 \ V$ for $Cl_2 \ | \ Cl^{-}$,$1.07 \ V$ for $Br_2 \ | \ Br^{-}$ and $0.54 \ V$ for $I_2 \ | \ I^{-}$. At $pH = 3$,permanganate is expected to oxidize $\left( \frac{RT}{F} = 0.059 \ V \right)$

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: $\frac{2.303 \ RT}{F} = 0.059$]

If $E^{\circ}_{\text{cell}}$ for $Cd_{(s)} | Cd_{(1M)}^{2+} || Ag_{(1M)}^{+} | Ag_{(s)}$ is $1.2 \ V$,what is the emf of the cell at $25^{\circ} C$ (in $V$)?

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)