$Zn_{(s)} + Cl_2(1 \ atm) \to Zn^{2+} + 2Cl^-$. The $E^0_{cell}$ of the cell is $2.12 \ V$. To increase $E_{cell}$:

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)

What is $E_{cell}$ (in $V$) of the following cell at $298 \ K$ ?
$(E^{\ominus}_{Zn^{2+}/Zn} = -0.76 \ V ; E^{\ominus}_{Ni^{2+}/Ni} = -0.25 \ V ; \frac{2.303 RT}{F} = 0.06 \ V)$
$Zn_{(s)} | Zn^{2+} (0.01 \ M) || Ni^{2+} (0.1 \ M) | Ni_{(s)}$

If $E_{cell} = 0.118 \, V$ for the following reaction,calculate $[H^{+}]$ and $pH$ at $298 \, K$ temperature.
$Pt \mid H_2(1 \, bar) \mid H^{+} (10^{-6} \, M) \parallel H^{+} (x \, M) \mid H_2 (1 \, bar) \mid Pt$

Consider the following cell reaction:
$2Fe_{(s)} + O_{2(g)} + 4H^{+}_{(aq)} \to 2Fe^{2+}_{(aq)} + 2H_2O_{(l)}$; $E^o = 1.67 \ V$
At $[Fe^{2+}] = 10^{-3} \ M$,$p(O_2) = 0.1 \ atm$ and $pH = 3$,the cell potential at $25 \ ^oC$ is .............. $V$.

Derive the Nernst equation for calculating $E_{cell}$ of a Daniell cell and explain the effect on $E_{cell}$ when there is a change in the concentration of $Zn^{2+}$ and $Cu^{2+}$ ions.