The standard e.m.f. of a cell,involving one electron change,is found to be $0.591 \ V$ at $25 \ ^oC$. The equilibrium constant of the reaction is: $(F = 96,500 \ C \ mol^{-1}; R = 8.314 \ J \ K^{-1} \ mol^{-1})$

Calculate the cell potential at $298 \ K$ for the following cell:
$Zn_{(s)} | Zn^{2+} (0.6 \ M) || Cu^{2+} (0.3 \ M) | Cu_{(s)} \quad [E_{cell}^{o} = 1.1 \ V]$

For the electrochemical cell,$Mg_{(s)} \mid Mg^{2+}(aq, 1 \ M) \parallel Cu^{2+}(aq, 1 \ M) \mid Cu_{(s)}$,the standard emf of the cell is $2.70 \ V$ at $300 \ K$. When the concentration of $Mg^{2+}$ is changed to $x$,the cell potential changes to $2.67 \ V$ at $300 \ K$. The value of $x$ is.
(Given: $\frac{F}{R} = 11500 \ K \ V^{-1}$,where $F$ is the Faraday constant and $R$ is the gas constant; $\ln(10) = 2.30$)

$Cu_{(s)} | Cu^{+2}(aq, 10^{-3} M) || Ag^{+}(aq, 10^{-5} M) | Ag_{(s)}$
If $E^{o}_{Cu^{+2}/Cu} = +0.34 \ V$
$E^{o}_{Ag^{+}/Ag} = +0.80 \ V$
$E_{cell}$ will be

Consider the following cell reaction:
$2 Fe^{3+}_{(aq)} + 2 I^{-}_{(aq)} \rightleftharpoons 2 Fe^{2+}_{(aq)} + I_{2(s)}$
At $298 \ K$,the cell emf is $0.237 \ V$. The equilibrium constant for the reaction is $10^x$. The value of $x$ is:
$(F = 96500 \ C \ mol^{-1}; R = 8.3 \ J \ K^{-1} \ mol^{-1})$

The cell potential for the following cell
$Pt \mid H_{2(g)} \mid H^{+}_{(aq)} \parallel Cu^{2+}(0.01 \, M) \mid Cu_{(s)}$
is $0.576 \, V$ at $298 \, K$. The $pH$ of the solution is $......$ (Nearest integer)