For the cell reaction $Sn_{(s)} + 2Ag^+_{(aq)} \rightarrow Sn^{2+}_{(aq)} + 2Ag_{(s)}$,what happens when the cell voltage increases?

Derive an equation for the equilibrium constant $K_C$ of any galvanic cell (redox reaction) and also state its uses.

If hydrogen electrodes dipped in two solutions of $pH=3$ and $pH=6$ are connected by a salt bridge,the $emf$ of the resulting cell is (in $V$)

For the reaction $Cr_{2}O_{7}^{2-} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_{2}O$,the standard electrode potential is $E^{\circ} = 1.33 \ V$. Given the concentrations $[Cr_{2}O_{7}^{2-}] = 4.5 \ mmol$,$[Cr^{3+}] = 15 \ mmol$,and the cell potential $E = 1.067 \ V$,the $pH$ of the solution is nearly equal to:

$Pt_{(s)} | H_{2(g)}(1 \ bar) | H^{+}_{(aq)}(1 \ M) || M^{3+}_{(aq)}, M^{+}_{(aq)} | Pt_{(s)}$
The $E_{cell}$ for the given cell is $0.1115 \ V$ at $298 \ K$ when $\frac{[M^{+}_{(aq)}]}{[M^{3+}_{(aq)}]} = 10^{a}$.
The value of $a$ is.
Given : $E^{\circ}_{M^{3+}/M^{+}} = 0.2 \ V$
$\frac{2.303 \ RT}{F} = 0.059 \ V$

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)