Consider the cell $Pt | H_2(P_1 \ atm) | H^{+}(X_1 \ M) || H^{+}(X_2 \ M) | H_2(P_2 \ atm) | Pt$. The cell reaction will be spontaneous if

Assume a cell with the following reaction:
$Cu_{(s)} + 2 Ag^{+} (1 \times 10^{-3} \, M) \rightarrow Cu^{2+} (0.250 \, M) + 2 Ag_{(s)}$
$E_{Cell}^{\ominus} = 2.97 \, V$
$E_{cell}$ for the above reaction is $.... \, V.$ (Nearest integer)
[Given: $\log 2.5 = 0.3979, T = 298 \, K]$

Under which of the following conditions is the $E$ value of the cell for the given reaction maximum?
$Zn_{(s)} + Cu^{2+}_{(aq)} \rightleftharpoons Cu_{(s)} + Zn^{2+}_{(aq)}$
$\left( \frac{2.303 RT}{F} \text{ at } 298 \ K = 0.059 \ V, E^{\circ}_{Zn^{2+}/Zn} = -0.76 \ V, E^{\circ}_{Cu^{2+}/Cu} = +0.34 \ V \right)$
Let $[Zn^{2+}] = C_2$ and $[Cu^{2+}] = C_1$.

For a cell involving a one-electron change at $25^o C$,$E^{o}_{cell} = 0.591 \ V$. The equilibrium constant for the reaction is .....

Fill in the blanks :
$1.$ The ratio of concentration of products to concentration of reactants is ........
$2.$ $\ln(\log(x)) =$ ............
$3.$ At equilibrium,between $E_{cell}$ and $E_{cell}^{o}$,......... will be zero.

The cell reaction involving the quinhydrone electrode is given by the following equation:
$C_6H_4(OH)_2 \rightleftharpoons C_6H_4O_2 + 2H^+ + 2e^-$,$E^{\circ} = 1.30 \ V$
What will be the electrode potential at $pH = 3$ (in $V$)?