What is the reduction electrode potential (in volts) of a copper electrode when $[Cu^{2+}]=0.01 \ M$ is in a solution at $25^{\circ} C$? $(E^{\circ}$ of $Cu^{2+}/Cu$ electrode is $+0.34 \ V)$

$Cu^{2+} + 2e^- \to Cu$. On increasing $[Cu^{2+}]$ concentration,electrode potential

Calculate the $emf$ of the cell at $25^{\circ} C$.
Cell notation: $M | M^{2+} (0.01 \ M) || M^{2+} (0.0001 \ M) | M$
Given: $E_{cell}^{o} = 4 \ V$ and $\frac{RT}{F} \ln 10 = 0.06$. (in $V$)

For a cell reaction involving a two-electron change,the standard $EMF$ of the cell is $0.295 \ V$ at $25 \ ^\circ C$. The equilibrium constant of the reaction at $25 \ ^\circ C$ will be:

The electrode potential of the following half cell at $298 \ K$ is given by the cell reaction:
$X | X^{2+}(0.001 \ M) || Y^{2+}(0.01 \ M) | Y$
The cell potential is $....... \times 10^{-2} \ V$ (Nearest integer).
Given: $E^0_{X^{2+} | X} = -2.36 \ V$,$E^0_{Y^{2+} | Y} = +0.36 \ V$,$\frac{2.303 \ RT}{F} = 0.06 \ V$.

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.