Calculate the potential of a hydrogen electrode in contact with a solution whose $pH = 10$.

What will be the reduction potential of $Cu$ in an aqueous solution with $pH = 12$? Given that the $K_{sp}$ of $Cu(OH)_2$ is $1 \times 10^{-19}$ and $E^o_{Cu^{2+}/Cu} = 0.34 \ V$.

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.

Which of the following will increase the voltage of the cell represented by the equation
$Cu_{(s)} + 2Ag^{+}_{(aq)} \to Cu^{2+}_{(aq)} + 2Ag_{(s)}$

If $E^{\circ}(Al_{(aq)}^{+3} \mid Al_{(s)}) = -1.66 \ V$. What is the potential of $Al_{(s)} \rightarrow Al_{(aq)}^{+3}(0.1 \ M) + 3e^-$ at $298 \ K$?

What is the potential of a cell containing two hydrogen electrodes,the negative one in contact with $10^{-8} \ M \ H^{+}$ and the positive one in contact with $0.025 \ M \ H^{+}$? (in $V$)