The reduction potential of a hydrogen electrode at $25^{\circ}C$ is ............... $V$ $(P_{H_2} = 1 \ atm; [H^+] = 0.1 \ M)$.

If a solution of $Cu^{+2}/Cu$ at $298 \, K$ is diluted $100$ times,how will the electrode potential change?

The reduction potential of a hydrogen half-cell will be negative if:

For the cell,$Zn_{(s)} | Zn^{2+} (1 \ M) || Ag^{+} (1 \ M) | Ag_{(s)}$. If the concentration of $Zn^{2+}$ decreases to $0.1 \ M$ at $298 \ K$,then the $EMF$ of the cell:

The equilibrium constant for the following general reaction is $10^{30}$. Calculate $E^{o}$ for the cell at $298 \ K$ ............ $V$
$2X_{2(s)} + 3Y^{2+}_{(aq)} \to 2{X_{2}}^{3+}_{(aq)} + 3Y_{(s)}$

For the cell at $298 \ K$:
$Ag_{(s)} | AgBr_{(s)} | Br^{-}(0.01 \ M) || I^{-}(0.02 \ M) | AgI_{(s)} | Ag_{(s)}$
The correct information is:
[Given: $K_{sp}(AgBr) = 4 \times 10^{-13}$,$K_{sp}(AgI) = 8 \times 10^{-17}$,$\frac{2.303 \ RT}{F} = 0.06 \ V$,$\log 2 = 0.3$]