For the concentration cell: $Cu | Cu^{2+} (0.01 \ M) || Cu^{2+} (0.1 \ M) | Cu$,calculate the $E_{cell}$.

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 .....

What will be the electrode potential of a $Cu$ electrode dipped in $0.025 \ M \ CuSO_4$ solution at $298 \ K$,given that the standard reduction potential of $Cu$ is $0.34 \ V$ (in $V$)?

At $298 \, K$ temperature,a hydrogen gas electrode is made by dipping platinum wire in a solution of $HCl$ of $pH = 10$ and by passing hydrogen gas around the platinum wire at $1 \, atm$ pressure. The potential of the electrode would be ......... $V$.

$Pt_{(s)} | Fe^{+2}(10^{-2} \ M, aq), Fe^{+3}(10^{-3} \ M, aq) || MnO_4^-(10^{-3} \ M, aq), Mn^{+2}(10^{-2} \ M, aq) | Pt_{(s)}$
At $298 \ K$,$E^o_{cell}$ for the cell is $-2.31 \ V$. What will be the $E_{cell}$ at $pH = 1$? .......... $Volt$

Calculate the emf of the half-cell given below: $Pt(s) | H_2(g, 2 \text{ atm}) | HCl(aq, 0.02 \text{ M})$, $E^\circ_{H^+/H_2} = 0 \text{ V}$. (Given: $\frac{2.303RT}{F} = 0.059$, $\log 2 = 0.3010$)