The standard emf for the cell $Cd_{(s)}|Cd^{2+}_{(aq)}(1 \ M)||Cu^{2+}_{(aq)}(1 \ M)|Cu_{(s)}$ is $0.74 \ V$. If the concentration of $Cd^{2+}_{(aq)}$ and $Cu^{2+}_{(aq)}$ both decrease by $10$ times at $298 \ K$,calculate the emf of the cell.

For the cell reaction,$Cu | Cu^{2+}(0.1 \ M) || Cu^{2+}(1.0 \ M) | Cu$,the emf of the cell at $25^{\circ}C$ is given that $E^{\circ}_{Cu^{2+}/Cu} = 0.34 \ V$. (in $V$)

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

Calculate the equilibrium constant of the reaction:
$Cu_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + 2Ag_{(s)}$
Given $E^{\Theta}_{cell} = 0.46 \ V$

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: $K_{sp}(AgBr) = 4 \times 10^{-13}$,$K_{sp}(AgI) = 8 \times 10^{-17}$,$\frac{2.303 \ RT}{F} = 0.06 \ V$,$\log 2 = 0.3$]

For a half-cell containing a $Pt$ rod immersed in a solution of $1 \ M$ $HA$,$O_{2(g)}$ is bubbled at $1 \ atm$. The standard reduction potential for water formation is $1.23 \ V$. Given a dissociation constant,$K_a = 1 \times 10^{-4}$ for $HA$,what is $E_{\text{Half-cell}}$ at $298 \ K$ in $V$?