By how much will the potential of the half-cell $Cu^{2+}/Cu$ change if the solution is diluted to $100$ times at $298 \ K$?

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

$E^0 = \frac{RT}{nF} \ln K_{eq}$. This is called

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.

The $\text{EMF}$ of a galvanic cell consisting of two hydrogen electrodes is $0.17 \ V$. If the solution of one of the electrodes has $[H^+] = 10^{-3} \ M$,the $\text{pH}$ at the other electrode is:

The following reaction takes place at $298 \, K$ in an electrochemical cell involving two metals $A$ and $B$,
$A^{2+}_{(aq)} + B_{(s)} \rightarrow B^{2+}_{(aq)} + A_{(s)}$
with $[A^{2+}] = 4 \times 10^{-3} \, M$ and $[B^{2+}] = 2 \times 10^{-3} \, M$ in the respective half-cells,the cell $EMF$ is $1.091 \, V$.
The equilibrium constant of the reaction is closest to