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

In the given electrochemical cell, $Ag_{(s)} | AgCl_{(s)} | Cl^-_{(aq)}, Fe^{2+}_{(aq)}, Fe^{3+}_{(aq)} | Pt_{(s)}$ at $298 \ K$, the cell potential $(E_{cell})$ will increase when :
$(A)$ Concentration of $Fe^{2+}$ is increased.
$(B)$ Concentration of $Fe^{3+}$ is decreased.
$(C)$ Concentration of $Fe^{2+}$ is decreased.
$(D)$ Concentration of $Fe^{3+}$ is increased.
$(E)$ Concentration of $Cl^-$ is increased.
Choose the correct answer from the options given below :

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

Write the Nernst equation for the anode and cathode of a Daniell cell.

For a cell reaction involving a two-electron change,the standard emf of the cell is found to be $0.295 \ V$ at $25 \ ^oC$. The equilibrium constant of the reaction at $25 \ ^oC$ will be

The e.m.f. of the cell $Zn | Zn^{2+} (0.01 \ M) || Fe^{2+} (0.001 \ M) | Fe$ at $298 \ K$ is $0.2905 \ V$. The value of the equilibrium constant for the cell reaction is: