What happens to the $emf$ of the cell $Zn_{(s)} | Zn^{+2} (1 \ M) || Ag^{+1} (1 \ M) | Ag_{(s)}$ if the concentration of $Ag^{+1}$ decreases to $0.1 \ M$?

Consider the following electrochemical cell,$Zn_{(s)} + 2Ag^{+}(0.04\, M) \longrightarrow Zn^{2+}(0.28\, M) + 2Ag_{(s)}$. If $E_{\text{cell}}^{\circ} = 2.57\, V$,then the emf of the cell at $298\, K$ is $......\, V$. (in $.5$)

Give the Nernst equation for the $E_{cell}$ of a Nickel-Copper galvanic cell.

If the $E^{\circ}_{cell}$ of an equilibrium reaction $A_{(s)} + 2B^{2+}_{(aq)} \rightleftharpoons A^{2+}_{(aq)} + 2B_{(s)}$ at $298 \ K$ is $0.59 \ V$,the equilibrium constant $K_c$ is

At $298 \ K$,some standard electrode potentials are given below:

$Pb^{2+} / Pb$	$-0.13 \ V$
$Ni^{2+} / Ni$	$-0.24 \ V$
$Cd^{2+} / Cd$	$-0.40 \ V$
$Fe^{2+} / Fe$	$-0.44 \ V$

Metal rods $X$ and $Y$ are inserted into a solution containing $0.001 \ M$ $X^{2+}$ and $0.1 \ M$ $Y^{2+}$ at $298 \ K$ and connected by a conducting wire. This results in the dissolution of $X$. The correct combination$(s)$ of $X$ and $Y$ are,respectively:
(Given: Gas constant,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$,Faraday constant,$F = 96500 \ C \ mol^{-1}$)
$(A) \ Cd$ and $Ni \ \ (B) \ Cd$ and $Fe \ \ (C) \ Ni$ and $Pb \ \ (D) \ Ni$ and $Fe$

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