Calculate the $E.M.F.$ of the following cell at $298 \ K$: $Zn_{(s)} | ZnSO_4(0.01 \ M) || CuSO_4(1.0 \ M) | Cu_{(s)}$ if $E^o_{cell} = 2.0 \ V$. (in $V$)

At $298 \ K$ the equilibrium constant for the reaction $M_{(s)} + 2 Ag^{+}_{(aq)} \rightarrow M^{2+}_{(aq)} + 2 Ag_{(s)}$ is $10^{15}$. What is the $E_{cell}^{\ominus}$ (in $V$) for this reaction? $\left(\frac{2.303 RT}{F}\right) = 0.06 \ V$

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

The name of the equation showing the relation between electrode potential $(E)$,standard electrode potential $(E^o)$,and concentration of ions in solution is

If $Zn/Zn^{+2}$ electrode is diluted $100$ times,then the change in electromotive force will be:

The $emf$ of a $Daniel$ cell at $298 \ K$ is ${E_1}$ for the cell reaction $Zn|ZnSO_4(0.01 \ M)||CuSO_4(1.0 \ M)|Cu$. When the concentration of $ZnSO_4$ is $1.0 \ M$ and that of $CuSO_4$ is $0.01 \ M$,the $emf$ changes to ${E_2}$. What is the relationship between ${E_1}$ and ${E_2}$?