At $298 \ K$, the $emf$ of the cell is ............ $V$.
$Pt | H_{2(2 \ atm)} | H_{(0.02 \ M)}^{+} || H_{(0.1 \ M)}^{+} | H_{2(1 \ atm)} | Pt$

The hydrogen electrode is dipped in a solution of $pH=3$ at $25^{\circ} C$. The potential of the electrode will be . . . . . . $\times 10^{-2} \ V$. $\left(\frac{2.303 RT}{F}=0.059 \ V\right)$

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 a cell reaction involving a two-electron change,the standard $EMF$ of the cell is $0.295 \ V$ at $25 \ ^\circ C$. The equilibrium constant of the reaction at $25 \ ^\circ C$ will be:

For a cell involving a one-electron change at $25^o C$,$E^{o}_{cell} = 0.591 \ V$. The equilibrium constant for the reaction is .....

For an electrochemical cell
$Sn_{(s)} | Sn^{2+}(aq, 1 \ M) || Pb^{2+}(aq, 1 \ M) | Pb_{(s)}$
the ratio $\frac{[Sn^{2+}]}{[Pb^{2+}]}$ when this cell attains equilibrium is
(Given $E^{0}_{Sn^{2+}/Sn} = -0.14 \ V$,$E^{0}_{Pb^{2+}/Pb} = -0.13 \ V$,$\frac{2.303 \ RT}{F} = 0.06$)