Calculate $E_{cell}$ for the following cell:
$Pt_{(s)} | H_{2(g)} | HA, 1 \ M \ (K_a = 10^{-7}) || HB, 1 \ M \ (K_a = 10^{-5}) | H_{2(g)} | Pt_{(s)}$ (in $V$)

$Pt | H_2 (1 \ atm) | H^{+} (0.001 \ M) || H^{+} (0.1 \ M) | H_2 (1 \ atm) | Pt$. What will be the value of $E_{cell}$ for this cell? ............. $V$

For the disproportionation reaction $2 Cu ^{+}( aq ) \rightleftharpoons Cu ( s ) + Cu ^{2+}( aq )$ at $298 \ K$,$\ln K$ (where $K$ is the equilibrium constant) is....... $\times 10^{-1}$.
Given: $(E _{ Cu ^{2+} / Cu ^{+}}^{0} = 0.16 \ V, E _{ Cu ^{+} / Cu }^{0} = 0.52 \ V, \frac{ RT }{ F } = 0.025 \ V)$

The cell potential for $Zn | Zn^{2+}_{(aq)} || Sn^{x+}| Sn$ is $0.801 \ V$ at $298 \ K$. The reaction quotient for the above reaction is $10^{-2}$. The number of electrons involved in the given electrochemical cell reaction is .... (Given $E^{0}_{Zn^{2+}|Zn} = -0.763 \ V, E^{0}_{Sn^{x+}|Sn} = +0.008 \ V$ and $\frac{2.303 \ RT}{F} = 0.06 \ V$)

The reduction potential of a hydrogen electrode at $25^{\circ} C$ in a neutral solution is $(P_{H_2} = 1 \ atm)$. (in $V$)

For the cell at $298 \ K$:
$Ag_{(s)} | AgBr_{(s)} | Br^{-}(0.01 \ M) || I^{-}(0.02 \ M) | AgI_{(s)} | Ag_{(s)}$
The correct information is:
[Given: $K_{sp}(AgBr) = 4 \times 10^{-13}$,$K_{sp}(AgI) = 8 \times 10^{-17}$,$\frac{2.303 \ RT}{F} = 0.06 \ V$,$\log 2 = 0.3$]