At $298 \ K$,if the $emf$ of the cell corresponding to the reaction,$Zn_{(s)} + 2H^+_{(aq)} \rightarrow Zn^{2+}(0.01 \ M) + H_{2(g)}(1 \ atm)$ is $0.28 \ V$,then the $pH$ of the solution at the hydrogen electrode is (Given: $\frac{2.303 \ RT}{F} = 0.06 \ V$,$E^o_{Zn^{2+}|Zn} = -0.76 \ V$)

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 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: $K_{sp}(AgBr) = 4 \times 10^{-13}$,$K_{sp}(AgI) = 8 \times 10^{-17}$,$\frac{2.303 \ RT}{F} = 0.06 \ V$,$\log 2 = 0.3$]

The $EMF$ of the cell $M | M^{n+} (0.02 \, M) || H^{+} (1 \, M) | H_{2(g)} (1 \, atm), Pt$ at $25 \, ^\circ C$ is $0.81 \, V$. Calculate the valency of the metal $(n)$ if the standard oxidation potential of the metal is $0.76 \, V$. (Use $\frac{2.303 \, RT}{F} = 0.06, \log \, 2 = 0.3$)

Considering the cell $Cu | Cu^{2+} || Ag^{+} | Ag$,what happens to the $emf$ if the concentrations of both $Cu^{2+}$ and $Ag^{+}$ ions are increased by a factor of $10$?

Consider the following electrochemical cell at $298 \ K$:
$Pt | HSnO_2^-(aq) | Sn(OH)_6^{2-}(aq) || Bi_2O_3(s) | Bi(s)$.
If the reaction quotient at a given time is $10^6$, then the cell $EMF$ $(E_{\text{cell}})$ is . . . . . . $\times 10^{-1} \ V$ (Nearest integer).
Given the standard half-cell reduction potential as
$E^0_{Bi_2O_3/Bi, OH^-} = -0.44 \ V$ and
$E^0_{Sn(OH)_6^{2-}/HSnO_2^-, OH^-} = -0.90 \ V$.