The standard cell potential of a cell reaction involving the transfer of $2$ electrons at $25^{\circ}C$ is $0.295 \ V$. What is its equilibrium constant at $25^{\circ}C$?

In acidic medium,$MnO_4^-$ acts as an oxidising agent: $MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O$. If the $H^+$ ion concentration is doubled,the electrode potential of the half-cell will:

If $E^{\circ}(Al_{(aq)}^{+3} \mid Al_{(s)}) = -1.66 \ V$. What is the potential of $Al_{(s)} \rightarrow Al_{(aq)}^{+3}(0.1 \ M) + 3e^-$ at $298 \ K$?

The logarithm of the equilibrium constant for the reaction $Pd^{2+}{(aq)} + 4Cl^{-}{(aq)} \rightleftharpoons PdCl_4^{2-}{(aq)}$ is (Nearest integer).
Given: $\frac{2.303 RT}{F} = 0.06 \ V$
$Pd^{2+}{(aq)} + 2e^{-} \rightleftharpoons Pd_{(s)} \quad E^{\circ} = 0.83 \ V$
$PdCl_4^{2-}{(aq)} + 2e^{-} \rightleftharpoons Pd_{(s)} + 4Cl^{-}{(aq)} \quad E^{\circ} = 0.65 \ V$

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 $E^{0}_{Sn^{2+}/Sn} = -0.14 \ V$,$E^{0}_{Pb^{2+}/Pb} = -0.13 \ V$,$\frac{2.303 \ RT}{F} = 0.06$)

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
$E^0_{Sn(OH)_6^{2-}/HSnO_2^-, OH^-} = -0.90 \ V$.