What will be the reduction potential of $Cu$ in an aqueous solution with $pH = 12$? Given that the $K_{sp}$ of $Cu(OH)_2$ is $1 \times 10^{-19}$ and $E^{\circ}_{Cu^{+2}/Cu} = 0.34 \ V$.

$A$ solution containing $4.5 \ mM$ of $MnO_4^{-}$ and $15 \ mM$ of $Mn^{2+}$ shows $pH$ of $2$. The potential of the half-cell reaction is $......$. (Given: $\log 15 = 1.176$,$\log 4.5 = 0.653$,and standard potential of $MnO_4^{-} \longrightarrow Mn^{2+}$ is $1.51 \ V$) (in $V$)

For a cell involving one electron $E_{cell}^{\ominus} = 0.59 \; V$ at $298 \; K$,the equilibrium constant for the cell reaction is.
Given that $\frac{2.303 \; RT}{F} = 0.059 \; V$ at $T = 298 \; K$.

Which one of the following has a potential more than zero?

The correct representation of Nernst's equation for the reduction of a metal ion $M^{n+}$ to metal $M$ is:

For the following electrochemical cell at $298 \ K$,
$Pt_{(s)} \mid H_2(g, 1 \ bar) \mid H^{+}(aq, 1 \ M) \parallel M^{4+}_{(aq)}, M^{2+}_{(aq)} \mid Pt_{(s)}$
$E_{\text{cell}} = 0.092 \ V$ when $\frac{[M^{2+}_{(aq)}]}{[M^{4+}_{(aq)}]} = 10^x$
Given : $E^0_{M^{4+}/M^{2+}} = 0.151 \ V$; $2.303 \frac{RT}{F} = 0.059 \ V$
The value of $x$ is