$A$ copper electrode is dipped in a $0.1 \, M$ copper sulfate solution at $25 \, ^\circ C$. Calculate the reduction potential of the copper electrode. $(E^o_{Cu^{2+}/Cu} = 0.34 \, V)$ (in $, V$)

Calculate the $emf$ of the cell at $25^{\circ} C$.
Cell notation: $M | M^{2+} (0.01 \ M) || M^{2+} (0.0001 \ M) | M$
Given: $E_{cell}^{o} = 4 \ V$ and $\frac{RT}{F} \ln 10 = 0.06$. (in $V$)

The half-cell reaction involving the quinhydrone electrode is shown below:
$HO-C_6H_4-OH_{(aq)} \rightleftharpoons O=C_6H_4=O_{(aq)} + 2H^+ + 2e^-$
If $E_{op}^o$ for this electrode is $1.30 \ V$,then what will be the oxidation electrode potential at $pH = 3$? ............ $V$

Calculate the equilibrium constant for the following reaction: $Ni_{(s)} + Cu_{(aq)}^{2+} \to Cu_{(s)} + Ni_{(aq)}^{2+}$. Given: $E_{Ni^{2+}|Ni}^o = -0.25 \ V$ and $E_{Cu^{2+}|Cu}^o = 0.34 \ V$.

$A$ 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 $V$)

The cell potential for the following cell
$Pt \mid H_{2(g)} \mid H^{+}_{(aq)} \parallel Cu^{2+}(0.01 \, M) \mid Cu_{(s)}$
is $0.576 \, V$ at $298 \, K$. The $pH$ of the solution is $......$ (Nearest integer)