What is the value of $E_{cell}$ at $298 \ K$ for the reaction,$Zn_{(s)} + Cu^{+2}(0.1 \ M) \rightarrow Zn^{+2}(0.1 \ M) + Cu_{(s)}$ if $E^{\circ}_{cell} = 1.1 \ V$ (in $V$)?

The magnitude of the change in oxidising power of the $MnO_4^- / Mn^{2+}$ couple is $x \times 10^{-4} \, V$,if the $H^{+}$ concentration is decreased from $1 \, M$ to $10^{-4} \, M$ at $25^{\circ} C$. (Assume concentration of $MnO_4^-$ and $Mn^{2+}$ to be same on change in $H^{+}$ concentration). The value of $x$ is ....... .
(Rounded off to the nearest integer)
$[\text{Given} : \frac{2.303 RT}{F} = 0.059]$

Calculate the cell potential for a $Cu$ plate kept in $0.2 \ M$ $CuSO_4$ solution. Given: $E_{Cu^{2+} \mid Cu}^o = 0.34 \ V$. (in $V$)

For a $Daniel$ cell $Zn | ZnSO_{4(0.01 \ M)} || CuSO_{4(1 \ M)} | Cu$ at $298 \ K$,the cell potential is $E_1$. When the concentrations of $ZnSO_4$ and $CuSO_4$ are changed to $1 \ M$ and $0.01 \ M$ respectively,the cell potential becomes $E_2$. Determine the relationship between $E_1$ and $E_2$.

At $298 \ K$,the standard reduction potentials are $1.51 \ V$ for $MnO_4^- \ | \ Mn^{2+}$,$1.36 \ V$ for $Cl_2 \ | \ Cl^{-}$,$1.07 \ V$ for $Br_2 \ | \ Br^{-}$ and $0.54 \ V$ for $I_2 \ | \ I^{-}$. At $pH = 3$,permanganate is expected to oxidize $\left( \frac{RT}{F} = 0.059 \ V \right)$

The cell reaction involving the quinhydrone electrode is given by the following equation:
$C_6H_4(OH)_2 \rightleftharpoons C_6H_4O_2 + 2H^+ + 2e^-$,$E^{\circ} = 1.30 \ V$
What will be the electrode potential at $pH = 3$ (in $V$)?