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 $V$)?

Write a note on the relation between Gibbs free energy and cell potential for a cell reaction.

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 $E_2$.

For the electrochemical cell shown below:
$Pt \mid H_{2}(p=1 \, atm) \mid H^{+}(aq., x \, M) \mid\mid Cu^{2+}(aq., 1.0 \, M) \mid Cu_{(s)}$
The potential is $0.49 \, V$ at $298 \, K$. The $pH$ of the solution is closest to:
[Given: Standard reduction potential,$E^{\circ}$ for $Cu^{2+}/Cu$ is $0.34 \, V$; Gas constant,$R = 8.31 \, J \, K^{-1} \, mol^{-1}$; Faraday constant,$F = 9.65 \times 10^{4} \, J \, V^{-1} \, mol^{-1}$]

$1 \ F$ electricity was passed through $Cu^{2+} (1.5 \ M, 1 \ L) / Cu$ and $0.1 \ F$ was passed through $Ag^{+} (0.2 \ M, 1 \ L) / Ag$ electrolytic cells. After this,the two cells were connected to make an electrochemical cell. The $emf$ of the cell thus formed at $298 \ K$ is:
Given: $E^0_{Cu^{2+} / Cu} = 0.34 \ V$,$E^0_{Ag^{+} / Ag} = 0.8 \ V$,$\frac{2.303 \ RT}{F} = 0.06 \ V$ (in $V$)

Consider the following cell reaction:
$2 Fe^{3+}_{(aq)} + 2 I^{-}_{(aq)} \rightleftharpoons 2 Fe^{2+}_{(aq)} + I_{2(s)}$
At $298 \ K$,the cell emf is $0.237 \ V$. The equilibrium constant for the reaction is $10^x$. The value of $x$ is:
$(F = 96500 \ C \ mol^{-1}; R = 8.3 \ J \ K^{-1} \ mol^{-1})$