At $298 \ K$ the equilibrium constant for the reaction $M_{(s)} + 2 Ag^{+}_{(aq)} \rightarrow M^{2+}_{(aq)} + 2 Ag_{(s)}$ is $10^{15}$. What is the $E_{cell}^{\ominus}$ (in $V$) for this reaction? $\left(\frac{2.303 RT}{F}\right) = 0.06 \ V$

$A$ $Daniel$ cell is made at $25\,^oC$ by connecting $Zn/Zn^{2+} (0.1\ M, 1\ L)$ and $Cu/Cu^{2+} (0.9\ M, 1\ L)$ electrodes. The cell is discharged until its $emf$ reaches $1.10\ V$,then it is charged (reversing the discharge process) by passing $0.6\ F$ charge.
[Given: ${E^0}_{Zn^{2+}/Zn} = -0.76\ V, {E^0}_{Cu^{2+}/Cu} = 0.34\ V, \frac{2.303RT}{F} = 0.06, \log 2 = 0.30]$
Select the incorrect option.

The standard $emf$ of a galvanic cell involving $2$ moles of electrons in its redox reaction is $0.59 \ V$. The equilibrium constant for the redox reaction of the cell is

If $E^{\circ}(Ag^{+}_{(aq)} \mid Ag_{(s)}) = +0.80 \ V$,what is the potential developed for $Ag_{(s)} \rightarrow Ag^{+}_{(aq)} (0.01 \ M) + e^{-}$ at $298 \ K$?

Write the Nernst equation and calculate the $emf$ of the following cells at $298 \, K$:
$(i) \; Mg_{(s)} | Mg^{2+}(0.001 \, M) || Cu^{2+}(0.0001 \, M) | Cu_{(s)}$
$(ii) \; Fe_{(s)} | Fe^{2+}(0.001 \, M) || H^{+}(1 \, M) | H_{2(g)}(1 \, bar) | Pt_{(s)}$
$(iii) \; Sn_{(s)} | Sn^{2+}(0.050 \, M) || H^{+}(0.020 \, M) | H_{2(g)}(1 \, bar) | Pt_{(s)}$
$(iv) \; Pt_{(s)} | Br_{2(l)} | Br^{-}(0.010 \, M), H^{+}(0.030 \, M) || H_{2(g)}(1 \, bar) | Pt_{(s)}$

What is the potential of the cell consisting of two hydrogen electrodes as shown below (in $V$)?
$Pt | H_2(g) | H^+_{(aq)} (10^{-8} \ M) || H^+_{(aq)} (0.001 \ M) | H_2(g) | Pt$