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

Consider the following Galvanic cell. By what value does the cell voltage change when the concentration of ions in both the anodic and cathodic compartments is increased by a factor of $10$ at $298 \ K?$

The cell potential for the given cell at $298 \, K$ is $Pt \mid H_2 (g, 1 \, bar) \mid H^{+}_{(aq)} \parallel Cu^{2+}_{(aq)} \mid Cu_{(s)}$. The cell potential is $0.31 \, V$. The $pH$ of the acidic solution is $3$,and the concentration of $Cu^{2+}$ is $10^{-x} \, M$. The value of $x$ is $.....$ (Given: $E^{\ominus}_{Cu^{2+}/Cu} = 0.34 \, V$ and $\frac{2.303 RT}{F} = 0.06 \, V$)

Which Nernst equation is correct for the following cell? $Al_{(s)}|Al_{(aq)}^{3+} || Zn_{(aq)}^{2+}| Zn_{(s)}$

Cell potential for the following reaction at $298 \ K$ is $0.096 \ V$. Calculate the $pH$ of the $HCl$ solution for the cell: $Sn_{(s)} | Sn^{2+} (0.05 \ M) || H^{+} (x \ M) | H_{2(g)} (1 \ bar) | Pt$ given $E_{Sn^{2+}|Sn}^{\circ} = -0.14 \ V$ and $E_{H^{+}|H_2}^{\circ} = 0.0 \ V$.

The $\text{EMF}$ of a galvanic cell consisting of two hydrogen electrodes is $0.17 \ V$. If the solution of one of the electrodes has $[H^+] = 10^{-3} \ M$,the $\text{pH}$ at the other electrode is: