The hydrogen electrode is dipped in a solution of $pH = 3$ at $25\,^oC$. The potential of the cell would be ............. $V$ (the value of $2.303\,RT/F$ is $0.059\,V$).

Represent the cell in which the following reaction takes place:
$Mg_{(s)} + 2Ag^{+}(0.0001 \, M) \rightarrow Mg^{2+}(0.130 \, M) + 2Ag_{(s)}$
Calculate its $E_{cell}$ if $E^{\Theta}_{cell} = 3.17 \, V$.

Calculate the cell potential for the following cell: ............... $V$
$Cr_{(s)} | Cr^{3+}_{(0.1 \, M)} || Fe^{2+}_{(0.01 \, M)} | Fe_{(s)}$
Given: $E^0_{Cr^{3+}|Cr} = -0.72 \, V$,$E^0_{Fe^{2+}|Fe} = -0.42 \, V$ (in $, V$)

The equilibrium constant of the reaction $Cu_{(s)} + 2Ag^{+}_{(aq)} \to Cu^{2+}_{(aq)} + 2Ag_{(s)}$ with $E^{\circ} = 0.46 \ V$ at $298 \ K$ is approximately:

Calculate $E_{cell}$ for the following cell:
$Pt_{(s)} | H_{2(g)} | HA, 1 \ M \ (K_a = 10^{-7}) || HB, 1 \ M \ (K_a = 10^{-5}) | H_{2(g)} | Pt_{(s)}$ (in $V$)

What is the potential of the cell containing two hydrogen electrodes as represented below $V$:
$Pt | \frac{1}{2} H_{2(g)} | H^{+} (10^{-8} M) || H^{+} (10^{-3} M) | \frac{1}{2} H_{2(g)} | Pt$