Calculate the $EMF$ of the cell: $Cr | Cr^{+3}(0.1 \, M) || Fe^{+2}(0.01 \, M) | Fe$
(Given: $E^o_{Cr^{+3}|Cr} = -0.75 \, V$,$E^o_{Fe^{+2}|Fe} = -0.45 \, V$) (in $, V$)

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

Calculate the $pH$ of an $HCl$ solution at $298 \ K$ for the following cell:
$Pt_{(s)} \mid H_2 \ (1 \ bar) \mid HCl \ (xM) \parallel Ag^+ \ (0.01 \ M) \mid Ag_{(s)}$
Given that the standard cell potential $E^\circ_{cell} = 1.05 \ V$. (in $.73$)

For the cell reaction,$Zn_{(s)} + 2 Ag_{(aq)}^{+} \longrightarrow Zn_{(aq)}^{2+} + 2 Ag_{(s)}$,the cell potential is less than $E^{\circ}_{cell}$ by $0.0592 \ V$ at $298 \ K$ when:

Assume a cell with the following reaction:
$Cu_{(s)} + 2 Ag^{+} (1 \times 10^{-3} \, M) \rightarrow Cu^{2+} (0.250 \, M) + 2 Ag_{(s)}$
$E_{Cell}^{\ominus} = 2.97 \, V$
$E_{cell}$ for the above reaction is $.... \, V.$ (Nearest integer)
[Given: $\log 2.5 = 0.3979, T = 298 \, K]$

For the cell, $Pt | Cl_{2(g)} (0.4 \ bar) | Cl^{-} (aq.) (0.1 \ M) || Cl^{-} (aq.) (0.01 \ M) | Cl_{2(g)} (0.2 \ bar) | Pt$, the measured potential at $298 \ K$ is .............. $V$.