Refer to the following standard reduction half$-$cell potentials at $25C$ :

$N{i}^{2+}(aq)+2e^{-},Ni(s),E=-0\mathrm{.}23V$

$Cl{O}_2(g)+e^{-},Cl{O}_2^{-}(aq),E=0\mathrm{.}95V$

Part A

An electrochemical cell is based on these two half$-$reactio...

$Ox$ :

$Ni(s)N{i}^{2+}(aq,5\mathrm{.}02(mol)L^{-1})+2e$

Red: $Cl{O}_2(g,\frac{0\mathrm{.}170}{b}ar)+e^{-}Cl{O}_2^{-}(aq,1\mathrm{.}20(mol)L^{-1})$

Calculate the cell potential at $25C.$

Express your answer in volts to two decimal places.

View Available Hint$($s$)$

$E_{\text{c}\text{e}\text{l}\text{l}}=$