- ✓osmium
- Bchromium
- Cplatinum
- Dgold
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$E_{\frac{1}{2}Cl_2/Cl^- }^o = 1.36\,V\,,\,E_{C{r^{3 + }}/Cr}^o = - 0.74\,V$
$E_{C{r_2}O_7^{2 - }/C{r^{3 + }}}^o = 1.33\,V\,,\,E_{MnO_4^ - /M{n^{2 + }}}^o = 1.51\,V$
The correct order of reducing power of the species $(Cr, Cr^{3+}, Mn^{2+}$ and $Cl^-)$ will be
($F = 96,500\;C\;mo{l^{ - 1}}; \,\, R = 8.314\;J{K^{ - 1}}mo{l^{ - 1}})$
$\begin{array}{*{20}{c}}
{C{H_3}\,\,\,\,\,\,\,O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3} - CH - C - C{H_2} - C{H_2}OH}
\end{array}$
$Cu ( s )+ Sn ^{2+}( aq ) \rightarrow Cu ^{2+}( aq )+ Sn ( s )$
$\left( E _{ Sn ^{2+} \mid Sn }^{0}=-0.16\, V , E _{ Cu ^{2+} \mid Cu }^{0}=0.34\, V \right.$ Take $F=96500\, C\, mol ^{-1}$ )