What will be the spontaneous reaction between the following half cell reactions ? (i) Cr^(3+)(aq)+3e^(-) rarr Cr(s) ""E^@=-0.74V (ii) MnO_2(s)+4H^(+)+2e^(-)rarr Mn^(2+)(aq)+2H_2O(l)""E^@=1.28V Calculate E_("cell")^@

What will be the spontaneous reaction between the following half cell

1.	What will be the spontaneous reaction between the following half cell reactions ? (i) Cr^(3+)(aq)+3e^(-) rarr Cr(s) ""E^@=-0.74V (ii) MnO_2(s)+4H^(+)+2e^(-)rarr Mn^(2+)(aq)+2H_2O(l)""E^@=1.28V Calculate E_("cell")^@
Answer» Solution : Since the reduction POTENTIAL REACTION (ii) is more than that of reaction (i), reaction (ii) will occur as reduction . Therefore , reaction (i) should be written as oxidation . To obtain the net reaction , we multiply by appropriate COEFFICIENTS so that electrons get cancelled . `MnO_2(s)+4H^(+)(AQ)_+2E^(-)rarrMn^(2+)(aq)+2H_2O(l)]xx3` `""Cr(s)rarrCr^(3+)(aq)+3e^(-)]xx2` `2Cr(s)+3MnO_2(s)+12H^(+)(aq)rarr2Cr^(3+)(aq)+3Mn^(2+)(aq)+6H_2O` `E_("cell")^@=E_("Substance reduced")-E_("Substance oxidised")` `=1.28-(-0.74)=2.02V`