When sodium and chromium are the two available cations, using the activity series for oxidation half-reactions,...
When sodium and chromium are the two available cations, using the activity series for oxidation half-reactions, determine and write the chemical equation.
Na(s)---->Na^+ (aq) + e^ -
Cr(s)----> Cr^3+ (aq) + 3 e^ -
Homework Answers
The standard reduction potential values for the reduction of Na+(aq) and Cr3+(aq) are
Na+(aq) + 1e- ----------------> Na(s), E0 = - 2.71V
Cr3+(aq) + 3e- ---------------> Cr(s), E0 = - 0.74 V
Since the standard reduction potential value is more for Cr3+(aq) , Cr3+(aq) will be reduced to Cr(s) and acts as an oxidizing agent, whereas Na(s) will be oxidized to Na+(aq) and acts as reducing agent.
Hence the half-cell rectionsa are
Oxidation half-cell: 3Na(s) --------------> 3Na+(aq) + 3e- , E0(oxi) = -(- 2.71V) = +2.71V
Reduction half-cell: Cr3+(aq) + 3e- ---------------> Cr(s), E0(red) = - 0.74 V
Hence the complete cell reaction can be written as
3Na(s) + Cr3+(aq) + 3e- -------------> 3Na+(aq) + 3e- +Cr(s), E0 (cell) = E0(oxi) + E0(red) = +2.71V+ ( - 0.74 V)
or
3Na(s) + Cr3+(aq)-------------> 3Na+(aq) +Cr(s), E0 (cell) = 1.97V (answer)
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