Question

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^ -

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Answer #1

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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