How many moles of electrons pass through the circuit when 0.60 mole of Hg2+ and 0.30 mole of HNO2 are produced in the cell that contains 0.50 mole of Hg2^2+ and 0.40 mole of NO3^– at the beginning of

Question

How many moles of electrons pass through the circuit when 0.60 mole of Hg2+ and 0.30 mole of HNO2 are produced in the cell that contains 0.50 mole of Hg2^2+ and 0.40 mole of NO3^– at the beginning of the reaction?

Prashant Kotian · Accepted Answer

Correct answer: B. The oxidation half-reaction is Hg2^2+ -> 2Hg^2+ + 2e^–. Thus, producing 2 moles of Hg^2+ involves 2 moles of electrons, meaning 0.60 moles of Hg^2+ involves 0.60 moles of electrons. Alternatively, for the reduction: NO3^– + 3H3O^+ + 2e^– -> HNO2 + 4H2O. Producing 1 mole of HNO2 requires 2 moles of electrons. Thus, producing 0.30 moles of HNO2 requires 0.30 × 2 = 0.60 moles of electrons. Therefore, 0.60 moles of electrons pass through the circuit.

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