Zadanie 4b
Wzorowane na: Chemistry HL Paper 2, May 2023, TZ1, pytanie 4(b). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
Darmowe konto pozwoli wrócić do niego później.
Informacja do zadań 4b-4c.
A voltaic cell was constructed from an iron(II) sulfate/iron half-cell and a copper(II) sulfate/copper half-cell, connected by a salt bridge and, externally, by a wire.
Polecenie
(i) Outline why electrons flow from the iron half-cell to the copper half-cell when the two are connected by a wire. Use section 19 of the data booklet.
(ii) Formulate equations for the reactions taking place at the anode (negative electrode) and at the cathode (positive electrode).
Sprawdź rozwiązanieUkryj rozwiązanieKlucz i punktacja
Odpowiedź z klucza
(i) iron is the more reactive metal / the stronger reducing agent / is oxidized more easily / loses electrons more readily
OR
E⊖(Fe2+/Fe) = −0.45 V is more negative than E⊖(Cu2+/Cu) = +0.34 V ✔
(ii) Anode (negative electrode): Fe(s) → Fe2+(aq) + 2 e− ✔
Cathode (positive electrode): Cu2+(aq) + 2 e− → Cu(s) ✔
Schemat punktowania
(i) [1]: accept 'iron is higher in the activity series'. (ii) [2]: one mark per electrode; state symbols not required; award [1 max] if equilibrium arrows are used; award [1 max] if the equations are written at the wrong electrodes.