Question

Use the standard half-cell potentials listed below to calculate the standard cell potential for the following reaction occurring in an electrochemical cell at 25°C. (The equation is balanced.) Pb(s) + Br2(l) → Pb2+(aq) + 2Br(aq) Pb2+(aq) + 2 e → Pb(s) E° = -0.13 V Br2(l) + 2 e → 2 Br(aq) E° = +1.07 V

Answer 1

Answer:

1.20 V

Explanation:

$Pb(s) + Br_2(l)\rightarrow Pb^{2+}(aq) + 2Br^-(aq)$

Here Pb undergoes oxidation by loss of electrons, thus act as anode. Bromine undergoes reduction by gain of electrons and thus act as cathode.

$E^0=E^0_{cathode}- E^0_{anode}$

Where both $E^0$ are standard reduction potentials.

Given,

$Pb^{2+}(aq) + 2 e^-\rightarrow Pb(s)$

$E^0_{[Pb^{2+}/Pb]}= -0.13\ V$

$Br_2(l) + 2 e^-\rightarrow 2 Br(aq)$

$E^0_{[Br_2/Br^{-}]}=+1.07\ V$

$E^0=E^0_{[Br_2/Br^{-}]}- E^0_{[Pb^{2+}/Pb]}$

$E^0=+1.07- (-0.13)\ V=1.20\ V$