a) The reaction at the left-hand electrode is the reduction of Br2 to 2 Br-:
Br2 + 2 e- → 2 Br-
The reaction at the right-hand electrode is the oxidation of Fe2+ to Fe3+:
Fe2+ → Fe3+ + e-
The overall cell reaction is the combination of these two half-reactions:
Br2 + 2 Fe2+ → 2 Br- + 2 Fe3+
b) To calculate E° for the cell, we need to calculate the standard reduction potentials for each half-reaction and use the Nernst equation:
E°red(Br2/2Br-) = 1.09 V
E°ox(Fe2+/Fe3+) = -0.77 V
E°cell = E°red(cathode) - E°ox(anode)
E°cell = 1.09 - (-0.77)
E°cell = 1.86 V
To calculate ∆G for the cell reaction, we can use the equation:
∆G° = -nF E°cell
where n is the number of electrons transferred (in this case, 2) and F is the Faraday constant (96,485 C/mol).
∆G° = -2 x 96,485 x 1.86
∆G° = -357,819 J/mol
Since ∆G is negative, the reaction is spontaneous.
For the galvanic cell ,Pt |Br2||Fe2+, Fe3+| Pt
a).write equation for the reaction at the left- hand and right - hand electrode and the overall cell reaction.
b). Calculate E° for the cell and ∆G for the cell reaction and comment on the spontaneity of the reaction.
1 answer
1 answer