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.
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.
a)
At the left-hand electrode: Pt | Br2:
The reaction at the left-hand electrode is the reduction of bromine (Br2) to bromide ions (2Br-) according to the half-reaction:
Br2 + 2e- -> 2Br-
At the right-hand electrode: Fe2+, Fe3+ | Pt:
The reaction at the right-hand electrode involves both the oxidation of Fe2+ to Fe3+ and the reduction of Fe3+ back to Fe2+. The overall reaction can be split into two half-reactions:
Oxidation half-reaction:
Fe2+ -> Fe3+ + e-
Reduction half-reaction:
Fe3+ + e- -> Fe2+
Overall cell reaction:
By combining the reduction half-reaction of bromine and the oxidation half-reaction of Fe2+, we get the overall reaction of the galvanic cell:
2Br- + Fe2+ -> Br2 + Fe3+
b)
To calculate E° for the cell, we need to look up the standard reduction potentials for the half-reactions involved:
Standard reduction potential for the reduction of bromine (Br2):
E°(Br2/2Br-) = +1.087 V (It is positive because it is a reduction half-reaction)
Standard reduction potential for the oxidation of Fe2+ to Fe3+:
E°(Fe2+/Fe3+) = +0.771 V (It is positive because it is an oxidation half-reaction)
The standard cell potential (E°cell) can be calculated by subtracting the reduction half-reaction potential from the oxidation half-reaction potential:
E°cell = E°(Fe2+/Fe3+) - E°(Br2/2Br-)
E°cell = 0.771 V - 1.087 V = -0.316 V
To calculate ∆G for the cell reaction, we can use the equation:
∆G = -nF E°cell
Where:
∆G is the change in Gibbs free energy
n is the number of electrons transferred (in this case, n = 2 because two electrons are transferred in the overall cell reaction)
F is Faraday's constant (F = 96,485 C/mol)
Plugging in the values, we get:
∆G = -2 * 96,485 C/mol * (-0.316 V)
∆G = +61,057 J/mol
The positive value of ∆G indicates that the reaction is not spontaneous under standard conditions.