The standard electrode potential for Zn/Zn2+ and cu/cu2+ are -0.96volts and +0.3volts respectively at 293k.(a) calculate the free energy change when a zinc rod dips into 1M of zinc tetraoxosulphate(vi) solution and a copper rod dips into a 1M of copper (ii) tetraoxosulphate(vi) solution.both are connected by a salt bridge.if the heat of the reaction; Zn+Cu2+(AQ)_zn2+(AQ)+cu(s) is -224k/j (b) calculate the entropy change.
I don't believe the Eo values you have but we'll go with what you posted.
Zn ==> Zn^+ 2e Eo = +0.96
Cu^2+ + 2e ==> Cu Eo = +-.3
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Zn + Cu^2+ ==> Zn^2+ + Cu Ecell = 0.96+0.3 = ?
Then dG = -nEF
Calculate dG. n is 2; F is 96,485 coulombs.
For part b.
dG = dH - TdS.
You know dG, you're given dH (make sure you use J for dG and dH (not kJ) and calculate dS in joules. Post your work if you get stuck.
To calculate the free energy change (ΔG) for the given electrochemical reaction, you can use the equation:
ΔG = -nFΔE
Where:
- ΔG: Free energy change
- n: Number of electrons transferred in the balanced equation
- F: Faraday's constant (96,485 C/mol)
- ΔE: Difference in standard electrode potentials (E1 - E2)
First, let's balance the given reaction:
Zn + Cu2+ -> Zn2+ + Cu
Balanced equation: Zn + Cu2+ -> Zn2+ + Cu
Since the reaction involves the transfer of 2 electrons, n = 2.
Now, we calculate ΔE:
ΔE = E2 - E1
ΔE = (+0.30 V) - (-0.96 V)
ΔE = 1.26 V
Now we can calculate ΔG using the values we obtained:
ΔG = -nFΔE
ΔG = -(2)(96,485 C/mol)(1.26 V)
ΔG = -2 * (96,485 C/mol) * (1.26 J/C)
(a) To get the value in kilojoules (kJ), divide the result by 1000:
ΔG = -2 * (96,485 C/mol) * (1.26 J/C) / 1000
ΔG = -243.948 kJ
(b) To calculate the entropy change (ΔS), you can use the equation:
ΔG = ΔH - TΔS
Where:
- ΔH: Enthalpy change (heat of the reaction)
- T: Temperature in Kelvin
- ΔS: Entropy change
Rearrange the equation to solve for ΔS:
ΔS = (ΔH - ΔG) / T
Given:
ΔH = -224 kJ
T = 293 K
ΔG = -243.948 kJ
Plug in the values:
ΔS = (-224 kJ - (-243.948 kJ)) / 293 K
ΔS = (-224 kJ + 243.948 kJ) / 293 K
ΔS = 19.948 kJ / 293 K
Simplify the calculation:
ΔS = 0.068 kJ/K
Therefore, the entropy change (ΔS) is 0.068 kJ/K.