Asked by Queeneth
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.
Answers
Answered by
DrBob222
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
-----------------------------
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.
Zn ==> Zn^+ 2e Eo = +0.96
Cu^2+ + 2e ==> Cu Eo = +-.3
-----------------------------
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.
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