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Calculating the cell emf from free-energ...

Calculating the cell `emf` from free-energy change: Suppose the reaction of zinc ions and chloride ions are formed in aqueous solution.
`Zn(s) + Cl_(2)(g) overset(H_(2)O)rarrZn^(2+)(aq.) + 2Cl^(-)(aq.)`
Calculate the standard `emf` for this cell at `25^(@)C` from standard free energies of formation.
Strategy: Calculate `Delta_(r)G^(@)` and substitute it along with the value of `n` into the equation `Delta_(r)G^(@) = -nfE_("cell")^(@)`. Solve for `E_("cell")^(@)`

Text Solution

Verified by Experts

Write the equation with `Delta_(r)G^(@)` beneath
`underset(0)(Zn(s))+underset(0)(Cl_(2))(g) overset(rarr)(larr) underset(-147)(Zn)^(2+)(aq.)+ underset(2xx(-131)kJ)(2Cl^(-)(aq.))`
Hence `Delta_(r)G^(@) = sumv_(p)Delta_(f)G^(@)("products")-sumv_(R)Delta_(f)G^(@)("reactants")`
`= [-147 + 2 xx (-131)] kJ - [0 + 0] kJ`
`= -409 kJ`
`= -4.09 xx 10^(5)J`
We ontain `n` by splitting the cell reacton into half-cell reactions
`Zn(s)overset(rarr)(larr)Zn^(2+)(aq.)+2e^(-)`
`2e^(-)+Cl_(2)(g) overset(rarr)(larr)2Cl^(-)(aq.)`
Each half-cell reaction involves two electrons, so `n = 2`. Now we substitute into `Delta_(r)G^(@) = -nfE_("cell")^(@)`
`-4.09 xx 10^(5)J = -(2 "mole" e^(-))(9.65 xx 10^(4)J//V.mol e^(-))`
Solving for `E_("cell")^(@)`, we get
`E_("cell")^(@) = 2.12 V`
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