E Cell ∘ for the cell Pt(s)|Fe²⁺(aq)|Fe³⁺(aq)||MnO₄⁻(aq),H⁺(aq)|Mn²⁺(aq)|Pt is 0.8 Volt. Given: Δ G f ∘ H 2 O ( l ) = − 230 k J m o l − 1 Δ G f ∘ F e 3 + ( a q ) = − 10.2 k J m o l − 1 Δ G f ∘

Question

E Cell ∘ ​ for the cell Pt(s)|Fe²⁺(aq)|Fe³⁺(aq)||MnO₄⁻(aq),H⁺(aq)|Mn²⁺(aq)|Pt is 0.8 Volt. Given: Δ G f ∘ ​ H 2 ​ O ( l ) = − 230 k J m o l − 1 Δ G f ∘ ​ F e 3 + ( a q ) = − 10.2 k J m o l − 1 Δ G f ∘ ​ M n 2 + ( a q ) = − 229 k J m o l − 1 Δ G f ∘ ​ F e 2 + ( a q ) = − 84 k J m o l − 1 Magnitude of Δ G f ∘ ​ M n O 4 − ​ ( a q ) in kJ mol⁻¹ is:

TestHub · Accepted Answer

The magnitude of Δ G f ∘ ​ [ M n O 4 − ​ ( a q )] is 394.00 k J m o l − 1 . Balanced Cell Reaction ( n = 5 electrons): 5 F e 2 + + M n O 4 − ​ + 8 H + → 5 F e 3 + + M n 2 + + 4 H 2 ​ O Calculate Gibbs Free Energy change ( Δ G ∘ ): Δ G ∘ = − n F E ce ll ∘ ​ = − 5 × 96.5 × 0.8 = − 386 k J m o l − 1 Using Δ G ∘ = ∑ Δ G f ∘ ​ ( p ro d u c t s ) − ∑ Δ G f ∘ ​ ( re a c t an t s ) (given Δ G f ∘ ​ [ H + ] = 0 ): − 386 = [ 5 ( − 10.2 ) + ( − 229 ) + 4 ( − 230 )] − [ 5 ( − 84 ) + Δ G f ∘ ​ ( M n O 4 − ​ )] Solve for Δ G f ∘ ​ ( M n O 4 − ​ ) : − 386 = − 1200 + 420 − Δ G f ∘ ​ ( M n O 4 − ​ ) ⟹ Δ G f ∘ ​ ( M n O 4 − ​ ) = − 394 k J m o l − 1

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