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rewona [7]
2 years ago
7

A galvanic cell whose cell reaction is 2Fe3+(aq) + Zn(s) → 2Fe2+(aq) + Zn2+(aq) has a cell potential of 0.72V. What is the maxim

um electrical work that can be obtained from this cell per mole of iron(III) ion?
Chemistry
1 answer:
sesenic [268]2 years ago
5 0

Answer:

138.96kJ is the maximum electrical work

Explanation:

The maximum electrical work that can be obtained from a cell is obtained from the equation:

W = -nFE

<em>Where W is work in Joules,</em>

<em>n are moles of electrons = 2mol e- because half-reaction of Zn is:</em>

Zn(s) → Zn²⁺(aq) + 2e⁻

F is faraday constant = 96500Coulombs/mol

E is cell potential = 0.72V

Replacing:

W = -2mol*96500Coulombs/mol*0.72V

W = - 138960J =

<h3>138.96kJ is the maximum electrical work</h3>

<em />

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What is the standard enthalpy of a reaction for which the equilibrium constant is (a) doubled, (b) halved when the temperature i
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0.693 = -ΔH°/8.314 (-0.00012)

0.693 = 0.00012ΔH°/8.314

0.693 * 8.314 = 0.00012ΔH°

ΔH° = 0.693 * 8.314/0.00012

ΔH° = 48KJ

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ln(K1/2/K1) =- ΔH°/R(1/T2 - 1/T1)

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-0.693 = -ΔH°/8.314  (-0.00012)

-0.693 = 0.00012ΔH°/8.314

-0.693 * 8.314 = 0.00012ΔH°

ΔH°= -0.693 * 8.314/0.00012

ΔH°= -48KJ

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3 years ago
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