The standard Gibbs free energy change (ΔG° in kJ mol^{-1}). in a Daniel cell (E^\circ_\cell = 1.1 V). when 2 moles of Zn(s) is oxidized at 298 K, is closest to

KVPY SB SX Exam 2011 Question Paper

Question : 54 of 120

Marks: +1, -0

The standard Gibbs free energy change (ΔG° in kJ mol

^{-1}

). in a Daniel cell

(E°_\cell = 1.1 V)

. when 2 moles of Zn(s) is oxidized at 298 K, is closest to

Solution:

Zn + Cu^{+2} → Zn^{+2} + Cu

2Zn + 2Cu^{+2} → 2Zn^{+2} + 2Cu

For 2 moles of Zn, n = 4

\∆\Gº = -\nFEº_\Cell

= - 4 × 96500 × 1.1

= -424.6 \kJ

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