Question
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Answer (Detailed Solution Below)
Option 3 : +146.7 kJ mol−1
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Detailed Solution
सही उत्तर +146.7 kJ mol−1 है।
Key Points
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prepsnap.in • Part -AQuestion:
For the reduction Zn2+ + 2e− → Zn, E° = −0.76 V. The standard Gibbs free energy change, $\Delta G^\circ = -nFE^\circ$, is / अपचयन Zn2+ + 2e− → Zn हेतु E° = −0.76 V है। मानक गिब्ज़ मुक्त ऊर्जा परिवर्तन, $\Delta G^\circ = -nFE^\circ$, है
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−146.7 kJ mol−1
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−73.3 kJ mol−1
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+146.7 kJ mol−1
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+73.3 kJ mol−1
Correct Answer: +146.7 kJ mol−1
ΔG° = −nFE° = −2 × 96500 × (−0.76) = +146.7 kJ mol−1; the negative reduction potential makes the reduction non-spontaneous under standard conditions. / ΔG° = −nFE° = −2 × 96500 × (−0.76) = +146.7 kJ mol−1; ऋणात्मक अपचयन विभव मानक परिस्थितियों में अपचयन को अस्वतःप्रवर्तित बनाता है।
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