Q61 · CSIR-NET Chemistry, December 2017
Paper: CSIR-NET December 2017 · Subject: Physical Chemistry · Chapter: Electrochemistry · Topic: Electrochemistry – General · Marks: 2 · Difficulty: Medium
The standard free energy of the reaction \[ \mathrm{AgBr}(\mathrm{s}) \rightarrow \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{Br}^{-}(\mathrm{aq}) \]
Is closest to. \[ \left[\mathrm{E}^{\circ}\left(\mathrm{AgBr} / \mathrm{Ag}, \mathrm{Br}^{-}\right)=0.07 \mathrm{V} ; \mathrm{E}^{\circ}\left(\mathrm{Ag} / \mathrm{Ag}^{+}\right)=0.80 \mathrm{V} ; \mathrm{F}=96500 \mathrm{C} \mathrm{mol}^{-1}\right] \]
(a)$7 \mathrm{kJ} \mathrm{mol}^{-1}$
(b)$70 \mathrm{J} \mathrm{mol}^{-1}$
(c)$70 \mathrm{kJ} \mathrm{mol}^{-1}$
(d)$7 \mathrm{J} \mathrm{mol}^{-1}$
Answer
Answer: C ✓ checked by 4AB · confidence high
Explanation
AgBr(s) → Ag⁺ + Br⁻: E° = E°(AgBr/Ag,Br⁻) − E°(Ag⁺/Ag) = 0.07 − 0.80 = −0.73 V; ΔG° = −nFE° = 96500 × 0.73 ≈ +70 kJ mol⁻¹.
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