Q69 · CSIR-NET Chemistry, June 2014

Paper: CSIR-NET June 2014 · Subject: Physical Chemistry · Chapter: Electrochemistry · Topic: Electrochemistry – General · Marks: 2 · Difficulty: Easy

Given;
\[ \begin{array}{l} \text{A. } \mathrm{Fe(OH)_2(s) + 2e^{-} \rightarrow Fe(s) + 2OH^{-}(aq)} \ ;\ E_0=-0.877\,\mathrm{V} \\ \text{B. } \mathrm{Al^{3+}(aq) + 3e^{-} \rightarrow Al(s)} \ ;\ E_0=1.66\,\mathrm{V} \\ \text{C. } \mathrm{AgBr(aq) + e^{-} \rightarrow Ag(s) + Br^{-}(aq)} \ ;\ E_0=0.071\,\mathrm{V} \end{array} \]
The overall reaction for the cells in the direction of spontaneous change would be
(a)Cell with A and B : Fe reduced ; Cell with A and C : Fe reduced
(b)Cell with A and B : Fe reduced ; Cell with A and C : Fe oxidized
(c)Cell with A and B : Fe oxidized ; Cell with A and C : Fe oxidized
(d)Cell with A and B : Fe oxidized ; Cell with A and C : Fe reduced
Answer
Answer: B ✓ checked by 4AB · confidence high
Explanation
With Al (E° = −1.66 V) the Fe(OH)₂ couple (−0.877 V) is the cathode — Fe is reduced; with AgBr (+0.071 V) the iron couple is the anode — Fe is oxidised.

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