Which of the following transformations can be carried out by using HI as a reducing agent, under acidic conditions? [Given:I2(s)→2I–E∘=0.54V] (i)Cu+→CuE∘=0.52V(ii)Cr3+→Cr2+E∘=–0.41V(iii)Fe3+→Fe2+E∘=0.77V(iv)Fe2+→FeE∘=–0.44V
A
(i) and (iii)
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B
(ii) and (iv)
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C
only (iii)
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D
only (ii)
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Solution
The correct option is C only (iii) Here only (iii) E∘ value is more than 0.54V so HI can be used to reduce Fe3+