In the cell reaction (i) Au→Au3++3e−ΔG01=−n1FE01 (ii) 3Au++3e−→3AuΔG02=−n2FE02 Overall reaction = ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯3Au+→Au3++2AuΔG03=−nFE0 What should be the value of n here?
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Solution
Since in both the half reactions, 3 electrons are involved, the value of n is 3.