please solve the question - electro chemistry
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cu is reduced and ag is oxidised... break it into two reaction.. that is oxidation reaction and reduction reaction then.. Emf of cu - Emf of ag
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Answer:
Cu(s)+2Ag²⁺(aq) ⟶ Cu²⁺(aq)+2Ag(s)
E° = 0.46V at 298K
log K꜀ = nFE° / RT
= 2 x 0.46 / 0.059
K꜀ = 4 x 10¹⁵
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