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Given : 2H2O → O2 + 4H+ + 4e– Eº = –1.23 V Calculate electrode potential at pH = 5.

Answer»

Answer:–00.93

E = –1.23  – 0.0591/4 log [H+]4

= –1.23 + 0.0591 × pH = –1.23 + 0.0591 × 5 

= –1.23 + 0.2955 = – 0.9345 V = –0.93 V



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