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lu243611904

Ìú³æ (³õÈëÎÄ̳)

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10Â¥: Originally posted by ¹ªÄª at 2013-09-13 10:40:32
²»ºÃÒâ˼£¬Ð´´íÁË£¬ÊÇÐýת»·Å̵缫£¬¸ù¾ÝH2O2µÄ²úÂʼÆËãµç×Ó×ªÒÆ¸öÊý¡£

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31Â¥2015-04-29 14:54:50
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¹ªÄª

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31Â¥: Originally posted by lu243611904 at 2015-04-29 14:54:50
´óÉñÄúºÃ£¬ÇëÎÊÄúÓÃÐýת»·Å̵缫Ôõô²âÑõ»¹Ô­µÄµç×Ó×ªÒÆÊýµÄ£¿¿ÉÒ԰ѹ«Ê½¸øÎÒÒ»ÏÂô£¿ÎÒ´ÓÊǵ绯ѧ·½ÃæµÄ֪ʶ»¹ÊÇС°×Ò»¸ö£¬»¹Çë´óÉñ¶à°ïæ...

RRDEÖ÷ÒªÊDzâH2O2²úÂÊ£¬Ô­ÀíÄã¿ÉÒÔ¿´¿´ORRÎÄÏ×»òÕߵ绯ѧ²âÊÔµÄÊé¡£ÏÂÃæÊÇÎÒ´ÓһƪÎÄÏ×ÖÐÕ³¹ýÀ´µÄ(0.1M KOH)¡£
The amount of H2O2produced during oxygen reduction cangenerally be calculated from the RRDE data using the followingequation:
%H2O2 = 200(Ir/N)/(Id +(Ir/N))                         n = 4ID/(ID + IR/N)
where Id, Ir, and N are the disk current, ring current, and collectionefficiency, respectively. Here, the value of N is taken to be 0.37.
32Â¥2015-04-30 09:08:09
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zhulw

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13Â¥: Originally posted by chere279 at 2013-09-13 15:57:13
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