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xl1987

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Show by calculation that E0=-1.662V for the reduction of Al3+ to Al(s),
regardless of whether the equation for the reaction is written:
(i)1/3 Al3+ +e=1/3Al(s)     ¡÷G0=160.4KJ/mol
or(ii)Al3+ +3e= Al(s)      ¡÷G0=481.2KJ/mol
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Òø³æ (СÓÐÃûÆø)

Õâ¸öÓÃÈÈÁ¦Ñ§Óëµç»¯Ñ§µÄÇÅÁº¹«Ê½  ¡÷G=-nEf  ¼´¿É
f= 9.65*10^4C/mol
ÓõÚÒ»¸öµç¶Ô   £¨160.4*1000/1£©/£¨ 9.65*10^4C/mol £©
Óõڶþ¸öµç¶Ô    £¨481.2*1000/3£©/£¨ 9.65*10^4C/mol £©
2Â¥2012-09-07 08:37:23
ÒÑÔÄ   »Ø¸´´ËÂ¥   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû

zhou-ck

½ð³æ (ÕýʽдÊÖ)

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xl1987: ½ð±Ò+20, ¡ï¡ï¡ï¡ï¡ï×î¼Ñ´ð°¸ 2012-09-09 12:16:01
ÀûÓÃÄÜË¹ÌØ¹«Ê½£¬ ¾ÍÊÇÂ¥ÉÏ˵µÄÄǸö£¡¡÷G=-nFE.
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3Â¥2012-09-07 11:14:54
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