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The cathodic reaction of a Zn - MnO_2dry cell is represented as2 MnO_2(s) + Zn^(2+) + 2e = Zn MnO_4(s) .If there is 8.0 g MnO_2present in the cathodic chamber, how long will the cell function to supply 4 xx 10^(-3)ampere of current? |
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Answer» Solution :`(4 XX 10^(-3) xx t(s))/(96500)= 8/87 , E_(MnO_2)= 87` 25.67 days |
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