1.

How much faraday current is required for reduction of 1.5 mol Cr_(2)O_(7)^(-2) to Cr^(3+) ?

Answer»

15F
9F
6F
3F

Solution :`Cr_(2)O_(7(aq))^(2-)+14H_((aq))^(+)+6e^(-) to 2Cr_((aq))^(3+)+7H_(2)O_((L))`
So, for 1 mol `Cr_(2)O_(7)^(-2)6F` current is REQUIRED
And for 1.5 mol `Cr_(2)O_(7)^(-2)` required `(6xx1.5)=9.0F`


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