How many faradays are needed for reduction of 5mole of cr2o7 into cr+3?
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Explanation:
Explanation: Cr_2O_7^{2-}+14H^+ +8e^-\rightarrow 2Cr^{2+}+7H_2OCr
2
O
7
2−
+14H
+
+8e
−
→2Cr
2+
+7H
2
O
Chromium in +6 oxidation state in Cr_2O_7^{2-}Cr
2
O
7
2−
changes to chromium in +2 oxidation state. Thus 4 mole of electrons are required for 1 mole of chromium an 8 mole of electrons for 2 moles of chromium.
1 electron carry charge=1.6\times 10^{-19}C1.6×10
−19
C
1 mole of electrons carry charge=\frac{1.6\times 10^{-19}}{1}\times 6.023\times 10^{23}=96500C=1 Faraday
1
1.6×10
−19
×6.023×10
23
=96500C=1Faraday
Thus 8 mole of electrons carry charge=\frac{1F}{1}\times 8=8Faraday
1
1F
×8=8Faraday
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