The amount of electricity required to deposit one mole of Copper from Copper sulphate solution is –
(A) 1 Faraday
(B) 2.33 Faraday
(C) 2 Faraday
(D) 1.33 Faraday
Answered Feb 05, 2023
Correct answer is (C) 2 Faraday
Correct Answer - C `Cu^(2+) + 2e^(-) rarr Cu (s)` 1 mole copper requires 2 mole electron i.e., 2 faraday charge
Correct Answer - B 1 faraday deposits 1 equivalent of nickel 1 equivalent of `Ni = 1//2` mole of nickel
`because m`=Zit and `Z=(E)/(F)=(M)/(pF) therefore m=(Mit)/(pF)` where m=2.5g. `F=96500C//"mol",M=63.5,i=10A` and p=2 `therefore t=(mpF)/(Mi)=(2.5xx2xx96500)/( 63.5xx10)= 760s`
`m=Zit=Z((V)/(R))t` `rARr V=(mR)/(Zt)=(mrhol)/(Zat)` `=(0.66xx1.2xx10^(-2)xx5xx10^(-2))/(3.33xx10^(-7)xx1xx60xx60)` =0.33 volt
Correct Answer - B Number of equivalents of copper deposited `=0.16/32=0.005` Volume of `H_(2)` gas at STP `=11.2xx0.005` `=0.056` litre `=56 cm^(3)`
Correct Answer - A `Al^(3+)+3e^(-)toAl` So mole of `e^(-)=3`
`R-overset(O)overset(||)C-NH_(2)+OH^(-)toR-overset(O)overset(||)C-O^(-)+NH_(3)` R+45=74orR=29i.e., `CH_(3)CH_(2)-` Monobsic acid=`CH_(3)CH_(2)COOH,` Mol. Mass=`CH_(3)CH_(2)CONH_(2)=73`
Correct Answer - 7711 s
Correct Answer - 1.713 A
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