In a classroom, the teacher has explained the quantitative aspects of electrolysis by stating the Faraday's laws of electrolysis.
Calculate the quantity of electricity required to deposit 0.09 g of aluminium during the following electrode reaction:
`Al^3 + 3e^(-) to Al ( Al=27u)`

Calculate the quantity of electricity required to deposit 0.09 g of Aluminium during the following electrode reaction. Al^(3+) + 3e^- rarr Al .

In a classroom, the teacher has explained the quantitative aspects of electrolysis by stating the Faraday's laws of electrolysis. Explain the term electrochemical equivalent

In class room , the teacher has explained the quantitative aspects of electrolysis by stating the Faraday's laws of electrolysis. Explain the term electrochemical equivalent.

Assertion : 1 Faraday of electricity deposits 1 g equivalent of Ag,Cu or Al. Reason : 1 mole of electrons are required to reduce 1 mole of Ag^(+) or 1/2 mole of Cu^(2+) or 1/3 mole of Al^(3+) ions.

Faraday's first law of electrolysis: Amount of a substance deposited or liberated on respective electrode is directly proportional to the quantity of electricity following through the circuit. W prop Q rArr W= ZQ W= Zit where Z= electrochemical equivalent of the substance. (i) In order to deposit or liberate 1 mole of substance, integeral no of Faradays are required. (ii) In order to deposit one gram equivalent of substance, 1 Faraday electricity is required. Current efficiency = ("Actual yield")/("Theoritical yield") xx 100% For the reaction, 2CH_(3)COO^(-) rarr C_(2)H_(6)(g) + 2CO_(2)(g) + 2e^(-) . If current of 20A is passed for 965sec, volume of liberated gases measured at NTP is found to be 3.36 lit, then calculate current efficiency of the reaction

The following galvanic cell Zn|Zn(NO_3)_2(aq)||Cu(NO_3)_2(aq)|Cu Anode (100mL,1M)(100mL,1M) cathode was operated as an electrolysis cell as Cu as the anode and Zn as the cathode. A current of 0.48 ampere was passed for 10 hours and then the cell was allowed to function to galvanic cell.What would be the emf of the cell at 25^@C Assume that the only electrode reactions occurring were those involving Cu//Cu^(2+) and Zn//Zn^(2+) (E_(Cu^(2+),Cu)^@=+0.34V, E_(Zn^(2+),Zn)^@=-0.76V)

Tarnished silver contains Ag_2 ~S . Can this tarnish. be removed by placing tarnished silver ware in an aluminium pan containing an inert electrolytic solution such as NaCl . The standard electrode potential for the half reactions are A g_2 S(s)+2 e^- 2 A g(s)^2+S^1--0.71 v and Al^4+3 e^- Al(s),-1.66 ~V

Show by calculation that E^@=-1.662V for the reduction of Al^(3+) to Al(s) , regardless of whether the equation for the reaction is written as Al^(3+)+3e to Al(s), Delta G^@=481.2 kJ //mol e

Delta G for the reaction (4)/(3) Al + O_(2) rarr (2)/(3) Al_(2)O_(3) is -772kJ mol^(-1) of O_(2) . Calculate the minimum EMF in volts required to carry out an electrolysis of Al_(2)O_(3)

Show by calculation that E^@=-1.662V for the reduction of Al^(3+) to Al(s) , regardless of whether the equation for the reaction is written as 1/3 Al^(3+)+e to 1/3 Al(s), Delta G^@=160.4 kJ//mol e