A galvanic cell has `E_(cell)^(@)=1.5" V"`. If an opposing potential of `1.5" V"` is applied to cell, what will happen to the cell reaction and current flowing through the cell ?
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The wire of potentiometer has resistance `4 Omega` and length 1m. It is connected to a cell of e.m.f. 2V and internal resistance `1 Omega`. The curren
Correct Answer - C `I=(E)/(R+r)=(2)/(4+1)=0.4A`
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A rectangular coil of 100 turns each of area `10 cm^2` hangs freely in radial magnetic field. The coil deflects through an angle of `30^@` when curren
Correct Answer - C `B=(K theta )/(n AI)=(2.5xx10^-9xx5)/(10^2xx10^-3xx5xx6xx10^-4)`
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At what rate the potential difference between the plates of a parallel plate capacitance `2 mu F` should be changed to establish a displacement curren
Correct Answer - C `I_(D) = C(dV)/(dt) :. (I_D)/C = (2 xx 10^(-3))/(2 xx 10^(-6))` `:. (dV)/(dt) = 10^(3) V//s`.
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The reaction, `1//2 H_(2) (g) + AgCl(s) = H^(+) (aq) + Cl^(-) (aq) + Ag(s)` occurs in the galvanic cell . The anode is
Correct Answer - C Anode is `H_(2) | H^(+)` , Cathode is `Ag^(+) |Ag`.
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If in a galvanic cell, the cell reaction is reversed as `Cu(s)|Cu^(2_(aq)||Zn^(2)aq|Zn` the cell potential will be
Correct Answer - c `Cu^(2+)+2e^(-)rarrCu,E^(@)=+0.337V` `Zn^(2)++2e^(-)rarrZn,E^(@)=-0.763V` In the cell `Cu|Cu^(2+)||Zn^(2+)||Zn` `E_("cell")=E_("cathode")^(@)-E_("anode")^(@)` `=-0.763(+0.337)=-1.10`
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The reaction `(1)/(2)H_(2)(g)+AgCl(s)hArrH^(+)(aq)+Cl^(-)(aq)+Ag(s)` occurs in the galvanic cell
Correct Answer - B Here half cell reaction are `(1)/(2)H_(2)rarrH^(+)+e^(-)` (At anode ) `Ag^(+)+e^(-)rarrAg` (At cathode) `therefore` Cell representations is `Pt|H_(2)(g)|HCl("soln")||AgC l(s)|Ag`
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The cell reaction of the galvanic cell: `Cu(s):Cu^(2+)(aq)||Hg^(2+)(aq)Hg|(l)` is
Correct Answer - D `CurarrCu^(2+)+2e^(-)` `Hg^(2+)+2e^(-)rarrHg` `Cu+Hg^(2+)rarrCu^(2+)+Hg`
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The standard e.m.f. of a galvanic cell involving 3 moles of electrons in a redox reaction is 0.59V. The equilibrium constnat for the reaction of the c
Correct Answer - D `E_("cell")^(@)=(0.591)/(n)"log"k` or log `k=(0.59xx3)/(0.0591)=30` `k=10^(30)`
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What is the relationship between Gibbs free energy of the cell reaction in a galvanic cell and the emf of the cell ? When will the maximum work be obt
The relation may be given as : `DeltaG=-nFE_(cell)` If the redox reaction is under standard conditions, then `DeltaG^(@)=-nFE_(cell)^(@)` Here nF is the quantity of charge passed. In case we want...
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Calculate the standard electrode potential of the `Ni^(2+)//Ni` electrode , if the cell potential potential of the cell, `Ni//N^(2+)(0.01 M)//Cu is 0.
Correct Answer - `-0.2205 V` Cell reaction : `Ni(s)+Cu^(2+)(0.1" M ") to Ni^(2+)(0.01" M ")+Cu(s)` `E_(cell)=E_(cell)^(@)-(0.0591)/(n)"log"([Ni^(2+)])/([Cu^(2+)])` `0.59=E_(cell)^(@)-(0.0591)/(n)"log"(0.01)/(0.1)=E_(cell)^(@)-0.02955" log "(10^(-1))=E_(cell)^(@)+0.02955` `E_(cell)^(@)=0.59-0.02955=0.5605` Now, `" " E_(cell)^(@)=E_((Cu^(2+)//Cu))^(@)-E_((Ni^(2+)//Ni))^(@)` `:. E_((Ni^(2+)//Ni))^(@)=E_((Cu^(2+)//Cu))^(@)-E_(cell)^(@)=0.34-0.5605=0.2205" V "`
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