Advertisements
Advertisements
प्रश्न
Why is anode in galvanic cell considered to be negative and cathode positive electrode?
Advertisements
उत्तर
A galvanic cell works basically in reverse to an electrolytic cell. The anode is the electrode where oxidation takes place, in a galvanic cell, it is the negative electrode, as when oxidation occurs, electrons are left behind on the electrode.
The anode is also the electrode where metal atoms give up their electrons to the metal and go into solution. The electron left behind on it renders it effectively negative and the electron flow goes from it through the wire to the cathode.
Positive aqueous ions in the solution are reduced by the incoming electrons on the cathode. This is why the cathode is a positive electrode because positive ions are reduced to metal atoms there.
APPEARS IN
संबंधित प्रश्न
Derive a relation between ΔH and ΔU for a chemical reaction. Draw neat labelled diagram of calomel electrode. Resistance and conductivity of a cell containing 0.001 M KCI solution at 298K are 1500Ω and 1.46x10-4 S.cm-1 respectively.
Define the following term:
Fuel cell
Describe the construction of Daniel cell. Write the cell reaction.
For the cell \[\ce{Mg_{(s)}|Mg^{2+}_{( aq)}||Ag^+_{( aq)}|Ag_{(s)}}\], calculate the equilibrium constant at 25°C and maximum work that can be obtained during operation of cell.
Given: \[\ce{E^0_{{Mg^{2+}|Mg}}}\] = −2.37 V and \[\ce{E^0_{{Ag^{+}|Ag}}}\] = 0.80 V
Use the data given in below find out the most stable ion in its reduced form.
`"E"_("Cr"_2"O"_7^(2-)//"Cr"^(3+))^⊖`= 1.33 V `"E"_("Cl"_2//"Cl"^-)^⊖` = 1.36 V
`"E"_("MnO"_4^-//"Mn"^(2+))^⊖` = 1.51 V `"E"_("Cr"^(3+)//"Cr")^⊖` = - 0.74 V
`E_(cell)^Θ` for some half cell reactions are given below. On the basis of these mark the correct answer.
(a) \[\ce{H^{+} (aq) + e^{-} -> 1/2 H_2 (g); E^Θ_{cell} = 0.00V}\]
(b) \[\ce{2H2O (1) -> O2 (g) + 4H^{+} (aq) + 4e^{-}; E^Θ_{cell} = 1.23V}\]
(c) \[\ce{2SO^{2-}_{4} (aq) -> S2O^{2-}_{8} (aq) + 2e^{-}; E^Θ_{cell} = 1.96V}\]
(i) In dilute sulphuric acid solution, hydrogen will be reduced at cathode.
(ii) In concentrated sulphuric acid solution, water will be oxidised at anode.
(iii) In dilute sulphuric acid solution, water will be oxidised at anode.
(iv) In dilute sulphuric acid solution, \[\ce{SO4^{2-}}\] ion will be oxidised to tetrathionate ion at anode.
For the given cell, \[\ce{Mg | Mg^{2+} || Cu^{2+} | Cu}\]
(i) \[\ce{Mg}\] is cathode
(ii) \[\ce{Cu}\] is cathode
(iii) The cell reaction is \[\ce{Mg^+ Cu^{2+} -> Mg^{2+} + Cu}\]
(iv) \[\ce{Cu}\] is the oxidising agent
How will the pH of brine (aq. \[\ce{NaCl}\] solution) be affected when it is electrolysed?
The electrochemical cell stops working after some time because
Can we construct an electrochemical cell with two half-cells composed of ZnSO4 solution and zinc electrodes? Explain your answer.
