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प्रश्न
Write a note on sacrificial protection.
उत्तर
In this method, the metallic structure to be protected is made cathode by connecting it with more active metal (anodic metal). So that all the corrosion will concentrate only on the active metal. The artificially made anode thus gradually gets corroded protecting the original metallic structure. Hence this process is otherwise known as sacrificial anodic protection. Al, Zn and Mg are used as sacrificial anodes.
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संबंधित प्रश्न
Arrange the following metals in the order in which they displace each other from the solution of their salts.
\[\ce{Al, Cu, Fe, Mg}\] and \[\ce{Zn}\]
Construct a labelled diagram for the following cell:
`Zn|Zn^(2+)(1M)||H^+(1M)|H_(2(g,1atm))|Pt`
How many electrons flow through a metllic wire if a current of 0·5 A is passed for 2 hours? (Given : 1 F = 96,500 C mol−1)
What is the SI unit tor electrochemical equivalent?
A gas X at 1 atm is bubbled through a solution containing a mixture of 1MY− and 1MZ− at 25°C. If the reduction potential of Z > Y > X, then ____________.
Can Fe3+ oxidises bromide to bromine under standard conditions?
Given: \[\ce{E^0_{{Fe^{3+}|Fe^{2+}}}}\] = 0.771 V
\[\ce{E^0_{{Br_{2}|Br^-}}}\] = −1.09 V
Reduction potential of two metals M1 and M2 are \[\ce{E^0_{{M_1^{2+}|M_1}}}\] = −2.3 V and \[\ce{E^0_{{M_2^{2+}|M_2}}}\] = 0.2 V. Predict which one is better for coating the surface of iron.
Given: \[\ce{E^0_{{Fe^{2+}|Fe}}}\] = −0.44 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.
Calculate the λ0m for Cl- ion from the data given below:
∧0m MgCl2 = 258.6 Scm2 mol-1 and λ0m Mg2+ = 106 Scm2 mol-1
Why is anode in galvanic cell considered to be negative and cathode positive?
