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प्रश्न
Write a note on sacrificial protection.
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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
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.
What happens if external potential applied becomes greater than E°cell of electrochemical cell?
For the electrochemical cell:
\[\ce{M | M+ || X- | X}\];
\[\ce{E^{\circ}_{{M^{+}/{M}}}}\] = 0.44 V,
\[\ce{E^{\circ}_{X/X^-}}\] = 0.33 V
Which of the following is TRUE for this data?
Cell equation: \[\ce{A + 2B^- -> A^{2+} + 2B}\]
\[\ce{A^{2+} + 2e^- -> A}\] E0 = +0.34 V and log10 k = 15.6 at 300 K for cell reactions find E0 for \[\ce{B^+ + e^- -> B}\]
Why is anode in galvanic cell considered to be negative and cathode positive electrode?
Use the data given in below find out which option the order of reducing power is correct.
`"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
Match the terms given in Column I with the units given in Column II.
| Column I | Column II |
| (i) Λm | (a) S cm-¹ |
| (ii) ECell | (b) m-¹ |
| (iii) K | (c) S cm2 mol-¹ |
| (iv) G* | (d) V |
Match the items of Column I and Column II.
| Column I | Column II |
| (i) K | (a) I × t |
| (ii) Λm | (b) `Λ_m/Λ_m^0` |
| (iii) α | (c) `K/c` |
| (iv) Q | (d) `G^∗/R` |
Which of the following is incorrect?
In a solution of CuSO4, how much time will be required to precipitate 2 g copper by 0.5 ampere current?
