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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
Arrange the following metals in the order in which they displace each other from the solution of their salts.
Al, Cu, Fe, Mg and Zn.
Zinc can be coated on iron to produce galvanized iron but the reverse is not possible. It is because ____________.
A gas X at 1 atm is bubbled through a solution containing a mixture of 1 MY− and 1 MZ− at 25°C. If the reduction potential of Z > Y > X, then ______.
Describe the construction of Daniel cell. Write the cell reaction.
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
An electrochemical cell can behave like an electrolytic cell when ______.
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
Use the data given in below find out the most stable oxidised species.
`E^0 (Cr_2O_1^(2-))/(Cr_(3+))` = 1.33 V `E^0 (Cl_2)/(Cl^-)` = 1.36 V
`E^0 (MnO_4^-)/(MN^(2+))` = 1.51 V `E^0 (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 |
Given the data at 25°C
\[\ce{Ag + I- -> AgI + e-}\]; E° = – 0.152 V
\[\ce{Ag -> Ag+ + e-}\]; E° = – 0.800 V
The value of log Ksp for AgI is ______.
