English
Tamil Nadu Board of Secondary EducationHSC Science Class 12

Two metals M1 and M2 have reduction potential values of −xV and +yV respectively. Which will liberate H2 and H2SO4. - Chemistry

Advertisements
Advertisements

Question

Two metals M1 and M2 have reduction potential values of −xV and +yV respectively. Which will liberate H2 and H2SO4.

Short Note

Solution

Metals having negative reduction potential act as powerful reducing agents. Since M1 has – xV, therefore M1 easily liberates H2 in H2SO4. Metals having higher oxidation potential will liberate H2 from H2SO4. Hence, the metal M1 having +xV, oxidation potential will liberate H2 from H2SO4.

shaalaa.com
  Is there an error in this question or solution?
Chapter 9: Electro Chemistry - Evaluation [Page 66]

APPEARS IN

Samacheer Kalvi Chemistry - Volume 1 and 2 [English] Class 12 TN Board
Chapter 9 Electro Chemistry
Evaluation | Q 16. | Page 66

RELATED QUESTIONS

At 25°C, the emf of the following electrochemical cell.

\[\ce{Ag_{(s)} | Ag^+ (0.01 M) | | Zn^{2+} {(0.1 M)} | Zn_{(s)}}\] will be:

(Given \[\ce{E^0_{cell}}\] = −1.562 V)


In the electrochemical cell: Zn|ZnSO4 (0.01 M)||CuSO4 (1.0 M)|Cu, the emf of this Daniel cell is E1. When the concentration of ZnSO4 is changed to 1.0 M and that CuSO4 changed to 0.01 M, the emf changes to E2. From the above, which one is the relationship between E1 and E2?


Describe the construction of Daniel cell. Write the cell reaction.


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


Can absolute electrode potential of an electrode be measured?


Calculate the standard EMF ofa cell which involves the following cell reactions

\[\ce{Zn + 2 Ag+ -> Zn^{2+} + 2 Ag}\]

Given that \[\ce{E^{o}_{Zn/Zn^{2+}}}\] = 0.76 volt and \[\ce{E^{o}_{Ag/Ag^{+}}}\] = – 0.80 volt.


If the half-cell reaction A + e → A has a large negative reduction potential, it follow that:-


The two half cell reaction of an electrochemical cell is given as 

\[\ce{Ag+ + e- -> Ag}\], `"E"_("Ag"^+//"Ag")^circ` = - 0.3995 V

\[\ce{Fe^{2+} -> Fe^{3+} + e-}\], `"E"_("Fe"^{3+}//"Fe")^{2+}` = - 0.7120 V

The value of EMF will be ______.


Why is anode in galvanic cell considered to be negative and cathode positive?


State the term for the following:

Two metal plates or wires through which the current enters and leaves the electrolytic cell.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×