Advertisements
Advertisements
प्रश्न
Two metals M1 and M2 have reduction potential values of −xV and +yV respectively. Which will liberate H2 and H2SO4.
Advertisements
उत्तर
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.
APPEARS IN
संबंधित प्रश्न
Can you store copper sulphate solutions in a zinc pot?
Write cathode and anode reaction in a fuel cell.
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
Which of the following statement is not correct about an inert electrode in a cell?
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 |
Assertion: Mercury cell does not give steady potential.
Reason: In the cell reaction, ions are not involved in solution.
If the half-cell reaction A + e– → A– has a large negative reduction potential, it follow that:-
In a Daniel cell, ______.
Which of the following is incorrect?
Explain the types of electrochemical cells.
