मराठी

What happens if external potential applied becomes greater than E°cell of electrochemical cell? - Chemistry

Advertisements
Advertisements

प्रश्न

What happens if external potential applied becomes greater than E°cell of electrochemical cell?

टीपा लिहा
Advertisements

उत्तर

If the external potential applied becomes greater than the E°cell of electrochemical cell, then the reaction gets reversed and the electrochemical cell acts as an electrolytic cell and vice-versa.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
2021-2022 (March) Term 2 - Outside Delhi Set 1

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

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.


If 'I' stands for the distance between the electrodes and 'a' stands for the area of cross-section of the electrode, `"l"/"a"` refers to ____________.


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)


Assertion: pure iron when heated in dry air is converted with a layer of rust.

Reason: Rust has the compositionFe3O4.


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 ______.


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


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 value of Ksp for Hg2Cl2 (s) is X then the value of X will be ____ where pX = - log X.
Given:

\[\ce{Hg2Cl2 + 2e- -> 2Hg(l) + 2Cl-}\],  E° = 0.27 V

\[\ce{Hg+2 + 2e- -> 2Hg(l)}\]   E° = 0.81 V


Explain why the anode is of negative polarity in a galvanic cell.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×