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EMF of a Cell MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

  • Electrode Potential and EMF of Cells is considered one the most difficult concept.

  • 40 Questions around this concept.

Solve by difficulty

EMF of a cell in terms of reduction potential of its left and right electrodes is

The standard electrode potentials\mathrm{\left ( E_{M^{+}|M}^{0} \right )} of four metals A, B, C, and D are - 1.2 V, 0.6 V, 0.85 V, and - 0.76 V, respectively.  The sequence of deposition of metals on applying potential is :

The E^{0}{_{M^{3+}/M^{2+}}} values of Cr, Mn, Fe, and Co are -0.41, +1.57, 0.77, and +1.97 V respectively. For which one of these metals the change in oxidation state from +2 to +3 is easiest?

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A button cell used in watches functions as following

\mathrm{Zn ( s )+ Ag _2 O ( s )+ H _2 O (l) \rightleftharpoons 2 Ag ( s )+ Zn ^{2+}( aq )+2 OH ^{-}( aq )}

If half cell potentials are

\mathrm{Zn ^{2+}( aq )+2 e ^{-} \rightarrow Zn ( s ) ; E ^{\circ}=-0.76 V}

\mathrm{Ag _2 O ( s )+ H _2 O (l)+2 e ^{-} \longrightarrow 2 Ag ( s )+2 OH ^{-}( aq ); E ^{\circ}=0.34 V}

The cell potential will be:

Standard electrode potential of three metals X, Y and Z are -1.2 V, +0.5 V and -3.0 V, respectively. The reducing power of these metals will be:

Standard reduction electrode potentials of three metals A, B and C are +0.5 V,-3.0 V and -1.2V respectively. The reducing power of these metals are

Standard reduction potential of the half reaction are given below

F_{2}\left ( g \right )+2e^{-}\rightarrow 2F^{-}\left ( aq \right );E^{0}= +2.85V

Cl_{2}\left ( g \right )+2e^{-}\rightarrow 2Cl^{-}\left ( aq \right );E^{0}= +1.36V

Br_{2}\left ( 1 \right )+2e^{-}\rightarrow 2Br^{-}\left ( aq \right );E^{0}= +1.06V

I_{2}\left ( s \right )+2e^{-}\rightarrow 2I^{-}\left ( aq \right );E^{0}= +0.53V

The strongest oxidising and reducing agents respectively are:

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Given

E_{Cr^{3+}/Cr}^{0}= -0.74 V;E_{MnO_{4}^{-}/Mn^{2+}}^{0}= 1.51 V

E_{Cr_{2}O_{7}^{2-}/Cr^{3+}}^{0}= 1.33 V;E_{Cl/Cl^{-}}^{0}= 1.36 V

Based on the data given above ,strongest oxidising agent will be:

The correct order of E^{0}_{M^{2+}/M}values with negative sign for the four successive elements Cr ,Mn,Fe \: and \: Co is

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Concepts Covered - 1

Electrode Potential and EMF of Cells

It is the potential difference between the two terminals of the cell when no current is drawn from it. It is measured with the help of potentiometer or vacuum tube voltmeter.

Calculation of the EMF of the Cell
Mathematically, it may be expressed as

\begin{array}{l}{\mathrm{E}_{\text {cell }} \text { or } \mathrm{EMF}=\left[\mathrm{E}_{\text {red }}(\text { cathode })-\mathrm{E}_{\text {red }}(\text { anode })\right]} \\ {\mathrm{E}_{\text {eell }}^{\circ} \text { or } \mathrm{EMF}^{\circ}} \\ {=\left[\mathrm{E}_{\text {red }}^{\circ}(\text { cathode })-\mathrm{E}_{\text {red }}^{\circ}(\text { anode })\right]}\end{array}

  • For cell reaction to occur the Ecell should be positive. This can happen only if Ered (cathode) > Ered(anode).
  • Eo cell must be positive for a spontaneous reaction.
  • Rate of reaction is directly proportional to the emf of the cell.
  • The emf of the cell depends on the intensity of the reaction in the cell.
  • It measures free energy change for maximum convertibility of heat into useful work.
  • It causes flow of current from higher value of Eo electrode to lower Eo value electrode.

Difference between EMF and Cell Potential

EMF Cell Potential

It is measured by potentiometer.

It is measured by voltmeter.

It is potential difference between two electrodes when no current is flowing in the circuit.

It is potential difference between two electrodes when current is flowing through the circuit.

It is maximum voltage obtained from cell.

It is less than maximum voltage.

It corresponds to maximum useful work obtained from galvanic cell.

It does not correspond.

 

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Electrode Potential and EMF of Cells

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