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This is one of our many quizzes on Cell potentials. Click on this to access more such competitive quizzes on Electrochemistry.

Cell potentials Quiz 2

chemistry

 The potential of a chemical can be measured by taking which of the following reference electrodes?

 To find the potential of a chemical we have to connect a glass electrode which is dipped in the chemical solution to the reference electrodes. 

At the determination of cell potentials of the chemical there is a need of reference electrodes. The commonly used reference electrodes are the Calomel electrode and hydrogen electrode.

 Lead-acid batteries are examples of electrochemical cells. What can we observe in the lead- acid electrochemical cell?

 Electrochemical cells are very useful in the field of automobile industries. 

In electrochemical cells chemical energy is going to convert into electrical energy and can be used as batteries in automobiles frequently.  

 Identify the oxidation half reaction of the zinc in the Daniel cell among the given options.

  Oxidation means loss of electrons and reduction means gain of electrons. 

The zinc undergoes oxidation and forms zinc ion as the product. In oxidation, zinc liberates electrons. The chemical reaction of oxidation of zinc is as follows. $Zn\to Z{{n}^{2+}}+2{{e}^{-}}$

 The cell reaction  $Zn+C{{u}^{2+}}\to ~~Z{{n}^{2+}}+Cu$ is correctly represented by…………

 Always the cell should be represented as anodic part | cathodic part. Oxidation occurs at anode and reduction occurs at cathode.

In the given chemical reaction the anodic chemical reaction is $Zn\to Z{{n}^{2+}}+2{{e}^{-}}$ and the cathodic chemical reaction is $C{{u}^{2+}}+2{{e}^{-}}\to Cu$. Therefore the total cell reaction is going to be represented as $Zn|Z{{n}^{2+}}||C{{u}^{\text{2+}}}\text{ }|\text{ Cu}$.

 The potential of two metal electrodes used in a cell are 0.20V and 1.1V. The e.m.f of the cell formed by combining them is								

  The potential of a cell should not be negative. Always the potential of the cell should be positive or zero. 

The given cell potentials are $0.20 V$ and $1.1 V.$ The formula to calculate the cell potential is $E={{E}_{cathode}}-{{E}_{anode}}$ . If we are going to take $0.20 V$ as a cathode potential and $1.1 V$ as anode potential. Then cell potential will be $= 0.20 – 1.1 = - 0.90 V.$ The cell potential should not be negative. Therefore we have to take $1.1 V$ as a cathode potential and $0.20 V$ as anode potential. Then the cell potential will be $= 1.1 – 0.20 = 0.90 V.$