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In our last post, we looked at the overview of electrode potentials, where we discussed metal ions/metal systems or half-cells, standard electrode potential and electrochemical cells in depth. Here, we will focus on the calculations involving electrode potentials, which include calculations of  the electromotive force (e.m.f) of electrochemical cells ,  the relationship between e .m.f & free energy and the relationship between e .m.f & equilibrium constant. Half-Cell Reactions                      Std Reduction Potential , E° (V) K+(aq) + e- <----> K(s)                              -2.92 Ca2+(aq) + 2e- <----> Ca(s)               ...

Metal Ions / Metal Systems If a metal plate or rod is placed in a solution of its own salt, two processes - oxidation and reduction, take place simultaneously. The atoms from the metal's surface, lose electrons and go into solution as ions, as the electrons lost are left on the surface of the metal plate, thereby making it electrons surplus and negatively charged.                    M(s) ----> M+(aq) + e- Conversely, the metallic ions in the salt solution gain electrons from the surface of the metal plate and get deposited as metallic atoms. The electrons gained from the surface of the electrode renders it electron deficient and positively charged.                     M+(aq) + e- ----> M(s) Depending on the nature of the metal, a particular process will predominate. Hence, a potential differe...

The oxidation number of an element in a compound or radical   is the valency or combining power of the element in that compound or radical. It is also defined as the electrical charge an element appears to have in a given molecule or ion. The importance of oxidation number in chemistry cannot be overemphasized, as it can be used to predict the nature of a reaction. A change in oxidation number is used to reflect an electron gain or loss in a reaction. If the oxidation number increases, it means there is an electron loss (oxidation), while a decrease in oxidation number shows an electron gain (reduction). Determining the oxidation numbers of element follows a set of rules as discussed below: Rules For Determining Oxidation Numbers 1) The oxidation number of all elements in the free state is zero. By free state, we mean their uncombined state with any other element. e.g., F2, Mg, O3, Cl2, S, Ca, K, Na, Zn etc. 2) The oxidation number of a simple ion is equal to the charge...

Just like any other chemical equation, a redox reaction must be balanced for the law of conservation of mass (matter) to be obeyed. Unlike other chemical equations, balancing redox equations goes beyond making sure that the atom counts of the respective elements on both sides of the equation are the same. It also involves the balancing of charges, because the fundamental process in redox reactions is the transfer of electrons. Now, talking about charges, before you think of balancing a redox equation, we must first learn how to change the given equation into an ionic form. Let's consider the reaction between Zn and CuSO4 to form ZnSO4 and Cu as an example:                         Zn(s) + CuSO4(aq) ---> ZnSO4(aq) + Cu(s) ……(i) In the above reaction, zinc metal reacts with an aqueous solution of copper(II)tetraoxosulphate(VI) to form a mixture of aqueous solutio...

A redox reaction is a type of chemical reaction that involves two opposite, yet complementary processes - reduction and oxidation (redox). By being complementary, it means that one cannot occur without the other; while one is giving out, the other is receiving. We know that the act of giving can never be complete if there is nothing no one to receive. Therefore, while oxidation is the loss of electrons or increase in oxidation number, reduction is said to be gain of electrons or decrease (reduction) in oxidation number. Example :                       Zn(s) + Cu2+(aq) ---> Zn(aq) + Cu(s) In the above equation, the zinc metal displaces the copper(II)ion out of solution to form zinc ion and Cu metal. The oxidation state/number of zinc changes from 0 to +2 (increase) due to the loss of 2 electrons, while the oxidation state/number of copper changes from +2 to 0 (decrease) beca...