Question

\[{{E}^{\circ }}\] value of \[N{{i}^{2+}}/Ni\] is -0.25V and \[A{{g}^{+}}/Ag\] is +0.80V. If a cell is made by taking the two electrodes. What is the feasibility of the reaction?
a.) Since \[{{E}^{\circ }}\] value for the cell will be positive, a redox reaction is feasible.
b.) Since \[{{E}^{\circ }}\] value for the cell will be negative, a redox reaction is not feasible.
c.) Ni cannot reduce \[A{{g}^{+}}\] to Ag hence reaction is not feasible
d.) Ag can reduce \[N{{i}^{2+}}\] to Ni hence reaction hence reaction is feasible.

Answer 1

Hint: To solve the given question, calculate the overall value of standard electrode potential by using individual values of oxidation potential and reduction potential. Redox reaction is the type of reaction, where oxidation and reduction occur simultaneously.

Complete step by step answer:
Given,
\[{{E}^{\circ }}\] value (Standard electrode potential) of \[N{{i}^{2+}}/Ni\] = -0.25V
\[{{E}^{\circ }}\] value (Standard electrode potential) of \[A{{g}^{+}}/Ag\] = +0.80V
For a reaction to be feasible, the electrode potential of the cell should be positive.
As we can see, the electrode potential for reduction of Silver (Ag) is positive. So, Ag will get reduced and oxidize Nickel.
\[{{E}^{\circ }}\] value (Standard electrode potential) of \[Ni/N{{i}^{2+}}\] = +0.25V

The net electrode potential will be –
\[E_{net}^{\circ }=E_{oxidation}^{\circ }+E_{reduction}^{\circ }\]
\[E_{net}^{\circ }\]= 0.25 + 0.80 V
\[E_{net}^{\circ }\]= 1.05 v
Therefore, the answer is – Since \[{{E}^{\circ }}\] value for the cell will be positive, a redox reaction is feasible.
So, the correct answer is “Option A”.

Note: The standard electrode potential of fluorine is the highest in electrochemical series.
 Standard electrode potential is defined as, “the measures the individual potential of reversible electrode at standard state with ions at an effective concentration of 1\[mol/d{{m}^{3}}\] at the pressure of 1 atm”.