Question

The oxidation potentials of Zn, Cu, Ag, ${H_2}$ and Ni are 0.76, - 0.34,- 0.80, 0 and 0.25 volt respectively. Which of the following reactions will provide maximum voltage?
(A) $Zn + C{u^{2 + }} \to Cu + Z{n^{2 + }}$
(B) $Zn + 2A{g^ + } \to 2Ag + Z{n^{2 + }}$
(C) ${H_2} + C{u^{2 + }} \to 2{H^ + } + Cu$
(D) ${H_2} + N{i^{2 + }} \to 2{H^ + } + Ni$
(E) $Zn\left( s \right) + 2{H^ + }\left( {aq.} \right) \rightleftharpoons Z{n^{2 + }}\left( {aq} \right) + {H_2}\left( g \right)$

Answer 1

A chemical reaction having more E.M.F. value has the maximum potential.
$E.M.F. =$ ${E_{cathode}} - {E_{anode}}$if we consider the reduction potential.
Oxidation potential and reduction potential are opposite to each other.

Complete step by step answer:
When an atom is oxidized, it loses electrons. Oxidation takes place at anode of an electrochemical cell. When an atom or ion is reduced, it gains electron. Reduction takes place at cathode of the cell. The difference between the E.M.F at cathode and anode determines the total E.M.F of the cell.
If we consider the reduction potential, then E.M.F. = ${E_{cathode}} - {E_{anode}}$ and if we consider the oxidation potential, then E.M.F. = ${E_{anode}} - {E_{cathode}}$.
According to the question, oxidation potential of Zn = 0.76 V
Reduction potential of Zn = - 0.76 V
Oxidation potential of Cu = - 0.34 V
Reduction potential of Cu = 0.34 V
Oxidation potential of Ag = - 0.80 V
Reduction potential of Ag = 0.80 V
oxidation potential of ${H_2}$ = 0 V
Reduction potential of ${H_2}$ = 0 V
oxidation potential of Ni = 0.25 V
Reduction potential of Ni = - 0.25V
As we know that, the reduction potential values are opposite in sign of oxidation potential.
For proceeding to the problem, reduction potential is taken into consideration here.
(A) In the reaction $Zn + C{u^{2 + }} \to Cu + Z{n^{2 + }}$, zinc is oxidized at anode and copper is reduced at cathode. So,the E.M.F. of this reaction is given by,
E.M.F = ${E_{cathode}} - {E_{anode}}$ = 0.34 - ( - 0.76) = 1.10 V
(B) In the reaction $Zn + 2A{g^ + } \to 2Ag + Z{n^{2 + }}$, zinc is oxidized at anode and silver is reduced at cathode. So, the E.M.F. of this reaction is given by,
E.M.F = ${E_{cathode}} - {E_{anode}}$ = 0.80 - ( - 0.76) = 1.56 V
(C) In the reaction ${H_2} + C{u^{2 + }} \to 2{H^ + } + Cu$, copper is reduced and hydrogen is oxidized. So, the E.M.F of the reaction is given by,
E.M.F = ${E_{cathode}} - {E_{anode}}$ = 0.34 - 0 = 0.34 V
(D) In the reaction ${H_2} + N{i^{2 + }} \to 2{H^ + } + Ni$, reduction of nickel is taking place and oxidation of hydrogen is taking place. So, the E.M.F of this reaction is given by,
E.M.F = ${E_{cathode}} - {E_{anode}}$ = - 0.25 - 0 = - 0.25 V
(E) In the reaction $Zn\left( s \right) + 2{H^ + }\left( {aq.} \right) \rightleftharpoons Z{n^{2 + }}\left( {aq} \right) + {H_2}\left( g \right)$, zinc is oxidized and hydrogen is reduced. So, the E.M.F of the reaction is given by,
E.M.F = ${E_{cathode}} - {E_{anode}}$ = 0 - ( - 0.76) = 0.76 V
Since, the E.M.F of the reaction $Zn + 2A{g^ + } \to 2Ag + Z{n^{2 + }}$ is more, it has maximum potential.

Hence, the correct answer is (B) $Zn + 2A{g^ + } \to 2Ag + Z{n^{2 + }}$.

Note: Oxidation potential means that the potential at which oxidation occurs at the anode in an electrochemical cell.
Reduction potential is the potential at which reduction occurs at the cathode of an electrochemical cell.
When both oxidation and reduction take place simultaneously, it is called a redox reaction.
We should be more careful regarding the sign of the oxidation or reduction potential during solving a problem.