Question

The electrode potential, $E^{\circ}$ , for the reduction of $MnO^{-}_{4}$ to $Mn^{2+}$ in acidic medium is $+1.51\, V$ . Which of the following metal(s) will be oxidised? The reduction reactions and standard electrode potentials for $Zn^{2+}, Ag^{+}$ , and $Au^{+}$ are given as $Zn^{2+} (aq) +2e \to Zn(s), E^{\circ} = -0.762\, V$ $Ag+ (aq) +e {\rightleftharpoons} Ag (s), E^{\circ}=+0.80 \,V$ $Au^{+} (aq) +e {\rightleftharpoons} Au(s), E^{\circ}+1.69\,V$

• A. $Zn$ and $Au$
• B. $Ag$ and $Au$
• C. $Au$
• D. $Zn$ and $Ag$

Answer 1

Answer: (D)

$Zn$ and $Ag$

Answer 2

The reduction potential for $MnO^{-}_{4}$ to $Mn^{2+}$ is $+1.51\,V$
It can oxidise only those metals, which have reduction potential lower than $+1.51\,V$ For $Zn^{2+}/Zn$ and $Ag^{+}/Ag$, reduction potential given is $-0.762\,V$ and $+0.80\,V$ respectively. These values are lower than reduction potential of $MnO^{-}_{4}/Mn^{2+}$
Hence, $Zn$ and Ag will be oxidised. Since, for $Au^{+}/Au$, reduction potential is higher, hence it will not be oxidised by Mn.

Both the electrode potential and the reduction potential are referred to as the oxidation potential. If oxidation occurs at the electrode under these circumstances, An increase in electrons is a component of reduction. The reduction potential of an electrode is therefore defined as its propensity to create electrons. The electrode potential is the difference in equilibrium potential between a solution electrode and a metal electrode. It is also defined as an electrode's propensity to produce more and less electrons.

Calculating the individual's reversible electrode potential at the standard state with ions at an altered concentration of 1 mol dm-3 atmospheric pressure is an element (E°) in electrochemistry. The standard potential for reduction is then frequently expressed as the typical electrode potential. The standard reduction potential may be calculated by subtracting the standard reduction potential for the cathode reaction from the standard reduction potential for the reaction occurring at the anode. Put the cell potential together to obtain the smallest simple cell potential. A potential distinction is framed at the metal interface and the solution when a metal piece is Solutions in its own response particles. The amount of the potential difference is a function of the cathode's tendency to experience either an increase in oxidation or a reduction in oxidation, or its propensity to either lose or gain electrons. The answer is the half-reaction, and the particle and metal stand for the half-cell. An electrode is the flooded metal, and the potential resulted from a reaction at the electrode contact. The electrode potential is the term for the answer. Accordingly, the terminal potential is represented as the cathode's propensity to either lose or gain electrons.

The oxidation potential is referred to as a result of the oxidation occurring at the electrode centre.

Potential for oxidation minus that of reduction.