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Chapter 19: Problem 15

Consider the following half-reactions: \(\mathrm{MnO}_{4}^{-}(a q)+8 \mathrm{H}^{+}(a q)+5 e^{-} \longrightarrow$$\mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l)\) \(\mathrm{NO}_{3}^{-}(a q)+4 \mathrm{H}^{+}(a q)+3 e^{-} \longrightarrow \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) Predict whether \(\mathrm{NO}_{3}^{-}\) ions will oxidize \(\mathrm{Mn}^{2+}\) to \(\mathrm{MnO}_{4}^{-}\) under standard-state conditions.

Short Answer

Expert verified
No, under standard-state conditions, \(NO_{3}^{-}\) ions will not oxidize \(Mn^{2+}\) to \(MnO_4^-\).

Step by step solution

01

Write down the reactions in the form of reduction half-reactions

The given reactions are already in the form of reduction half-reactions. \nFor \(MnO_4^-\), the half-reaction is: \[MnO_{4}^{-}(a q)+8 H^{+}(a q)+5 e^{-} \longrightarrow Mn^{2+}(a q)+4 H_{2}O(l)\]And for \(NO_{3}^{-}\), the half-reaction is:\[NO_{3}^{-}(a q)+4 H^{+}(a q)+3 e^{-} \longrightarrow NO(g)+2 H_{2}O(l)\]
02

Determine the Standard Cell Potentials

The standard electrode potentials or E° values for these half-reactions can be obtained from a table of standard reduction potentials. From such a table, the E° values are:For the \(MnO_4^-\) to \(Mn^{2+}\) half-reaction, E° = +1.51 VFor the \(NO_{3}^{-}\) to \(NO(g)\) half-reaction, E° = +0.96 V
03

Compare the Cell Potentials

The element with the higher standard reduction potential is more likely to gain electrons (be reduced), whereas the element with the lower standard reduction potential is more likely to lose electrons (be oxidised). As the E° value for \(MnO_4^-\) to \(Mn^{2+}\) half-reaction (+1.51 V) is higher than the E° value for the \(NO_{3}^{-}\) to \(NO(g)\) half-reaction (+0.96 V), it means that \(MnO_4^-\) has a greater tendency to be reduced to \(Mn^{2+}\) than \(NO_{3}^{-}\) being reduced to \(NO(g)\). Hence, under standard-state conditions, \(NO_{3}^{-}\) ions will not oxidize \(Mn^{2+}\) to \(MnO_4^-\).

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Most popular questions from this chapter

In the electrolysis of an aqueous \(\mathrm{AgNO}_{3}\) solution, \(0.67 \mathrm{~g}\) of \(\mathrm{Ag}\) is deposited after a certain period of time. (a) Write the half-reaction for the reduction of \(\mathrm{Ag}^{+}\). (b) What is the probable oxidation halfreaction? (c) Calculate the quantity of electricity used, in coulombs.

Steel hardware, including nuts and bolts, is often coated with a thin plating of cadmium. Explain the function of the cadmium layer.

What is the difference between a galvanic cell (such as a Daniell cell) and an electrolytic cell?

The magnitudes (but not the signs) of the standard reduction potentials of two metals \(X\) and \(Y\) are $$ \begin{array}{ll} \mathrm{Y}^{2+}+2 e^{-} \longrightarrow \mathrm{Y} & \left|E^{\circ}\right|=0.34 \mathrm{~V} \\ \mathrm{X}^{2+}+2 e^{-} \longrightarrow \mathrm{X} & \left|E^{\circ}\right|=0.25 \mathrm{~V} \end{array} $$ where the \(\|\) notation denotes that only the magnitude (but not the sign) of the \(E^{\circ}\) value is shown. When the half-cells of \(X\) and \(Y\) are connected, electrons flow from \(X\) to \(Y\). When \(X\) is connected to a SHE, electrons flow from \(X\) to SHE. (a) Are the \(E^{\circ}\) values of the half- reactions positive or negative? (b) What is the standard emf of a cell made up of \(X\) and \(Y ?\)

To remove the tarnish \(\left(\mathrm{Ag}_{2} \mathrm{~S}\right)\) on a silver spoon, a student carried out the following steps. First, she placed the spoon in a large pan filled with water so the spoon was totally immersed. Next, she added a few tablespoonfuls of baking soda (sodium bicarbonate), which readily dissolved. Finally, she placed some aluminum foil at the bottom of the pan in contact with the spoon and then heated the solution to about \(80^{\circ} \mathrm{C}\). After a few minutes, the spoon was removed and rinsed with cold water. The tarnish was gone and the spoon regained its original shiny appearance. (a) Describe with equations the electrochemical basis for the procedure. (b) Adding \(\mathrm{NaCl}\) instead of \(\mathrm{NaHCO}_{3}\) would also work because both compounds are strong electrolytes. What is the added advantage of using \(\mathrm{NaHCO}_{3}\) ? (Hint: Consider the \(\mathrm{pH}\) of the solution.) (c) What is the purpose of heating the solution? (d) Some commercial tarnish removers containing a fluid (or paste) that is a dilute \(\mathrm{HCl}\) solution. Rubbing the spoon with the fluid will also remove the tarnish. Name two disadvantages of using this procedure compared to the one described here.
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