Chapter 19, Problem 1E

From the observations listed, estimate the value of $E°$for the half-cell reaction ${\text{M}}^{\text{2+}}\left(\text{aq}\right){\text{+2e}}^{\text{-}}\to \text{M}\left(\text{s}\right)$ .

a. The metal M reacts with ${\text{HNO}}_{\text{2}}\left(\text{aq}\right)$ , but not with $\text{HCl}\left(\text{aq}\right)$ , M displaces Ag(aq), but not ${\text{Cu}}^{2+}\left(\text{aq}\right)$

b. The metal M reacts with HCI(aq), producing ${\text{H}}_{\text{2}}\left(\text{g}\right)$ , but displaces neither ${\text{Zn}}^{\text{2+}}\left(\text{aq}\right)$ nor ${\text{Fe}}^{2+}\left(\text{aq}\right)$ .

Interpretation Introduction

(a)

Interpretation:

The value of E°_cellshould be estimated for the half reaction of ${\text{M}}^{\text{2+}}(\text{aq})+2e^{-}\to \text{M(s)}$ where M reacts with HNO₃(aq), but not with HCl and M displaces Ag+(aq), but not Cu²⁺(aq).

Concept introduction:

The solid metal should have higher reactivity than the aqueous metal in order to observe the displacement.

In order to react with an acid, the acid anion should have higher reduction potential while oxidizing the metal to metal cation in the acidic medium.

Answer to Problem 1E

$0.340<E^{\circ }{}_{\text{cell}}<0.800$

Explanation of Solution

If M displaces the Ag⁺(aq) the $E{°}_{cell}$ of given M reduction half-cell should be lower than the $E{°}_{cell}$ value of the Ag+ reduction and if M doesn’t displaces the Cu²⁺(aq) the $E{°}_{cell}$ of given M reduction half-cell should be higher than the $E{°}_{cell}$ value of the Cu²+ reduction.

Therefore, lower limit of the estimation for the $E{°}_{cell}$ of given M reduction half-cell should be the value of $E{°}_{cell}$ of the Cu²⁺(aq) reduction and the upper limit of the estimation for the $E{°}_{cell}$ of given M reduction half-cell should be the value of $E{°}_{cell}$ of the Ag⁺(aq).

$\begin{array}{l}{E}^{\circ }{}_{{\text{Cu}}^{\text{+}}\text{ Red}}<E^{\circ }{}_{\text{cell}}<E^{\circ }{}_{{\text{Ag}}^{\text{+}}\text{ Red}}\\ 0.340<E^{\circ }{}_{\text{cell}}<0.800\end{array}$

Interpretation Introduction

(b)

Interpretation:

The value of $E{°}_{cell}$ should be estimated for the half reaction of ${\text{M}}^{\text{2+}}(\text{aq})+2e^{-}\to \text{M(s)}$ where M reacts with HCl, producing H₂(g), but displaces neither Zn+(aq) nor Fe²⁺(aq).

Concept introduction:

The solid metal should have higher reactivity than the aqueous metal in order to observe the displacement.

In order to react with an acid, the acid anion should have higher reduction potential while oxidizing the metal to metal cation in the acidic medium.

Answer to Problem 1E

$-0.440<E^{\circ }{}_{\text{cell}}<0.00$

Explanation of Solution

If displaces neither Zn⁺(aq) nor Fe²⁺(aq), the $E{°}_{cell}$ of given M reduction half-cell should be higher than the both $E{°}_{cell}$ values while it should be lower than the $E{°}_{cell}$ value of H+ reduction to H₂(g).

Therefore, lower limit of the estimation for the $E{°}_{cell}$ of given M reduction half-cell should be the value of $E{°}_{cell}$ of the Fe²⁺(aq) reduction and the upper limit of the estimation for the $E{°}_{cell}$ of given M reduction half-cell should be the value of $E{°}_{cell}$ of the H⁺(aq).

$\begin{array}{l}{E}^{\circ }{}_{{\text{Fe}}^{\text{2+}}\text{ Red}}<E^{\circ }{}_{\text{cell}}<E^{\circ }{}_{{\text{Zn}}^{\text{2+}}\text{ Red}}\\ -0.440<E^{\circ }{}_{\text{cell}}<0.00\end{array}$