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Chapter 20: Problem 42

A voltaic cell consists of a strip of cadmium metal in a solution of \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\) in one beaker, and in the other beaker a platinum electrode is immersed in a NaCl solution, with \(\mathrm{Cl}_{2}\) gas bubbled around the electrode. A salt bridge connects the two beakers. (a) Which electrode serves as the anode, and which as the cathode? (b) Does the Cd electrode gain or lose mass as the cell reaction proceeds? (c) Write the equation for the overall cell reaction. (d) What is the emf generated by the cell under standard conditions?

Short Answer

Expert verified
In this voltaic cell, the anode is the Cd electrode, and the cathode is the Pt electrode. The Cd electrode loses mass as the cell reaction proceeds. The overall cell reaction is: \( \mathrm{Cd} + \mathrm{Cl}_{2} \rightarrow \mathrm{Cd}^{2+} + 2\mathrm{Cl}^{-} \). Under standard conditions, the emf generated by the cell is 1.76 V.

Step by step solution

01

Identify anode and cathode

A voltaic cell contains two half-cells. In each half-cell, a redox reaction occurs. The electrode where oxidation occurs is called the anode, and the electrode where reduction occurs is called the cathode. In this cell, we have a Cd electrode in a Cd(NO3)2 solution and a Pt electrode in a NaCl solution with Cl2 gas. Cadmium half-cell (Cd to Cd2+): \[ \mathrm{Cd} \rightarrow \mathrm{Cd}^{2+} + 2\mathrm{e}^{-} \] Chlorine half-cell (Cl2 to 2Cl-): \[ \mathrm{Cl}_{2} + 2\mathrm{e}^{-} \rightarrow 2\mathrm{Cl}^{-} \] Here, Cd is being oxidized and Cl2 is being reduced. Thus: - Anode: Cd electrode - Cathode: Pt electrode
02

Determine mass change in Cd electrode

Since the Cd electrode serves as the anode, it undergoes oxidation, losing electrons and converting into Cd2+. Therefore, the Cd electrode loses mass as the cell reaction proceeds.
03

Write the overall cell reaction

Next, we combine the half-reactions to write the balanced overall cell reaction. \[ \mathrm{Cd} + \mathrm{Cl}_{2} \rightarrow \mathrm{Cd}^{2+} + 2\mathrm{Cl}^{-} \]
04

Calculate the emf under standard conditions

To calculate the emf under standard conditions, we use the following equation: \[ E_{cell} = E_{cathode} - E_{anode} \] Using the standard reduction potentials for the two half-reactions: - Cd2+ + 2e → Cd: \( E^{\circ} = -0.40V \) (anode) - Cl2 + 2e → 2Cl-: \( E^{\circ} = +1.36V \) (cathode) By substituting the values, \[ E_{cell} = (1.36) - (-0.40) = 1.76V \] The emf generated by the voltaic cell under standard conditions is 1.76 V.

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Most popular questions from this chapter

A disproportionation reaction is an oxidation-reduction reaction in which the same substance is oxidized and reduced. Complete and balance the following disproportionation reactions: (a) $\mathrm{Fe}^{2+}(a q) \longrightarrow \mathrm{Fe}(s)+\mathrm{Fe}^{3+}(a q)$ (b) $\mathrm{Br}_{2}(l) \longrightarrow \mathrm{Br}^{-}(a q)+\mathrm{BrO}_{3}^{-}(a q)$ (acidic solution) (c) $\mathrm{Cr}^{3+}(a q) \longrightarrow \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{Cr}(s)$ (acidic solution) (d) $\mathrm{NO}(g) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{NO}_{3}^{-}(a q)$ (acidic solution)

(a) Write the reactions for the discharge and charge of a nickel-cadmium (nicad) rechargeable battery. (b) Given the following reduction potentials, calculate the standard emf of the cell: $$ \begin{aligned} \mathrm{Cd}(\mathrm{OH})_{2}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cd}(s)+2 \mathrm{OH}^{-}(a q) & \\ E_{\mathrm{red}}^{\circ} &=-0.76 \mathrm{~V} \\ \mathrm{NiO}(\mathrm{OH})(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{e}^{-} \longrightarrow \mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{OH}^{-}(a q) \\ E_{\mathrm{red}}^{\circ} &=+0.49 \mathrm{~V} \end{aligned} $$ (c) A typical nicad voltaic cell generates an emf of \(+1.30 \mathrm{~V}\). Why is there a difference between this value and the one you calculated in part (b)? (d) Calculate the equilibrium constant for the overall nicad reaction based on this typical emf value.

Complete and balance the following half-reactions. In each case indicate whether the half-reaction is an oxidation or a reduction. (a) \(\mathrm{Mo}^{3+}(a q) \longrightarrow \mathrm{Mo}(s)\) (acidic solution) (b) $\mathrm{H}_{2} \mathrm{SO}_{3}(a q) \longrightarrow \mathrm{SO}_{4}^{2-}(a q)$ (acidic solution) (c) \(\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)\) (acidic solution) (d) \(\mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\) (acidic solution) (e) \(\mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\) (basic solution) (f) \(\mathrm{Mn}^{2+}(a q) \longrightarrow \mathrm{MnO}_{2}(s)\) (basic solution) (g) $\mathrm{Cr}(\mathrm{OH})_{3}(s) \longrightarrow \mathrm{CrO}_{4}^{2-}(a q)$ (basic solution)

For each of the following reactions, write a balanced equation, calculate the standard emf, calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K},\) and calculate the equilibrium constant \(K\) at \(298 \mathrm{~K}\). (a) Aqueous iodide ion is oxidized to \(\mathrm{I}_{2}(s)\) by \(\mathrm{Hg}_{2}^{2+}(a q) .\) (b) In acidic solution, copper(I) ion is oxidized to copper(II) ion by nitrate ion. (c) In basic solution, \(\mathrm{Cr}(\mathrm{OH})_{3}(s)\) is oxidized to \(\mathrm{CrO}_{4}^{2-}(a q)\) by \(\mathrm{ClO}^{-}(a q)\)

A cell has a standard cell potential of \(+0.257 \mathrm{~V}\) at $298 \mathrm{~K}$. What is the value of the equilibrium constant for the reaction \((\mathbf{a})\) if \(n=1 ?(\mathbf{b})\) if \(n=2 ?(\mathbf{c})\) if \(n=3 ?\)
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