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Chapter 20: Problem 99

Predict whether the following reactions will be spontaneous in acidic solution under standard conditions: (a) oxidation of \(S n\) to \(S n^{2+}\) by \(I_{2}(\) to form I \(),\) (b) reduction (a) oxidation of \(\mathrm{Sn}\) to \(\mathrm{Sn}^{2+}\) by \(\mathrm{I}_{2}\) \(( \text { to form } \mathrm{I})\); (b) reduction of \(\mathrm{Ni}^{2+}\) to \(\mathrm{Ni}\) by \(\mathrm{I}^{-}(\) to form \(\mathrm{I}_{2}),(\mathbf{c})\) reduction of \(\mathrm{Ce}^{4+}\) to \(\mathrm{Ce}^{3+}\) by \(\mathrm{H}_{2} \mathrm{O}_{2}\) (d) reduction of \(\mathrm{Cu}^{2+}\) to Cu by \(\operatorname{Sn}^{2+}(\) to form \( \mathrm{Sn}^{4+} )\).

Short Answer

Expert verified
In summary, based on the standard cell potential (E°) calculations, we can predict the following reactions in acidic solution under standard conditions: (a) The oxidation of Sn to Sn²⁺ by I₂ to form I is spontaneous (E° = 0.68 V, positive). (b) The reduction of Ni²⁺ to Ni by I⁻ to form I₂ is non-spontaneous (E° = -0.79 V, negative).

Step by step solution

01

Write the half-reactions of the reaction

We can write the half-reactions as follows: Oxidation of Sn: Sn -> Sn²⁺ + 2e⁻ Reduction of I2: I₂ + 2e⁻ -> 2I⁻
02

Look up the standard reduction potentials for each half-reaction

We can use the standard reduction potentials table to find the values for each half-reaction: E°(Sn²⁺/Sn) = -0.14 V E°(I₂/I⁻) = +0.54 V
03

Calculate the standard cell potential (E°) for the reaction

To calculate E° for the reaction, we subtract E° of the reduction (oxidation) half-reaction from E° of the oxidation (reduction) half-reaction: E° = E°(I₂/I⁻) - E°(Sn²⁺/Sn) = 0.54 - (-0.14) = 0.68 V Since E° is positive, the reaction is spontaneous in acidic solution under standard conditions. (b) Reduction of Ni²⁺ to Ni by I⁻ to form I₂
04

Write the half-reactions of the reaction

We can write the half-reactions as follows: Oxidation of I⁻: 2I⁻ -> I₂ + 2e⁻ Reduction of Ni²⁺: Ni²⁺ + 2e⁻ -> Ni
05

Look up the standard reduction potentials for each half-reaction

We can use the standard reduction potentials table to find the values for each half-reaction: E°(Ni²⁺/Ni) = -0.25 V E°(I₂/I⁻) = +0.54 V
06

Calculate the standard cell potential (E°) for the reaction

To calculate E° for the reaction, we subtract E° of the reduction (oxidation) half-reaction from E° of the oxidation (reduction) half-reaction: E° = E°(Ni²⁺/Ni) - E°(I₂/I⁻) = -0.25 - 0.54 = -0.79 V Since E° is negative, the reaction is non-spontaneous in acidic solution under standard conditions. We will follow the same process for reactions (c) and (d).

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Most popular questions from this chapter

At \(900^{\circ} \mathrm{C},\) titanium tetrachloride vapor reacts with molten magnesium metal to form solid titanium metal and molten magnesium chloride. (a) Write a balanced equation for this reaction. (b) What is being oxidized, and what is being reduced? (c) Which substance is the reductant, and which is the oxidant?

At 298 \(\mathrm{K}\) a cell reaction has a standard cell potential of \(+0.17 \mathrm{V} .\) The equilibrium constant for the reaction is \(5.5 \times 10^{5} .\) What is the value of \(n\) for the reaction?

For each of the following balanced oxidation-reduction reactions, (i) identify the oxidation numbers for all the elements in the reactants and products and (ii) state the total number of electrons transferred in each reaction. $$ \begin{array}{l}{\text { (a) } 2 \mathrm{MnO}_{4}^{-}(a q)+3 \mathrm{S}^{2-}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 3 \mathrm{S}(s)+} \\ {\quad 2 \mathrm{MnO}_{2}(s)+8 \mathrm{OH}^{-}(a q)} \\ {\text { (b) } 4 \mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{Cl}_{2} \mathrm{O}_{7}(g)+2 \mathrm{OH}^{-}(a q) \longrightarrow 2 \mathrm{ClO}_{2}^{-}(a q)+} \\ {\quad 5 \mathrm{H}_{2} \mathrm{O}(l)+4 \mathrm{O}_{2}(g)} \\\\{\text { (c) } \mathrm{Ba}^{2+}(a q)+2 \mathrm{OH}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q)+2 \mathrm{ClO}_{2}(a q) \longrightarrow} \\ {\quad \mathrm{Ba}\left(\mathrm{ClO}_{2}\right)_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)}\end{array} $$

A voltaic cell is constructed that is based on the following reaction: $$ \mathrm{Sn}^{2+}(a q)+\mathrm{Pb}(s) \longrightarrow \mathrm{Sn}(s)+\mathrm{Pb}^{2+}(a q) $$ (a) If the concentration of \(\mathrm{Sn}^{2+}\) in the cathode half-cell is 1.00\(M\) and the cell generates an emf of \(+0.22 \mathrm{V},\) what is the concentration of \(\mathrm{Pb}^{2+}\) in the anode half-cell? (b) If the anode half-cell contains \(\left[\mathrm{SO}_{4}^{2-}\right]=1.00 M\) in equilibrium with \(\mathrm{PbSO}_{4}(s),\) what is the \(K_{s p}\) of \(\mathrm{PbSO}_{4} ?\)

Complete and balance the following half-reactions. In each case, indicate whether the half-reaction is an oxidation or a reduction. $$ \text { (a)} \mathrm{Sn}^{2+}(a q) \longrightarrow \mathrm{Sn}^{4+}(a q) \text {(acidic solution)} \\ \text {(b)} \mathrm{TiO}_{2}(s) \longrightarrow \mathrm{Ti}^{2+}(a q) \text {(acidic solution)} \\ \text {(c)} \mathrm{ClO}_{3}^{-}(a q) \longrightarrow \mathrm{Cl}^{-}(a q) \text {(acidic solution)} \\ \text {(d)} \mathrm{N}_{2}(g) \longrightarrow \mathrm{NH}_{4}^{+}(a q) \text {(acidic solution)} \\ \text {(e)} \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{O}_{2}(g) \text {(acidic solution)} \\ \text {(f)} \operatorname{SO}_{3}^{2-}(a q) \longrightarrow \mathrm{SO}_{4}^{2-}(a q) \text {(acidic solution)} \\\\(\mathrm{g}) \mathrm{N}_{2}(g) \longrightarrow \mathrm{NH}_{3}(g) \text {(acidic solution)} $$
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