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Chapter 18: Problem 95

Consider a voltaic cell in which the following reaction occurs. $$ \mathrm{Zn}(s)+\mathrm{Sn}^{2+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{Sn}(s) $$ (a) Calculate \(E^{\circ}\) for the cell. (b) When the cell operates, what happens to the concentration of \(\mathrm{Zn}^{2+}\) ? The concentration of \(\mathrm{Sn}^{2+}\) ? (c) When the cell voltage drops to zero, what is the ratio of the concentration of \(\mathrm{Zn}^{2+}\) to that of \(\mathrm{Sn}^{2+}\) ? (d) If the concentration of both cations is \(1.0 \mathrm{M}\) originally, what are the concentrations when the voltage drops to zero?

Short Answer

Expert verified
In summary, the process for solving this exercise involved the following steps: (a) The E掳 for the cell was calculated as 0.62V, based on the standard reduction potentials of the half-reactions. (b) As the cell operates, the concentration of Zn虏鈦� increases while the concentration of Sn虏鈦� decreases. (c) The ratio of the concentrations of Zn虏鈦� to Sn虏鈦� when the voltage drops to zero was found using the Nernst equation: [Zn虏鈦篯/[Sn虏鈦篯 = e^((0.62V * 2F) / RT). (d) The equilibrium concentrations of Sn虏鈦� and Zn虏鈦� can be determined by solving the equation numerically. The resulting concentrations will depend on the given temperature and initial concentrations, which were not provided in the question.

Step by step solution

01

Find the standard reduction potentials

To calculate the E掳 for the given cell, you'll first need the standard reduction potentials for the half-reactions: Zn (s) 鈫� Zn虏鈦�(aq) + 2e鈦� (oxidation) Sn虏鈦�(aq) + 2e鈦� 鈫� Sn (s) (reduction) You can find these values in a table of standard reduction potentials. The standard reduction potentials are: E掳(Zn虏鈦�/Zn) = -0.76 V (a negative value denotes oxidation) E掳(Sn虏鈦�/Sn) = -0.14 V (a negative value denotes reduction)
02

Calculate E掳 for the cell

Now that you have the standard reduction potentials for both half-reactions, you can calculate the E掳 for the cell by adding the two potentials together. E掳cell = E掳(reduction) - E掳(oxidation) E掳cell = (-0.14V) - (-0.76V) E掳cell = 0.62V (a) The E掳 for the cell is 0.62V.
03

Determine changes in concentration

As the cell operates, Zn (s) is oxidized and Zn虏鈦� (aq) is produced, so the concentration of Zn虏鈦� increases. Concurrently, Sn虏鈦�(aq) gains electrons and is reduced to Sn(s), so the concentration of Sn虏鈦� decreases. (b) Concentration of Zn虏鈦� increases; concentration of Sn虏鈦� decreases.
04

Calculate the ratio of concentrations when voltage drops to zero

When the cell voltage drops to zero, the reaction reaches equilibrium. You will use the Nernst equation to determine the ratio of concentrations: 螖E = E掳 - (RT/nF) lnQ Here, 螖E is the cell voltage, E掳 is the standard cell potential, R is the gas constant, T is the temperature in Kelvin, n is the number of electrons transferred (2 in this case), F is the Faraday constant, and Q is the reaction quotient. When the cell voltage drops to zero: 0 = 0.62V - (RT/2F) ln([Zn虏鈦篯/[Sn虏鈦篯) You can now rearrange the equation to find the ratio of concentrations: ln([Zn虏鈦篯/[Sn虏鈦篯) = (0.62V * 2F) / RT (c) The ratio of the concentration of Zn虏鈦� to that of Sn虏鈦�: [Zn虏鈦篯/[Sn虏鈦篯 = e^((0.62V * 2F) / RT)
05

Calculate the concentrations when the voltage drops to zero

Given that the initial concentrations of both cations (Zn虏鈦� and Sn虏鈦�) were 1.0M: [Zn虏鈦篯/[Sn虏鈦篯 = x / (1 - x) Where x is the concentration of Sn虏鈦� at equilibrium and (1 - x) is the concentration of Zn虏鈦� at equilibrium. You can now substitute the ratio of concentrations previously calculated: x / (1 - x) = e^((0.62V * 2F) / RT) Now, solve for x and (1 - x) numerically to find the equilibrium concentrations of Sn虏鈦� and Zn虏鈦�: (d) The concentrations of Zn虏鈦� and Sn虏鈦� when the voltage drops to zero are obtained by solving the above equation numerically.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Reduction Potential
In a voltaic cell, reactions occur in two separate half-cells, each featuring its own half-reaction. The standard reduction potential is a key concept in understanding these processes. It refers to the tendency of a substance to gain electrons, measured in volts under standard conditions (25掳C, 1M concentration, and 1 atm pressure). Each half-cell has a standard reduction potential, which helps in determining the direction of electron flow. For instance, in the exercise provided:
  • The zinc half-reaction ( Zn (s) 鈫� Zn虏鈦�(aq) + 2e鈦� ) has a standard reduction potential of -0.76 V, denoting oxidation.
  • The tin half-reaction ( Sn虏鈦�(aq) + 2e鈦� 鈫� Sn (s) ) has a standard reduction potential of -0.14 V, indicating reduction.
The more positive the standard reduction potential, the greater the substance's tendency to be reduced. This driving force determines which half-reaction will proceed as a reduction and which as oxidation.
Cell Voltage
In a voltaic cell, the cell voltage is the difference in potential between the two half-cells. It represents the cell鈥檚 power to perform electrical work when discharging. Calculating the cell voltage involves using the standard reduction potentials of the half-reactions involved:
  • The overall cell potential ( E掳_{cell} ) is found by subtracting the oxidation potential from the reduction potential.
  • It can be written as E掳_{cell} = E掳_{reduction} - E掳_{oxidation} . In our example, this translates to E掳_{cell} = -0.14V - (-0.76V) = 0.62V .
A positive cell voltage indicates a spontaneous reaction, suggesting that the cell can energetically drive the reaction forward and produce an electric current. The calculated voltage (0.62 V) shows a functioning voltaic cell at standard conditions.
Nernst Equation
The Nernst equation is used to calculate the cell potential under non-standard conditions by accounting for concentrations of the reactants and products. It is an important tool for assessing how cell voltage changes as the reaction proceeds toward equilibrium:
  • The Nernst equation is given by 螖E = E掳 - (RT/nF) lnQ , where 螖E is the cell voltage, E掳 is the standard cell potential, R is the gas constant, T is temperature in Kelvin, n is the number of moles of electrons transferred, and F is the Faraday constant.
  • Q is the reaction quotient, represented by the ratio of the concentrations of products to reactants, each raised to the power of their stoichiometric coefficients.
When the cell voltage reaches zero, the equation demonstrates the relationship between concentration changes and voltage change, balancing the driving forces of the reaction.
Reaction Equilibrium
In a voltaic cell, the reaction equilibrium occurs when the forward and reverse reactions proceed at the same rate, resulting in no net change in reactant or product concentration. This state is achieved when the cell voltage drops to zero:
  • At equilibrium, the reaction quotient Q equals the equilibrium constant K , signifying that the concentrations of reactants and products have stabilized.
  • In our exercise, with a cell voltage of zero, the Nernst equation comes into play to establish the equilibrium ratios. We use it to determine the ratio of concentrations ( [Zn虏鈦篯/[Sn虏鈦篯 ) by solving it for ln([Zn虏鈦篯/[Sn虏鈦篯) = (0.62V * 2F) / RT .
  • In equilibrium, both the concentration of oxidized zinc increases and reduced tin decreases until these forces are equally opposed by their respective reaction tendencies.
Understanding equilibrium in voltaic cells is crucial in designing efficient batteries and predicting how they'll behave over time.

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Most popular questions from this chapter

Which species in each pair is the stronger reducing agent? (a) \(\mathrm{Cr}\) or \(\mathrm{Cd}\) (b) \(\mathrm{I}^{-}\) or \(\mathrm{Br}^{-}\) (c) \(\mathrm{OH}^{-}\) or \(\mathrm{NO}_{2}^{-}\) (d) NO in acidic solution or NO in basic solution

Calculate \(E^{\circ}\) for the following cells: (a) \(\mathrm{Pb}\left|\mathrm{PbSO}_{4} \| \mathrm{Pb}^{2+}\right| \mathrm{Pb}\) (b) \(\mathrm{Pt}\left|\mathrm{Cl}_{2}\right| \mathrm{ClO}_{3}^{-} \| \mathrm{O}_{2}\left|\mathrm{H}_{2} \mathrm{O}\right| \mathrm{Pt}\) (c) \(\mathrm{Pt}\left|\mathrm{OH}^{-}\right| \mathrm{O}_{2} \| \mathrm{ClO}_{3}^{-}, \mathrm{Cl}^{-} \mid \mathrm{Pt} \quad\) (basic medium)

For the following half-reactions, answer the questions below. $$ \begin{array}{cc} \mathrm{Co}^{3+}(a q)+e^{-} \longrightarrow \mathrm{Co}^{2+}(a q) & E^{\circ}=+1.953 \mathrm{~V} \\ \mathrm{Fe}^{3+}(a q)+e^{-} \longrightarrow \mathrm{Fe}^{2+}(a q) & E^{\circ}=+0.769 \mathrm{~V} \\ \mathrm{I}_{2}(a q)+2 e^{-} \longrightarrow 2 \mathrm{I}^{-}(a q) & E^{o}=+0.534 \mathrm{~V} \\ \mathrm{~Pb}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Pb}(s) & E^{\circ}=-0.127 \mathrm{~V} \\ \mathrm{Cd}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Cd}(s) & E^{\circ}=-0.402 \mathrm{~V} \\ \mathrm{Mn}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Mn}(s) & E^{\circ}=-1.182 \mathrm{~V} \end{array} $$ (a) Which is the weakest reducing agent? (b) Which is the strongest reducing agent? (c) Which is the strongest oxidizing agent? (d) Which is the weakest oxidizing agent? (e) Will \(\mathrm{Pb}(s)\) reduce \(\mathrm{Fe}^{3+}(a q)\) to \(\mathrm{Fe}^{2+}(a q) ?\) (f) Will \(\mathrm{I}^{-}(a q)\) reduce \(\mathrm{Pb}^{2+}(a q)\) to \(\mathrm{Pb}(s) ?\) (g) Which ion(s) can be reduced by \(\mathrm{Pb}(s)\) ? (h) Which if any metal(s) can be oxidized by \(\mathrm{Fe}^{3+}(a q)\) ?

Calculate voltages of the following cells at \(25^{\circ} \mathrm{C}\) and under the following conditions. (a) \(\mathrm{Fe}\left|\mathrm{Fe}^{2+}(0.010 \mathrm{M}) \| \mathrm{Cu}^{2+}(0.10 \mathrm{M})\right| \mathrm{Cu}\) (b) \(\mathrm{Pt}\left|\mathrm{Sn}^{2+}(0.10 \mathrm{M}), \mathrm{Sn}^{4+}(0.010 \mathrm{M}) \| \mathrm{Co}^{2+}(0.10 \mathrm{M})\right| \mathrm{Co}\)

Which species in each pair is the stronger oxidizing agent? (a) \(\mathrm{NO}_{3}^{-}\) or \(\mathrm{I}_{2}\) (b) \(\mathrm{Fe}(\mathrm{OH})_{3}\) or \(\mathrm{S}\) (c) \(\mathrm{Mn}^{2+}\) or \(\mathrm{MnO}_{2}\) (d) \(\mathrm{ClO}_{3}^{-}\) in acidic solution or \(\mathrm{ClO}_{3}^{-}\) in basic solution
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