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(a) In the Nernst equation, what is the numerical value of the reaction quotient, \(Q,\) under standard conditions? (b) Can the Nernst equation be used at temperatures other than room temperature?

Short Answer

Expert verified
(a) Under standard conditions, the numerical value of the reaction quotient (Q) is 1. (b) Yes, the Nernst equation can be used at temperatures other than room temperature as it contains the temperature term (T) in Kelvin and is valid at all temperatures if other variables are known.

Step by step solution

01

The Nernst Equation

The Nernst equation is a fundamental equation in electrochemistry that relates the potential (voltage) of an electrochemical cell to the concentration of its components. Mathematically, the Nernst equation is given by: \[E = E^0 - \frac{RT}{nF} lnQ\] where - \(E\) is the potential of the cell, - \(E^0\) is the standard potential (also known as the standard reduction potential) of the cell, - \(R\) is the universal gas constant, \(8.314 J \cdot mol^{-1} \cdot K^{-1}\), - \(T\) is the temperature in Kelvin, - \(n\) is the number of electrons transferred in the redox reaction, and - \(F\) is Faraday's constant, \(96500 C \cdot mol^{-1}\), - \(Q\) is the reaction quotient. Now let us address each part of the problem. Part (a): We are asked to find the numerical value of the reaction quotient (Q) under standard conditions.
02

Standard Conditions

Under standard conditions, the concentrations of all aqueous species are set to 1 M, and the pressure of all gaseous species is set to 1 atm. In this case, the reaction quotient (Q) equals the equilibrium constant (K). However, at standard conditions, the cell potential, \(E\), equals the standard potential, \(E^0\). Now we will use the Nernst equation to find the numerical value of Q under standard conditions.
03

Applying the Nernst Equation

Since \(E = E^0\) under standard conditions, we can rewrite the Nernst equation as: \[E^0 = E^0 - \frac{RT}{nF} lnQ\] Now, we need to solve for Q. Rearrange the equation to isolate Q: \[0 = - \frac{RT}{nF} lnQ\] \[lnQ = 0\] To find the numerical value of Q, take the exponent on both sides: \[Q = e^0\] \[Q = 1\] Under standard conditions, the numerical value of the reaction quotient (Q) is 1. Part (b): We are asked if the Nernst equation can be used at temperatures other than room temperature.
04

The Nernst Equation and Temperature

Since the Nernst equation contains the temperature term \(T\) in Kelvin, there is no restriction on using the Nernst equation only at room temperature. The Nernst equation is valid at all temperatures as long as other variables such as the concentration of the species and standard electrode potentials are known. So, yes, the Nernst equation can be used at temperatures other than room temperature.

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Key Concepts

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

Understanding Reaction Quotient
When dealing with electrochemical cells, the reaction quotient, denoted as \(Q\), plays a crucial role. It quantifies the ratio of product concentrations to reactant concentrations at any point in time during a chemical reaction. In the context of the Nernst equation, \(Q\) helps determine the cell potential relative to the standard condition where equilibrium is not necessarily achieved.
Under standard conditions, which we'll discuss in more detail later, every species is at 1 M concentration, or 1 atm pressure for gases. This simplifies the reaction quotient to \(Q = 1\), because all terms cancel out when divided by themselves. At these conditions, the cell potential \(E\) equals the standard cell potential \(E^0\), making the Nernst equation straightforward. This is due to the logarithm natural value \(\ln(1) = 0\), meaning no additional potential energy adjustment is needed. Understanding \(Q\) is essential for calculating the cell potential when conditions deviate from the standard, which is often the case in practical applications.
Role of Standard Conditions
Standard conditions are a benchmark used to create consistency in thermodynamics and electrochemistry calculations. They involve setting solutions to 1 M concentration and gases to 1 atm pressure, along with a temperature of 298 K (25°C).
This uniform frame of reference allows chemists to compare different reactions' thermodynamic properties reliably.
When under these particular conditions, it simplifies calculations significantly as the potential provided by the Nernst equation, represented as \(E\), equals the standard potential \(E^0\). This gives us a useful baseline or point of comparison.
  • Aqueous Solutions: Concentrations at 1 M.
  • Gases: Pressure at 1 atm.
  • Temperature: Set at 298 K.
Standard conditions are theoretical, but they serve as a crucial starting point for understanding real-world deviations that occur in labs and industry.
Temperature's Impact on Electrochemical Cells
Temperature is a pivotal factor in electrochemical reactions, and it directly influences cell potentials through the Nernst equation. The term \(T\) in the equation denotes the temperature in Kelvin, reflecting how reaction spontaneity changes with temperature.
As temperature rises or falls, the movement and energy of molecules change, impacting the cell potential \(E\). This means different temperatures can lead to different calculated potentials even if concentration remains unchanged. Thus, by adjusting \(T\) in the Nernst equation, it's possible to model or predict cell behavior under a variety of temperature conditions.
This adaptability means that the Nernst equation is not limited to standard room temperature (298 K). It supports a wide array of experimental conditions, making it highly versatile and practical for diverse scientific inquiries and industrial processes where temperatures may significantly differ from the standard condition.

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Most popular questions from this chapter

(a) Write the anode and cathode reactions that cause the corrosion of iron metal to aqueous iron(II). \((\mathbf{b})\) Write the balanced half-reactions involved in the air oxidation of \(\mathrm{Fe}^{2+}(a q)\) to \(\mathrm{Fe}_{2} \mathrm{O}_{3} \cdot 3 \mathrm{H}_{2} \mathrm{O}(s)\).

A voltaic cell utilizes the following reaction: $$ 2 \mathrm{Fe}^{3+}(a q)+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Fe}^{2+}(a q)+2 \mathrm{H}^{+}(a q) $$ (a) What is the emf of this cell under standard conditions? (b) What is the emf for this cell when \(\left[\mathrm{Fe}^{3+}\right]=3.50 \mathrm{M}, P_{\mathrm{H}_{2}}=\) \(96.3 \mathrm{kPa},\left[\mathrm{Fe}^{2+}\right]=0.0010 \mathrm{M},\) and the \(\mathrm{pH}\) in both half-cells is \(4.00 ?\)

A voltaic cell is constructed that uses the following half-cell reactions: $$ \begin{array}{l} \mathrm{Ag}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s) \\ \mathrm{I}_{2}(s)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{I}^{-}(a q) \end{array} $$ The cell is operated at \(298 \mathrm{~K}\) with \(\left[\mathrm{Ag}^{+}\right]=0.15 \mathrm{M}\) and \(\left[\mathrm{I}^{-}\right]=0.035 \mathrm{M}\). (a) Determine \(E\) for the cell at these concentrations. (b) Which electrode is the anode of the cell? (c) Is the answer to part (b) the same as it would be if the cell were operated under standard conditions? (d) With \(\left[\mathrm{Ag}^{+}\right]\) equal to \(0.15 \mathrm{M}\), at what concentration of \(\mathrm{I}^{-}\) would the cell have zero potential?

In a galvanic cell the cathode is an \(\mathrm{Ag}^{+}(1.00 \mathrm{M}) / \mathrm{Ag}(s)\) half-cell. The anode is a standard hydrogen electrode immersed in a buffer solution containing \(0.10 \mathrm{M}\) benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\) and \(0.050 \mathrm{M}\) sodium benzoate \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-} \mathrm{Na}^{+}\right)\). The measured cell voltage is \(1.030 \mathrm{~V}\). What is the \(\mathrm{p} K_{\mathrm{a}}\) of benzoic acid?

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. (a) \(14 \mathrm{H}^{+}(a q)+2 \mathrm{Mn}^{2+}(a q)+5 \mathrm{NaBiO}_{3}(s)\) \(\quad \longrightarrow 7 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{MnO}_{4}^{-}+5 \mathrm{Bi}^{3+}(a q)+5 \mathrm{Na}^{+}(a q)\) (b) \(2 \mathrm{KMnO}_{4}(a q)+3 \mathrm{Na}_{2} \mathrm{SO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) \(\quad \longrightarrow 2 \mathrm{MnO}_{2}(s)+3 \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{KOH}(a q)\) (c) \(\mathrm{Cu}(s)+2 \mathrm{AgNO}_{3}(a q) \longrightarrow 2 \mathrm{Ag (s)+\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)\)

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