Medium Nernst Equation Practice Questions

Concept Explanation

The Nernst equation is a mathematical relationship that calculates the reduction potential of an electrochemical cell or a half-cell under non-standard conditions by relating it to the standard electrode potential, temperature, and the activities of the chemical species involved. While standard potentials ($E^\circ$) are measured at 298 K, 1 atm, and 1.0 M concentrations, real-world applications often involve varying concentrations and temperatures. This equation is fundamental in electrochemistry for understanding how battery voltage changes as it discharges or how biological membranes maintain electrical gradients. According to the IUPAC definition, the equation accounts for the reaction quotient ($Q$), which represents the ratio of product activities to reactant activities at any given moment.

The general form of the Nernst equation at any temperature is:

$E = E^\circ - \frac{RT}{nF} \ln Q$

Where:

• $E$: Cell potential under non-standard conditions (V)

• $E^\circ$: Standard cell potential (V)

• $R$: Universal gas constant (8.314 J/(mol·K))

• $T$: Absolute temperature (K)

• $n$: Number of moles of electrons transferred in the redox reaction

• $F$: Faraday’s constant (96,485 C/mol)

• $Q$: Reaction quotient

At the standard temperature of 25°C (298.15 K), the equation is frequently simplified using the base-10 logarithm to:

$E = E^\circ - \frac{0.0592}{n} \log Q$

This simplified version is the most common tool for solving Medium Nernst Equation Practice Questions, as it allows for quick calculations of voltage shifts based on concentration changes. Understanding the relationship between concentration and potential is a key step before moving on to cell potential calculations in more complex systems.

Solved Examples

Below are three worked examples demonstrating how to apply the Nernst equation to different chemical systems.

Example 1: Calculating Half-Cell Potential
Calculate the electrode potential of a copper electrode immersed in a 0.01 M $Cu^{2+}$ solution at 25°C. ($E^\circ$ for $Cu^{2+}/Cu = +0.34$ V).

1. Identify the half-reaction: $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$. Here, $n = 2$.

2. Determine the reaction quotient $Q$: Since $Cu(s)$ is a pure solid, $Q = 1 / [Cu^{2+}] = 1 / 0.01 = 100$.

3. Apply the simplified Nernst equation: $E = 0.34 - (0.0592 / 2) \log(100)$.

4. Calculate the log value: $\log(100) = 2$.

5. Solve: $E = 0.34 - (0.0296 \times 2) = 0.34 - 0.0592 = 0.2808$ V.

Example 2: Zinc-Copper Voltaic Cell
A galvanic cell consists of a Zn electrode in 0.50 M $Zn^{2+}$ and a Cu electrode in 0.02 M $Cu^{2+}$. If $E^\circ_{cell} = 1.10$ V, find $E_{cell}$ at 25°C.

1. Write the net cell reaction: $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$.

2. Determine $n$: Two electrons are transferred ($n = 2$).

3. Set up $Q$: $Q = [Zn^{2+}] / [Cu^{2+}] = 0.50 / 0.02 = 25$.

4. Apply the equation: $E = 1.10 - (0.0592 / 2) \log(25)$.

5. Calculate: $E = 1.10 - (0.0296 \times 1.398) = 1.10 - 0.0414 = 1.0586$ V.

Example 3: Hydrogen Electrode at non-standard pH
Find the potential of a Standard Hydrogen Electrode (SHE) if the partial pressure of $H_2$ is 1.0 atm but the pH is 4.0 at 25°C.

1. Identify the reaction: $2H^+(aq) + 2e^- \rightarrow H_2(g)$. $E^\circ = 0.00$ V and $n = 2$.

2. Convert pH to concentration: $[H^+] = 10^{-pH} = 10^{-4}$ M.

3. Determine $Q$: $Q = P_{H2} / [H^+]^2 = 1.0 / (10^{-4})^2 = 10^8$.

4. Apply the equation: $E = 0 - (0.0592 / 2) \log(10^8)$.

5. Solve: $E = 0 - (0.0296 \times 8) = -0.2368$ V.

Practice Questions

Test your knowledge with these Medium Nernst Equation Practice Questions. Ensure you have a scientific calculator and a table of standard reduction potentials handy.

1. Calculate the potential of a silver electrode ($Ag^+/Ag$) in a 0.15 M $AgNO_3$ solution at 25°C. ($E^\circ = +0.80$ V).

2. A Daniel cell is operated with $[Zn^{2+}] = 2.0$ M and $[Cu^{2+}] = 0.05$ M. Calculate the cell potential at 25°C if $E^\circ_{cell} = 1.10$ V.

3. What is the potential of a $Mg^{2+}/Mg$ electrode at 25°C if the concentration of $Mg^{2+}$ is $4.5 \times 10^{-3}$ M? ($E^\circ = -2.37$ V).

4. Find the cell potential at 25°C for the reaction: $Fe(s) + Cd^{2+}(aq) \rightarrow Fe^{2+}(aq) + Cd(s)$ where $[Fe^{2+}] = 0.01$ M and $[Cd^{2+}] = 0.5$ M. ($E^circ_{Fe^{2+}/Fe} = -0.44$ V, $E^\circ_{Cd^{2+}/Cd} = -0.40$ V).

5. A concentration cell consists of two hydrogen electrodes. Electrode A has $P_{H2} = 1.0$ atm and $[H^+] = 1.0$ M. Electrode B has $P_{H2} = 1.0$ atm and $[H^+] = 0.001$ M. Calculate the cell potential.

6. Calculate the EMF of the following cell at 298 K: $Ni(s) | Ni^{2+}(0.01 M) || Cl^-(0.2 M) | Cl_2(0.5 atm) | Pt(s)$. ($E^\circ_{Ni^{2+}/Ni} = -0.25$ V, $E^\circ_{Cl_2/Cl^-} = +1.36$ V).

7. At what concentration of $Cu^{2+}$ will the potential of a $Cu^{2+}/Cu$ electrode be exactly 0.25 V at 25°C? ($E^\circ = +0.34$ V).

8. Determine the temperature (in Kelvin) if a cell with $n=1$, $E^\circ = 0.50$ V, and $Q = 10$ produces a potential of 0.45 V.

9. A cell reaction is $2Ag^+(aq) + Sn(s) \rightarrow 2Ag(s) + Sn^{2+}(aq)$. If $E_{cell} = 0.95$ V and $[Sn^{2+}] = 0.15$ M, what is the concentration of $Ag^+$? ($E^\circ_{cell} = 0.94$ V).

10. Calculate the pH of a solution in a hydrogen half-cell if the electrode potential is -0.118 V at 25°C and $P_{H2} = 1.0$ atm.

Answers & Explanations

1. Answer: 0.751 V
   Using $E = E^\circ - (0.0592/n) \log(1/[Ag^+])$. Here $n=1$. $E = 0.80 - 0.0592 \log(1/0.15) = 0.80 - 0.0592(0.824) = 0.80 - 0.0488 = 0.7512$ V.

2. Answer: 1.053 V
   $Q = [Zn^{2+}]/[Cu^{2+}] = 2.0 / 0.05 = 40$. $E = 1.10 - (0.0592/2) \log(40) = 1.10 - 0.0296(1.602) = 1.10 - 0.0474 = 1.0526$ V.

3. Answer: -2.439 V
   $Q = 1 / 4.5 \times 10^{-3} = 222.2$. $E = -2.37 - (0.0296) \log(222.2) = -2.37 - (0.0296 \times 2.347) = -2.37 - 0.0695 = -2.4395$ V.

4. Answer: 0.090 V
   $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = -0.40 - (-0.44) = +0.04$ V. $n=2$. $Q = [Fe^{2+}]/[Cd^{2+}] = 0.01 / 0.5 = 0.02$. $E = 0.04 - 0.0296 \log(0.02) = 0.04 - 0.0296(-1.699) = 0.04 + 0.0503 = 0.0903$ V.

5. Answer: 0.178 V
   In a concentration cell, $E^\circ = 0$. The reaction is $2H^+(1.0 M) \rightarrow 2H^+(0.001 M)$. $Q = (0.001)^2 / (1.0)^2 = 10^{-6}$. $E = 0 - (0.0592/2) \log(10^{-6}) = -0.0296(-6) = 0.1776$ V.

6. Answer: 1.66 V
   Net reaction: $Ni(s) + Cl_2(g) \rightarrow Ni^{2+}(aq) + 2Cl^-(aq)$. $E^\circ_{cell} = 1.36 - (-0.25) = 1.61$ V. $n=2$. $Q = ([Ni^{2+}][Cl^-]^2) / P_{Cl2} = (0.01 \times 0.2^2) / 0.5 = 0.0008$. $E = 1.61 - 0.0296 \log(0.0008) = 1.61 - 0.0296(-3.097) = 1.61 + 0.0917 = 1.70$ V. (Note: Slight variations in rounding may occur).

7. Answer: $9.1 \times 10^{-4}$ M
   $0.25 = 0.34 - (0.0296) \log(1/[Cu^{2+}])$. $-0.09 = -0.0296 \log(1/[Cu^{2+}])$. $3.04 = \log(1/[Cu^{2+}])$. $10^{3.04} = 1/[Cu^{2+}]$. $[Cu^{2+}] = 1/1096 = 9.12 \times 10^{-4}$ M.

8. Answer: 252 K
   Using $E = E^\circ - (RT/nF) \ln Q$: $0.45 = 0.50 - (R \times T / (1 \times 96485)) \ln(10)$. $-0.05 = - (8.314 \times T \times 2.303) / 96485$. $0.05 = 0.0001984 \times T$. $T = 0.05 / 0.0001984 = 252.0$ K.

9. Answer: 0.34 M
   $0.95 = 0.94 - (0.0296) \log(0.15 / [Ag^+]^2)$. $0.01 = -0.0296 \log(0.15 / [Ag^+]^2)$. $-0.3378 = \log(0.15 / [Ag^+]^2)$. $0.459 = 0.15 / [Ag^+]^2$. $[Ag^+]^2 = 0.15 / 0.459 = 0.326$. $[Ag^+] = 0.57$ M. (Recalculating with precise logs: $0.95 = 0.94 - 0.0296 \log Q \rightarrow \log Q = -0.3378 \rightarrow Q = 0.459$. $0.15/x^2 = 0.459 \rightarrow x = 0.57$).

10. Answer: pH 2.0
    $E = -0.0592 \times pH$ for SHE at 25°C. $-0.118 = -0.0592 \times pH$. $pH = -0.118 / -0.0592 = 1.993 \approx 2.0$.

Quick Quiz

Frequently Asked Questions

What is the difference between E and E° in the Nernst equation?

E represents the cell potential under specific, non-standard conditions of concentration, pressure, and temperature. E° is the standard cell potential measured under fixed conditions: 1.0 M concentrations, 1 atm pressure, and usually 25°C.

How do you determine the value of 'n' for the equation?

The value 'n' is the total number of moles of electrons transferred in the balanced redox reaction. You can find 'n' by looking at the balanced half-reactions or the overall balanced redox equation.

Can the Nernst equation be used for gases?

Yes, the Nernst equation applies to gases by using their partial pressures (in atmospheres) in the reaction quotient $Q$. For example, in a hydrogen electrode, the activity of $H_2$ is represented by its partial pressure $P_{H2}$.

What happens to the cell potential at equilibrium?

At chemical equilibrium, the cell potential (E) becomes zero because the system can no longer perform electrical work. In this state, the reaction quotient $Q$ is equal to the equilibrium constant $K$.

Why is the value 0.0592 used in the simplified equation?

The value 0.0592 is a constant derived from $(R \times T \times 2.303) / F$ at 298.15 K. The 2.303 factor converts the natural logarithm ($\ln$) to a base-10 logarithm ($\log$) for easier manual calculation.

Can I use the Nernst equation for a single half-cell?

Absolutely, the Nernst equation is frequently used to calculate the reduction potential of an individual half-cell relative to the standard hydrogen electrode. This is a common step in solving medium-level chemistry problems involving electrochemistry.

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