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\(E_{\text {cathode }}^{\circ}=(2.71-2.310) V=+0.40 \mathrm{V}\)

Short Answer

Expert verified
The reduction potential at the cathode is \(+0.40 V\).

Step by step solution

01

Substitute the known values into the equation

Given \(E_{\text {substance }}^{\circ}=2.71 V\) and \(E_{\text {reference }}^{\circ}=2.310 V\). Substitute these values into the equation: \(E_{\text {cathode }}^{\circ}=E_{\text {substance}}^{\circ}-E_{\text {reference}}^{\circ}\). This becomes: \(E_{\text {cathode }}^{\circ}=2.71 V - 2.310 V\)
02

Calculate the result

By implementing subtraction as per the previous step, this results in: \(E_{\text {cathode }}^{\circ}=+0.40 V\)

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Key Concepts

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

Standard Electrode Potential
In electrochemistry, the standard electrode potential, denoted as \( E^\circ \), is a fundamental concept used to understand the voltage potentials of electrodes. It represents the potential difference between an electrode and a reference electrode under standard conditions. These conditions include a concentration of 1 mol/L for any dissolved substances, a pressure of 1 atm, and a temperature of 25°C (298 K).
This potential is measured using a hydrogen electrode, which is assigned a potential of zero volts. By comparing other electrode potentials to this standard, chemists can determine the ability of an electrode to gain or lose electrons.
  • A positive \( E^\circ \) value indicates a stronger tendency to gain electrons (reduction).
  • A negative \( E^\circ \) value suggests a tendency to lose electrons (oxidation).

Understanding standard electrode potentials allows us to predict the direction of electron flow in electrochemical cells, where the electrode with the higher potential will act as the cathode (the site of reduction). The difference between potentials of two electrodes generates the electromotive force (emf) of the cell.
Cathode Reaction
The cathode is a crucial component in an electrochemical cell, where reduction reactions occur. It's the electrode where electrons are gained by the chemical species. This process is often simplified into the concise term 'cathode reaction.'
The electrode potential of the cathode is measured, as seen in the calculation \((E_{\text {cathode}}^{\circ}\) = +0.40 V), to determine the direction of electron flow in a cell. At the cathode, ions or molecules in solution gain electrons, converting them into more stable forms.
Consider these simple points about cathode reactions:
  • Electrons move from the anode to the cathode.
  • The cathode reaction typically forms a product that is stable in the chemical environment.
  • In galvanic cells, the cathode is where positive ions are discharged.

The reaction at the cathode is as important as the overall cell reaction because it drives the electrochemical processes necessary for energy conversion or storage.
Electrochemical Cell
An electrochemical cell is a device that converts chemical energy directly into electrical energy through redox reactions. It consists of two half-cells, each containing an electrode and a solution of ions. These cells can be broadly categorized into galvanic cells, which generate electricity, and electrolytic cells, which use electricity to induce chemical changes.
In a galvanic cell, chemical energy is converted into electrical energy through spontaneous redox reactions. The cell contains:
  • A cathode, where reduction takes place.
  • An anode, where oxidation occurs.
  • An electrolyte solution that facilitates the flow of ions.

An electrochemical cell shows the cell potential, a measure of the voltage or electric potential difference between the two electrodes. The cell potential can be calculated using the standard electrode potentials of the half-reactions occurring at the anode and cathode.
Understanding the structure and function of electrochemical cells allows us to harness chemical reactions to do electrical work, which is the principle behind batteries and fuel cells.

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Most popular questions from this chapter

Construct a concept map illustrating the relationship between batteries and electrochemical ideas.

Refer to standard reduction potentials, and predict which metal in each of the following pairs is the stronger reducing agent: (a) sodium or potassium (b) magnesium or barium

Only a tiny fraction of the diffusible ions move across a cell membrane in establishing a Nernst potential (see Focus On 20: Membrane Potentials), so there is no detectable concentration change. Consider a typical cell with a volume of \(10^{-8} \mathrm{cm}^{3},\) a surface area \((A)\) of \(10^{-6} \mathrm{cm}^{2},\) and a membrane thickness \((l)\) of \(10^{-6} \mathrm{cm}\) Suppose that \(\left[\mathrm{K}^{+}\right]=155 \mathrm{mM}\) inside the cell and \(\left[\mathrm{K}^{+}\right]=4 \mathrm{mM}\) outside the cell and that the observed Nernst potential across the cell wall is \(0.085 \mathrm{V}\). The membrane acts as a charge-storing device called a capacitor, with a capacitance, \(C,\) given by $$C=\frac{\varepsilon_{0} \varepsilon A}{l}$$ where \(\varepsilon_{0}\) is the dielectric constant of a vacuum and the product \(\varepsilon_{0} \varepsilon\) is the dielectric constant of the membrane, having a typical value of \(3 \times 8.854 \times 10^{-12}\) \(\mathrm{C}^{2} \mathrm{N}^{-1} \mathrm{m}^{-2}\) for a biological membrane. The SI unit of capacitance is the firad, \(1 \mathrm{F}=1\) coulomb per volt \(=1 \mathrm{CV}^{-1}=1 \times \mathrm{C}^{2} \mathrm{N}^{-1} \mathrm{m}^{-1}\) (a) Determine the capacitance of the membrane for the typical cell described. (b) What is the net charge required to maintain the observed membrane potential? (c) How many \(\mathrm{K}^{+}\) ions must flow through the cell membrane to produce the membrane potential? (d) How many \(\mathrm{K}^{+}\) ions are in the typical cell? (e) Show that the fraction of the intracellular \(K^{+}\) ions transferred through the cell membrane to produce the membrane potential is so small that it does not change \(\left[\mathrm{K}^{+}\right]\) within the cell.

Which of the following reactions occur spontaneously, and which can be brought about only through electrolysis, assuming that all reactants and products are in their standard states? For those requiring electrolysis, what is the minimum voltage required? (a) \(2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow 2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})\left[\text { in } 1 \mathrm{M} \mathrm{H}^{+}(\mathrm{aq})\right]\) (b) \(\mathrm{Zn}(\mathrm{s})+\mathrm{Fe}^{2+}(\mathrm{aq}) \longrightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Fe}(\mathrm{s})\) (c) \(2 \mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{I}_{2}(\mathrm{s}) \longrightarrow 2 \mathrm{Fe}^{3+}(\mathrm{aq})+2 \mathrm{I}^{-}(\mathrm{aq})\) (d) \(\mathrm{Cu}(\mathrm{s})+\mathrm{Sn}^{4+}(\mathrm{aq}) \longrightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Sn}^{2+}(\mathrm{aq})\)

Construct a concept map illustrating the principles of electrolysis and its industrial applications.

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