For the cell : $$ \text { Zn }\left|\begin{array}{c} \mathrm{Zn}_{\text {aq }}^{2+} \\ 1 M \end{array}\right|\left|\begin{array}{c|c} \mathrm{Cu}_{a q}^{2+} \\ 2 M \end{array}\right| \mathrm{Cu} $$ Calculate the values for ; (a) cell reaction, (b) \(E_{\text {cell }}^{\circ}\). (c) \(E_{\text {cell }}\) (d) the minimum concentration of \(\mathrm{Cu}^{2+}\) at which cell reaction is spontaneous if \(\mathrm{Zn}^{2+}\) is \(1 M\) (e) does the displacement of \(\mathrm{Cu}^{2+}\) goes almost to completion. Given : \(E_{R P_{\mathrm{Cu}^{2+} / \mathrm{Cu}}}^{\circ}=+0.35 \mathrm{~V}\) $$ E_{R P_{\mathrm{Zn}^{2+} / Z n}}^{\circ}=-0.76 \mathrm{~V} $$

Short Answer

Expert verified
The cell reaction is \(\text{Zn} + \text{Cu}^{2+} \to \text{Zn}^{2+} + \text{Cu}\), the standard cell potential \(E_{\text{cell}}^{\circ}\) is +1.11 V, the cell potential under non-standard conditions can be calculated with the Nernst equation, the minimum concentration of \(\text{Cu}^{2+}\) for spontaneity at 1 M \(\text{Zn}^{2+}\) is \(10^{-37.87} M\), and the displacement of \(\text{Cu}^{2+}\) by Zn is nearly complete.

Step by step solution

01

- Write the Cell Reaction

To determine the cell reaction, write the half-reactions for each electrode and then balance the electrons to get the overall cell reaction. For the zinc electrode (anode), which is being oxidized: \(\text{Zn} \to \text{Zn}^{2+} + 2e^-\). For the copper electrode (cathode), which is being reduced: \(\text{Cu}^{2+} + 2e^- \to \text{Cu}\). Combine these half-reactions to get the cell reaction: \(\text{Zn} + \text{Cu}^{2+} \to \text{Zn}^{2+} + \text{Cu}\).
02

- Calculate the Standard Cell Potential

To calculate the standard cell potential \(E_{\text{cell}}^{\circ}\), use the standard reduction potentials for each electrode and the formula \(E_{\text{cell}}^{\circ} = E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ}\). For copper, \(E_{\text{cathode}}^{\circ} = +0.35 \text{V}\) and for zinc, \(E_{\text{anode}}^{\circ} = -0.76 \text{V}\). Plug these values into the formula to get \(E_{\text{cell}}^{\circ} = +0.35 \text{V} - (-0.76 \text{V}) = +1.11 \text{V}\).
03

- Calculate the Cell Potential Under Non-Standard Conditions

To calculate the cell potential under non-standard conditions \(E_{\text{cell}}\), use the Nernst equation: \(E_{\text{cell}} = E_{\text{cell}}^{\circ} - \frac{0.0592}{n}\log Q\), where \(n\) is the number of moles of electrons exchanged (which is 2 in this case) and \(Q\) is the reaction quotient. Here \(Q = \frac{[\text{Cu}^{2+}]}{[\text{Zn}^{2+}]} = \frac{2}{1}\), so \(E_{\text{cell}} = 1.11V - \frac{0.0592}{2}\log 2\).
04

- Calculate Minimum Concentration of Cu²⁺ for Spontaneity

For the cell reaction to be spontaneous, \(E_{\text{cell}}\) must be positive. Using the Nernst equation, set \(E_{\text{cell}}\) to zero and solve for the minimum concentration of \(\text{Cu}^{2+}\) given that \(\text{Zn}^{2+}\) is \(1 M\): \(0 = E_{\text{cell}}^{\circ} - \frac{0.0592}{2}\log(\frac{[\text{Cu}^{2+}]}{[\text{Zn}^{2+}]})\). Rearrange and solve for \([\text{Cu}^{2+}]\): \([\text{Cu}^{2+}] = [\text{Zn}^{2+}] \times 10^{-\frac{E_{\text{cell}}^{\circ} \times 2}{0.0592}} = 1 \times 10^{-\frac{1.11 \times 2}{0.0592}}\).
05

- Assess Completion of Copper Ion Displacement

To determine if the displacement of \(\text{Cu}^{2+}\) goes to completion, compare the calculated cell potential under non-standard conditions with the standard cell potential. If \(E_{\text{cell}}\) is close to \(E_{\text{cell}}^{\circ}\), the reaction goes nearly to completion. Since step 3 results in \(E_{\text{cell}}\) being positive and considerably high, it suggests the reaction proceeds nearly to completion.

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Key Concepts

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

Standard Cell Potential
The standard cell potential (\(E_{\text{cell}}^{\text{o}}\)) is a crucial concept in electrochemistry that gauges the inherent voltage of an electrochemical cell at standard conditions. It reflects the energy difference between the oxidation and reduction reactions occurring at the two electrodes. This potential is the driving force behind the cell's ability to do electrical work and is quantified in volts (V).

To calculate the standard cell potential, you use the standard reduction potentials of the cathode and the anode. The standard cell potential is then given by the formula \(E_{\text{cell}}^{\circ} = E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ}\). A positive value suggests that the reaction is favorable at standard conditions, which is an indicator for spontaneous reaction direction under standard conditions.

The importance of understanding the standard cell potential lies in its use for predicting the feasibility of cell reactions and their ability to proceed spontaneously.
Nernst Equation
The Nernst equation is a fundamental relation used in electrochemistry to calculate the actual voltage (also known as the cell potential, \(E_{\text{cell}}\)) of an electrochemical cell under non-standard conditions. This equation adjusts the standard cell potential to account for temperature, pressure, and concentration differences.

Expressed mathematically, the Nernst equation is: \(E_{\text{cell}} = E_{\text{cell}}^{\circ} - \frac{0.0592}{n}\log Q\), where \(n\) is the number of electrons transferred in the cell reaction, and \(Q\) is the reaction quotient, representing the ratio of product activities to reactant activities. This calculation can provide insights into the reaction's direction and spontaneity under varying conditions. When the cell potential is positive, the reaction is spontaneous. Conversely, if the cell potential is negative, the reaction is non-spontaneous.
Electrode Reactions
Electrode reactions are at the heart of electrochemical cells, where oxidation occurs at the anode and reduction at the cathode. During these reactions, electrons are either gained or lost by chemical species. An understanding of electrode reactions includes writing half-cell reactions for both the anode and the cathode to determine the overall cell reaction.

For example, a common half-reaction at the anode is the oxidation of metallic zinc: \(\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^{-}\). At the cathode, copper ions can be reduced to copper metal: \(\text{Cu}^{2+} + 2e^{-} \rightarrow \text{Cu}\). The electrons transferred during these half-reactions are responsible for the flow of electricity within the circuit. Deciphering these reactions is key in predicting the behavior of the entire electrochemical cell, involving concepts like potential, energy changes, and electron transfer.
Reaction Spontaneity
Reaction spontaneity in electrochemistry is determined by the sign and magnitude of the cell potential (\(E_{\text{cell}}\)). A positive cell potential indicates that a chemical reaction is spontaneous, meaning it will proceed without any external intervention. Conversely, if the cell potential is negative, the reaction is not spontaneous and will not occur under the given conditions.

Spontaneity is fundamentally linked with the concept of Gibbs free energy, where a negative change in Gibbs free energy (\(\Delta G\)) signals a spontaneous process. The relationship between the cell potential and Gibbs free energy is given by the equation \(\Delta G = -nFE_{\text{cell}}\), where \(n\) is the number of moles of electrons exchanged, \(F\) is Faraday's constant, and \(E_{\text{cell}}\) is the cell potential. Studying reaction spontaneity is essential for understanding the natural direction of chemical processes and for the design of batteries and other electrochemical devices.

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Most popular questions from this chapter

The e.m.f. of a cell corresponding to the reaction : $$ \begin{aligned} \mathrm{Zn}(s)+2 \mathrm{H}^{+}(\mathrm{aq}) \longrightarrow \mathrm{Zn}^{2+}+& \mathrm{H}_{2}(\mathrm{~g}) \\ (0.1 \mathrm{M}) &(1 \mathrm{~atm}) \end{aligned} $$ is \(0.28 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\) and \(E_{\mathrm{Zn} / \mathrm{Zn}^{2+}}^{\circ}=0.76 \mathrm{~V}\) (i) Write half cell reactions. (ii) Calculate \(\mathrm{pH}\) of the solution at \(\mathrm{H}\) electrode.

A metal is known to form fluoride \(M \mathrm{~F}_{2}\). When 10 ampere electricity is passed throwgh i multen salt for \(330 \mathrm{sec}, 1.95 \mathrm{~g}\) metal is deposited. Find out the atomic weight of metal. What will be the quantity of charge required to deposit the same mass of \(\mathrm{Cu}\) from \(\mathrm{CuSO}_{4}\) (aq.) (At. wt. of \(\mathrm{Cu}=63.6\) )

$$ \begin{aligned} &\text { The e.m.f. of the cell } M \mid M^{n+}(0.02 M) \| \mathrm{H}^{+}(1 M) \mathrm{H}_{2(\mathrm{~g})}(1 \mathrm{~atm}) \mathrm{Pt} \text { at }\\\ &25^{\circ} \mathrm{C} \text { is } 0.81 \mathrm{~V} \text { . Calculate the valence of metal if } E^{\circ}{ }_{\ldots 1 / 2}^{n+}=0.76 \mathrm{~V} \end{aligned} $$

$$ \text { Write equations for the electrolysis of } \mathrm{CaH}_{2} \text { in fused state. } $$

Write the nernst equation and e.m.f. of the following cells at \(298 \mathrm{~K}\) : (a) \(\mathrm{Mg}(\mathrm{s})\left|\mathrm{Mg}^{2+}(0.001 \mathrm{M}) \| \mathrm{Cu}^{2+}(0.0001 \mathrm{M})\right| \mathrm{Cu}(\mathrm{s})\) (b) \(\mathrm{Fe}(\mathrm{s})\left|\mathrm{Fe}^{2+}(0.001 M) \| \mathrm{H}^{\mathrm{i}}(1 \mathrm{M})\right| \mathrm{H}_{2}(\mathrm{~g})(1 \mathrm{bar}) \mid \mathrm{Pt}(\mathrm{s})\) (c) \(\mathrm{Sn}(\mathrm{s})\left|\mathrm{Sn}^{2+}(0.050 \mathrm{M}) \| \mathrm{H}^{+}(0.020 \mathrm{M})\right| \mathrm{H}_{2}(\mathrm{~g})(1 \mathrm{bar}) \mid \mathrm{Pt}(\mathrm{s})\) (d) \(\mathrm{Pt}(\mathrm{s})\left|\mathrm{Br}_{2}(\mathrm{l})\right| \mathrm{Br}^{-}(0.010 \mathrm{M}) \| \mathrm{H}^{+}(0.030 \mathrm{M}) \mid \mathrm{H}_{2}(\mathrm{~g})(\mathrm{I} \mathrm{bar}) ! \mathrm{Pt}(\mathrm{s})\) Given : \(E_{\text {OP }}^{\circ} \mathrm{Mg}=2.36 \mathrm{~V}, E_{\mathrm{OP}}^{\circ} \mathrm{Cu}=-0.34 \mathrm{~V}, E_{\mathrm{OP}}^{\mathrm{O}} \mathrm{Fe}=0.44 \mathrm{~V}\) \(E_{\mathrm{OP}}^{\circ} \mathrm{Sn}=0.14 \mathrm{~V}\) and \(E_{\mathrm{OP}}^{\circ} \mathrm{Br}_{2}=-1.09 \mathrm{~V}\) respectively.

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