In: Chemistry

Standard reduction potentials

Reduction half-reaction | E∘ (V) |

Ag+(aq)+e−→Ag(s) | 0.80 |

Cu2+(aq)+2e−→Cu(s) | 0.34 |

Sn4+(aq)+4e−→Sn(s) | 0.15 |

2H+(aq)+2e−→H2(g) | 0 |

Ni2+(aq)+2e−→Ni(s) | −0.26 |

Fe2+(aq)+2e−→Fe(s) | −0.45 |

Zn2+(aq)+2e−→Zn(s) | −0.76 |

Al3+(aq)+3e−→Al(s) | −1.66 |

Mg2+(aq)+2e−→Mg(s) | −2.37 |

Part A

**Use the table of standard reduction potentials given above to calculate the equilibrium constant at standard temperature (25 ∘C) for the following reaction:**

Fe(s)+Ni2+(aq)→Fe2+(aq)+Ni(s)

Express your answer numerically.

Part B

**Calculate the standard cell potential (E∘) for the reaction**

X(s)+Y+(aq)→X+(aq)+Y(s)

if K = 3.04×10−3.

Express your answer to three significant figures and include the appropriate units.

Concepts and reason The question is divided into 2 parts:

First part is about calculating the equilibrium constant using the standard cell potential. It can be done using the following formula:

\(\mathrm{E}_{\mathrm{cell}}^{\mathrm{o}}=\frac{0.0591 \mathrm{~V}}{\mathrm{n}} \log \mathrm{K}\)

The second part is based on calculating the standard cell potential using the given equilibrium constant. It can also be done by the same formula given above.

Fundamentals

Standard cell potential \(\left(\mathrm{E}_{\mathrm{cell}}^{\circ}\right):\) It is the cell potential of the overall net reaction. It is calculated as follows:

\(\mathrm{E}_{\mathrm{cell}}^{\circ}=\mathrm{E}_{\mathrm{cathode}}^{\circ}-\mathrm{E}_{\mathrm{anode}}^{\circ}\)

Here, \(\mathrm{E}_{\text {cathode }}^{\circ}\) is the standard reduction potential of the reaction occurring at cathode; and \(\mathrm{E}_{\text {anode }}^{\circ}\) is the standard reduction potential of the reaction occurring at anode. Relation between standard cell potential and equilibrium constant:

It is known that the relation of standard free energy change, \(\Delta \mathrm{G}^{\circ}\) are given as follows:

\(\Delta \mathrm{G}^{\circ}=-\mathrm{nFE}_{\mathrm{cell}}^{\mathrm{o}} \ldots \ldots(1)\)

And, \(\Delta \mathrm{G}^{\circ}=-\mathrm{RT} \ln \mathrm{K} \ldots \ldots .(2)\)

Therefore, equating (1) and (2) as follows:

\(-\mathrm{nFE}_{\text {cell }}^{\mathrm{o}} =-\mathrm{RT} \ln \mathrm{K} \)

\(\mathrm{E}_{\text {cell }}^{\mathrm{o}} =\frac{\mathrm{RT}}{\mathrm{nF}} \ln \mathrm{K}\)

At \(298 \mathrm{~K}\), the equation can be rewritten as follows:

\(\mathrm{E}_{\text {cell }}^{\mathrm{o}}=\frac{0.0591 \mathrm{~V}}{\mathrm{n}} \log \mathrm{K}\)

Here, \(\mathrm{n}\) is the number of electrons participating in the overall reaction; \(\mathrm{K}\) is the equilibrium constant.

(A)

Consider the given reaction:

\(\mathrm{Fe}(\mathrm{s})+\mathrm{Ni}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Ni}(\mathrm{s})\)

At cathode: \(\mathrm{Ni}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Ni}(\mathrm{s}) \quad \mathrm{E}_{\text {cathode }}^{\mathrm{o}}=-0.26 \mathrm{~V}\)

At anode: \(\mathrm{Fe}(\mathrm{s}) \rightarrow \mathrm{Fe}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \quad \mathrm{E}_{\text {anode }}^{\mathrm{o}}=-0.45 \mathrm{~V}\)

Calculate the standard cell potential as follows:

\(\mathrm{E}_{\text {cell }}^{\circ}=\mathrm{E}_{\text {cathode }}^{\circ}-\mathrm{E}_{\text {anode }}^{\circ}\)

\(=-0.26 \mathrm{~V}-(-0.45 \mathrm{~V})\)

\(\quad=0.19 \mathrm{~V}\)

At cathode, reduction reaction occurs in which \(\mathrm{Ni}^{2+}\) reduces to \(\mathrm{Ni}\) and at anode, oxidation reaction occurs in which Fe oxidizes to \(\mathrm{Fe}^{2+}\)

Calculate equilibrium constant as follows:

\(\mathrm{E}_{\mathrm{cell}}^{\mathrm{o}}=\frac{0.0592}{\mathrm{n}} \log \mathrm{K}\)

\(0.19 \mathrm{~V}=\frac{0.0592}{2} \log \mathrm{K}\)

\(\log \mathrm{K}=6.42\)

\(\mathrm{~K}=2.63 \times 10^{6}\)

Part A The equilibrium constant for the given reaction is found to be .

The standard cell potential for the given reaction is found to be \(0.19 \mathrm{~V}\). The number of electrons, n, participating in the overall reaction is 2. These values are substituted in the formula, and the equilibrium constant is calculated.

(B) Consider the given reaction:

\(\mathrm{X}(\mathrm{s})+\mathrm{Y}^{+}(\mathrm{aq}) \rightarrow \mathrm{X}^{+}(\mathrm{aq})+\mathrm{Y}(\mathrm{s})\)

Calculate the standard cell potential as follows:

\(\mathrm{E}_{\mathrm{cell}}^{\mathrm{o}}=\frac{0.0592 \mathrm{~V}}{\mathrm{n}} \log \mathrm{K}\)

\(=\frac{0.0592 \mathrm{~V}}{1} \log \left(3.04 \times 10^{-3}\right)\)

\(=-0.149 \mathrm{~V}\)

Part B The standard cell potential for the given reaction is

The equilibrium constant for the reaction is given as \(3.04 \times 10^{-3}\). The number of electrons, n, participating in the overall reaction is \(1 .\) These values are substituted in the formula, and standard reduction potential is calculated.

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