Question

Consider a galvanic cell that uses the following half-reactions: \(\mathrm{MnO}_{4}^{-}(a q)+8 \mathrm{H}^{+}(a q)+5 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l)\) \(\mathrm{Sn}^{4+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Sn}^{2+}(a q)\) (a) Write a balanced equation for the overall cell reaction. (b) What is the oxidizing agent, and what is the reducing agent? (c) Calculate the standard cell potential.

Short answer

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Step-by-step solution

Balance Electron Transfer

To write the balanced equation, we need to balance the electrons transferred in each half-reaction. The reduction of \(\mathrm{MnO}_{4}^{-}\) to \(\mathrm{Mn}^{2+}\) transfers 5 electrons, and the reduction of \(\mathrm{Sn}^{4+}\) to \(\mathrm{Sn}^{2+}\) transfers 2 electrons. The least common multiple of 5 and 2 is 10. So, we multiply the first equation by 2 and the second by 5:\[2(\mathrm{MnO}_{4}^{-} + 8 \mathrm{H}^{+} + 5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+} + 4 \mathrm{H}_{2}\mathrm{O})\]\[5(\mathrm{Sn}^{4+} + 2 \mathrm{e}^{-} \rightarrow \mathrm{Sn}^{2+})\]

Key concepts

Half-Reactions

Half-reactions are the fundamental building blocks of a redox (reduction-oxidation) reaction. Each half-reaction represents either the loss or gain of electrons in a chemical process. In the context of a galvanic cell, these reactions occur in separate compartments, allowing for the conversion of chemical energy into electrical energy.
In the given galvanic cell:

• The first half-reaction is: \ \(\mathrm{MnO}_{4}^{-} + 8 \mathrm{H}^{+} + 5 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+} + 4 \mathrm{H}_{2}\mathrm{O}\) \ It involves the reduction of permanganate ion \(\mathrm{MnO}_{4}^{-}\) into manganese ion \(\mathrm{Mn}^{2+}\).
• The second half-reaction is: \ \(\mathrm{Sn}^{4+} + 2 \mathrm{e}^{-} \longrightarrow \mathrm{Sn}^{2+}\) \ This is a reduction process transforming tin \(\mathrm{Sn}^{4+}\) to tin \(\mathrm{Sn}^{2+}\).

Together, these reactions show the electron exchange that drives the overall cell process. Understanding half-reactions helps in predicting how elements and compounds will interact during an electrochemical reaction.

Oxidizing Agent

In a redox reaction, the oxidizing agent is the substance that gains electrons. Since it causes the oxidation of another substance by taking electrons away from it, the oxidizing agent itself is reduced. In our galvanic cell:The oxidizing agent is the substance in the reaction that receives electrons. Looking at the first half-reaction:

• \(\mathrm{MnO}_{4}^{-} + 8 \mathrm{H}^{+} + 5 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+} + 4 \mathrm{H}_{2}\mathrm{O}\) \ Here, \(\mathrm{MnO}_{4}^{-}\) is gaining electrons and forms \(\mathrm{Mn}^{2+}\). So, \(\mathrm{MnO}_{4}^{-}\) serves as the oxidizing agent.

By accepting electrons and reducing to \(\mathrm{Mn}^{2+}\), \(\mathrm{MnO}_{4}^{-}\) helps the system advance towards equilibrium.

Reducing Agent

Conversely, the reducing agent in a redox reaction is the species that donates electrons. It facilitates the reduction of another component and is itself oxidized in the process.
In the regarded galvanic cell:

• The reducing agent corresponds to the second half-reaction: \ \(\mathrm{Sn}^{4+} + 2 \mathrm{e}^{-} \longrightarrow \mathrm{Sn}^{2+}\). \ In this half-reaction, \(\mathrm{Sn}^{4+}\) becomes \(\mathrm{Sn}^{2+}\) by gaining electrons.

However, focusing on contributions, \(\mathrm{Sn}^{4+}\) doesn't donate electrons. Instead, it's the element or ion that lets go of electrons elsewhere that acts as the true reducing agent. The species giving its electrons to \(\mathrm{Sn}^{4+}\) indirectly could help identify this within the whole reaction context.

Standard Cell Potential

The standard cell potential (\(E^\circ_{\text{cell}}\)), provides a measure of the potential difference between two electrodes in their standard states. It essentially gauges the cell's ability to do electrical work.
To calculate the standard cell potential, you:

• Use the formula: \ \(E^\circ_{\text{cell}} = E^\circ_{ ext{cathode}} - E^\circ_{ ext{anode}}\)
• Identify the standard reduction potentials of each half-reaction from electrochemical series tables.
• For the given cell: \
  • Find the reduction potential for \(\mathrm{MnO}_{4}^{-} / \mathrm{Mn}^{2+}\) as it reduces at the cathode.
  • Determine the reduction potential for \(\mathrm{Sn}^{4+} / \mathrm{Sn}^{2+}\) as it involves oxidation at the anode.

Both values are essential to derive the \(E^\circ_{\text{cell}}\) value. This potential informs about the spontaneity of a reaction in a galvanic cell, establishing whether the setup can effectively generate electricity.