Question

You have the following materials that can be used to construct a galvanic cell: \(1.0 \mathrm{M} \mathrm{NiCl}_{2}, 1.0 \mathrm{M} \mathrm{AgNO}_{3}, \mathrm{Ni}(s), \mathrm{Ag}(s)\) and a salt bridge. (a) What is the overall reaction and cell potential for the galvanic cell? (b) What is shorthand notation for the galvanic cell? (c) Which metal is the cathode and which metal is the anode?

Short answer

(a) Reaction: \( \mathrm{Ni}(s) + 2\mathrm{Ag}^+ \rightarrow \mathrm{Ni}^{2+} + 2\mathrm{Ag}(s) \), \( E^\circ = 1.05\,\mathrm{V} \); (b) Shorthand: \( \mathrm{Ni}(s) | \mathrm{Ni}^{2+} || \mathrm{Ag}^+ | \mathrm{Ag}(s) \); (c) Cathode: Ag, Anode: Ni.

Step-by-step solution

Identify Half-Reactions

First, we need to determine the half-reactions involved when using nickel and silver. The reduction half-reaction for nickel is \( \mathrm{Ni}^{2+} + 2e^- \rightarrow \mathrm{Ni}(s) \) with a standard electrode potential \( E^\circ = -0.25 \mathrm{V} \). The reduction half-reaction for silver is \( \mathrm{Ag}^+ + e^- \rightarrow \mathrm{Ag}(s) \) with \( E^\circ = +0.80 \mathrm{V} \).

Determine Anode and Cathode

In a galvanic cell, the anode is the site of oxidation, and the cathode is the site of reduction. Silver, having a more positive standard electrode potential (+0.80 V), will undergo reduction and thus acts as the cathode. Nickel will therefore act as the anode, where it undergoes oxidation.

Write the Oxidation and Reduction Reactions

At the anode (oxidation): \( \mathrm{Ni}(s) \rightarrow \mathrm{Ni}^{2+} + 2e^- \). At the cathode (reduction): \( 2\mathrm{Ag}^+ + 2e^- \rightarrow 2\mathrm{Ag}(s) \).

Develop the Overall Cell Reaction

The overall cell reaction is obtained by combining the half-reactions: \( \mathrm{Ni}(s) + 2\mathrm{Ag}^+ \rightarrow \mathrm{Ni}^{2+} + 2 \mathrm{Ag}(s) \).

Calculate Cell Potential

The cell potential \( E_{\text{cell}}^\circ \) can be found using the formula \( E_{\text{cell}}^\circ = E_{\text{cathode}}^\circ - E_{\text{anode}}^\circ \). Substituting the values, we get \( E_{\text{cell}}^\circ = 0.80\,\mathrm{V} - (-0.25\,\mathrm{V}) = 1.05\,\mathrm{V} \).

Write Shorthand Notation

The shorthand notation for the cell is: \( \mathrm{Ni}(s) | \mathrm{Ni}^{2+}(1.0 \mathrm{M}) || \mathrm{Ag}^+ (1.0 \mathrm{M}) | \mathrm{Ag}(s) \). In this notation, the single vertical lines represent phase boundaries, and the double line signifies the salt bridge.

Key concepts

Electrode Potential

Electrode potential is a measure of the ability of an electrode to drive an electric current through a circuit. This is crucial in a galvanic cell, where it helps determine which substance will undergo reduction and oxidation. Each electrode has a standard electrode potential, expressed in volts (V), measured under standard conditions.

In the example of the galvanic cell featuring nickel and silver, nickel has a standard electrode potential of -0.25 V, while silver has a potential of +0.80 V. These values indicate that silver has a stronger tendency to gain electrons (undergo reduction) compared to nickel.

Understanding electrode potential helps predict the direction of electron flow in the cell. In this case, electrons will flow from nickel to silver, facilitating a spontaneous redox reaction. This capability to contrast the electrode potentials forms the backbone of calculating the overall cell potential (denoted as \(E_{\text{cell}}^\circ\)).

Half-Reactions

A half-reaction in a galvanic cell depicts either the oxidation or reduction process occurring at the respective electrodes. These equations demonstrate the movement of electrons, highlighting what is gained or lost.

For the nickel-silver galvanic cell:

• The oxidation half-reaction involves nickel: \( \mathrm{Ni}(s) \rightarrow \mathrm{Ni}^{2+} + 2e^- \). Here, solid nickel loses electrons, transforming into nickel ions.
• The reduction half-reaction involves silver: \( 2\mathrm{Ag}^+ + 2e^- \rightarrow 2\mathrm{Ag}(s) \). Silver ions are gaining electrons to become solid silver.

Half-reactions are combined to form the complete cell reaction. In this case: \( \mathrm{Ni}(s) + 2\mathrm{Ag}^+ \rightarrow \mathrm{Ni}^{2+} + 2\mathrm{Ag}(s) \). Easy evaluation of these components allows you to see clearly which molecules change and how.

Anode and Cathode

The anode and cathode are the two main electrodes in a galvanic cell. They are crucial because they determine where oxidation and reduction are occurring.

• **Anode**: This is the electrode where oxidation takes place. In the nickel-silver cell, nickel acts as the anode. During oxidation, nickel loses electrons which travel through the external circuit to the cathode.
• **Cathode**: This is where reduction happens. In this setup, silver acts as the cathode. Silver ions accept electrons from the anode to form silver metal.

By knowing the position of these electrodes, you can further understand the flow of current within the cell. Remember, aid from electrode potentials is vital here, as it highlights which material will be oxidized or reduced. Thus, the identification of the anode and cathode becomes easier when comparing electrode potentials.

Cell Notation

Cell notation provides a simplified way of representing the setup of a galvanic cell. It outlines which elements and ions are involved, along with their concentrations, indicating the path electrons take.

The notation comprises parts separated by vertical lines and double vertical lines:

• **Single vertical lines** (|) denote phase boundaries. For example, between a solid metal and an ion in solution.
• **Double vertical lines** (||) represent the salt bridge, crucial for maintaining electrical neutrality in the cell by allowing ion flow.

For the nickel and silver galvanic cell, the shorthand notation is: \( \mathrm{Ni}(s) | \mathrm{Ni}^{2+}(1.0 \mathrm{M}) || \mathrm{Ag}^+ (1.0 \mathrm{M}) | \mathrm{Ag}(s) \). Understanding this notation helps create a concise visual map of the cell's function and flow.