Question

Consider the concentration cell shown below. Calculate the cell potential at 25°C when the concentration ofNi²⁺ in the compartment on the right has each of the following values.

a. 1.0 M

b. 2.0 M

c. 0.10 M

d. 4.0 x 10^-5 M

e. Calculate the potential when bothsolutions are 2.5 M in Ni²⁺.

For each case, identify the cathode, the anode, and thedirection in which electrons flow.

Short answer

The cells with non-zero cell potential are cathodes. Therefore, the other compartment is the anode. Electrons always flow from anode to cathode.

(a) The cell potential is zero. The anode and cathode cannot be identified, and there is no current flow.

(b) The cell potential is $8.89\times {10}^{-3}V$. The compartment withNi²⁺=2M is the cathode, and the other compartment with Ni²⁺=1M is the anode.

(c) The cell potential is $0.03V$. The compartment withNi²⁺=1M is the cathode, and the other compartment with Ni²⁺=0.10M is the anode.

(d) The cell potential is $0.13V$. The compartment withNi²⁺=1M is the cathode, and the other compartment with Ni²⁺= $4.0\times {10}^{-5}V$is the anode.

(e) The cell potential is zero. The anode and cathode cannot be identified, and there is no current flow.

Step-by-step solution

Cell potential of an electrochemical cell at 1.0 M

The expression of cell potential for cell concentration can be expressed as:

$E_{cell}=-\frac{0.0591}{n}\log \left(\frac{\left[anode\right]}{\left[cathode\right]}\right)$

On substituting the values,

$$\begin{gathered} E_{cell}=-\frac{0.0591}{2}\log \left(\frac{1}{1}\right) \\ {E}_{cell}=0 \end{gathered}$$

The anode and cathode have not been identified, and there is no current flow.

Cell potential of an electrochemical cell at 2 M

The expression of cell potential for cell concentration can be expressed as:

$E_{cell}=-\frac{0.0591}{n}\log \left(\frac{\left[anode\right]}{\left[cathode\right]}\right)$

On substituting the values,

$$\begin{gathered} E_{cell}=-\frac{0.0591}{2}\log \left(\frac{1}{2}\right) \\ {E}_{cell}=8.89\times {10}^{-3}V. \end{gathered}$$

The cell potential is$8.89\times {10}^{-3}V$. The compartment withNi²⁺=2M is the cathode, and the other compartment with Ni²⁺=1M is the anode.

Cell potential of an electrochemical cell at 0.1 M

The expression of cell potential for cell concentration can be expressed as:

$E_{cell}=-\frac{0.0591}{n}\log \left(\frac{\left[anode\right]}{\left[cathode\right]}\right)$

On substituting the values,

$$\begin{gathered} E_{cell}=-\frac{0.0591}{2}\log \left(\frac{0.1}{1}\right)\backslash \mathrm{hfill} \\ {E}_{cell}=0.03V. \end{gathered}$$

The cell potential is$0.03V$. The compartment withNi²⁺=1M is the cathode, and the other compartment withNi²⁺=0.10 M is the anode.

Cell potential of an electrochemical cell at 4.0 x 10-5 M

The expression of cell potential for cell concentration can be expressed as:

$E_{cell}=-\frac{0.0591}{n}\log \left(\frac{\left[anode\right]}{\left[cathode\right]}\right)$

On substituting the values,

$$\begin{gathered} E_{cell}=-\frac{0.0591}{2}\log \left(\frac{4\times {10}^{-3}}{1}\right) \\ {E}_{cell}=0.13V. \end{gathered}$$

The cell potential is $0.13V$. The compartment withNi²⁺=1M is the cathode, and the other compartment with Ni²⁺= $4.0\times {10}^{-5}V$is the anode.

Cell potential of an electrochemical cell at 2.5 M in Ni2+

The expression of cell potential for cell concentration can be expressed as:

$E_{cell}=-\frac{0.0591}{n}\log \left(\frac{\left[anode\right]}{\left[cathode\right]}\right)$

On substituting the values,

$$\begin{gathered} E_{cell}=-\frac{0.0591}{2}\log \left(\frac{2.5}{2.5}\right) \\ {E}_{cell}=0 \end{gathered}$$

The anode and cathode have not been identified, and there is no current flow.