All Exams Test series for 1 year @ ₹349 only
Question

The E°(M+/M) of the cell, SHE||MX|M can be obtained from the plot of (E cell is the cell potential and m is the molality of ideal dilute solution of MX)

The correct answer is

Ecell against (T log m)

Electrochemical Cell Potential Analysis

The question asks how to obtain the standard electrode potential, E°(M+/M), from a plot involving the cell potential (Ecell) of the cell SHE||MX|M, where MX is an ideal dilute solution with molality m, and temperature T.

The cell notation SHE||MX|M represents an electrochemical cell where the Standard Hydrogen Electrode (SHE) acts as one electrode, and the M|MX electrode acts as the other. The double vertical line (||) indicates a salt bridge or porous membrane separating the two half-cells. The reaction at the SHE (anode) is typically:

\begin{equation*} \frac{1}{2}\text{H}_2\text{(g, 1 atm)} \rightarrow \text{H}^+\text{(aq, } a_{\text{H}^+} \text{)} + \text{e}^- \quad E^\circ = 0 \text{ V} \end{equation*}

Assuming a standard SHE where $a_{\text{H}^+} = 1$ and $P_{\text{H}_2} = 1$ atm, its potential is 0 V.

The reaction at the M|MX electrode (cathode) is:

\begin{equation*} \text{M}^+\text{(aq, } a_{\text{M}^+} \text{)} + \text{e}^- \rightarrow \text{M(s)} \end{equation*}

The potential of the M|MX electrode is given by the Nernst equation:

\begin{equation*} E_{\text{M}^+/\text{M}} = E^\circ_{\text{M}^+/\text{M}} - \frac{RT}{nF} \ln\left(\frac{a_{\text{M}}}{a_{\text{M}^+}}\right) \end{equation*}

For the reaction M$^+$ + e$^-$ → M(s), the number of electrons transferred, n, is 1. The activity of the solid metal, $a_{\text{M}}$, is taken as 1. Thus, the equation becomes:

\begin{equation*} E_{\text{M}^+/\text{M}} = E^\circ_{\text{M}^+/\text{M}} - \frac{RT}{1F} \ln\left(\frac{1}{a_{\text{M}^+}}\right) = E^\circ_{\text{M}^+/\text{M}} + \frac{RT}{F} \ln(a_{\text{M}^+}) \end{equation*}

For an ideal dilute solution of MX, the activity of the cation M$^+$ can be approximated by its molality, m. Therefore, $a_{\text{M}^+} \approx m$. Using the relationship $\ln(x) = 2.303 \log_{10}(x)$, we get:

\begin{equation*} E_{\text{M}^+/\text{M}} \approx E^\circ_{\text{M}^+/\text{M}} + \frac{2.303 RT}{F} \log_{10}(m) \end{equation*}

The cell potential, Ecell, is the difference between the cathode potential and the anode potential:

\begin{equation*} E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} = E_{\text{M}^+/\text{M}} - E_{\text{SHE}} \end{equation*}

Assuming the SHE is standard (potential 0 V):

\begin{equation*} E_{\text{cell}} \approx \left(E^\circ_{\text{M}^+/\text{M}} + \frac{2.303 RT}{F} \log_{10}(m)\right) - 0 \end{equation*}

So, the relationship between Ecell and molality m at temperature T is approximately:

\begin{equation*} E_{\text{cell}} \approx E^\circ_{\text{M}^+/\text{M}} + \left(\frac{2.303 R}{F}\right) T \log_{10}(m) \end{equation*}

This equation shows that Ecell is linearly related to the term $(T \log_{10} m)$, assuming $E^\circ_{\text{M}^+/\text{M}}$ and the constant $\frac{2.303 R}{F}$ are constant.

If we plot Ecell on the y-axis against $(T \log_{10} m)$ on the x-axis, the equation is in the form $y = mx + c$, where:

  • $y = E_{\text{cell}}$
  • $x = T \log_{10}(m)$
  • $m = \frac{2.303 R}{F}$ (slope)
  • $c = E^\circ_{\text{M}^+/\text{M}}$ (y-intercept)

Therefore, the standard electrode potential E°(M+/M) can be obtained as the intercept of the plot of Ecell against $(T \log m)$.

Let's check the given options based on this derivation:

  • Option 1: Ecell against (T log m). This matches our finding.
  • Option 2: Ecell against (1/T log m). This does not fit the derived linear relationship.
  • Option 3: Ecell against (T√m). This does not fit the derived linear relationship which involves log(m).
  • Option 4: Ecell against ($\sqrt m / T$). This does not fit the derived linear relationship.

The plot of Ecell versus (T log m) should yield a straight line, and the value of E°(M+/M) is the intercept on the Ecell axis (when T log m = 0, which happens when m=1 at any T, or when T=0 at any m, practically evaluated by extrapolating to log m = 0, i.e., m=1). However, often such plots are extrapolated to m=0 (infinite dilution) to account for deviations from ideal behavior, but the ideal dilute solution assumption simplifies it to a plot against T log m.

Thus, plotting Ecell against (T log m) allows for the determination of E°(M+/M).

Was this answer helpful?

Important Questions from Electrochemistry

  1. At $298 \, K$, given the standard electrode potentials: $E^\circ_{Cu^{2+}/Cu} = 0.34 \, V$, $E^\circ_{Zn^{2+}/Zn} = -0.76 \, V$, $E^\circ_{Fe^{2+}/Fe} = -0.44 \, V$, and $E^\circ_{Ag^{+}/Ag} = 0.80 \, V$.
    Based on these values, which of the following reactions is NOT expected to occur spontaneously under standard conditions?
  2. Which of the following processes is required for extracting metal from cinnabar ore?
  3. You are given three metals 'X', 'Y' and 'Z'. Metal 'X' is found to react with an aqueous solution of both YSO 4and ZSO 4whereas metal 'Z' is found to react only with aqueous solution of YSO 4. Based on these observations, select the correct statement from the following.

  4. The mobility of a divalent cation in water is 8 × 10-8 m2 V-1 s-1. The effective radius of the ion is (viscosity of water = 1 cP : c = 1.6 × 10-19 C)

  5. The electrical double layer model among the following that consists of both fixed and diffuse layers is

Need Expert Advice?

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App