Question

An electrochemical cell, consist of the following two redox couples, \(M^{x+}(aq)/M(s) [E_{red}^\circ = +0.15 V]\) and \(Fe^{3+}(aq)/Fe(s) [E_{red}^\circ = -0.036 V]\). The cell EMF (\(E_{cell}\)) is recorded to be 0.2057 V. If the reaction quotient of the electrochemical cell is found to be \(10^{-z}\), then the value of x is _________. (Nearest integer)
[Given : M is a p-block metal and \(\frac{2.303 RT}{F} = 0.059 V\)]

Answer 1

Correct Answer: 2

Step 1: Understanding the Concept:
By identifying the anode and cathode based on standard reduction potentials, we can establish the standard cell potential $E^\circ_{cell}$. Using the Nernst equation alongside the given non-standard $E_{cell}$ and the reaction quotient $Q$, we can back-calculate $n$, the total number of electrons exchanged. Analyzing the balanced redox equation reveals the specific oxidation state $x$.
Step 2: Key Formula or Approach:
Cathode (Reduction) = Higher $E^\circ_{red}$.
Anode (Oxidation) = Lower $E^\circ_{red}$.
$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$
Nernst Equation: $E_{cell} = E^\circ_{cell} - \frac{0.059}{n} \log Q$
Step 3: Detailed Explanation:
Identify the half-reactions:
$M^{x+}/M$ has higher reduction potential ($+0.15\text{ V}$), so it acts as the Cathode (Reduction).
$Fe^{3+}/Fe$ has lower reduction potential ($-0.036\text{ V}$), so it acts as the Anode (Oxidation).
Calculate $E^\circ_{cell}$:
$E^\circ_{cell} = 0.15 \text{ V} - (-0.036 \text{ V}) = 0.186 \text{ V}$.
Use the Nernst equation:
\[ E_{cell} = E^\circ_{cell} - \frac{0.059}{n} \log Q \] Substitute the given values ($E_{cell} = 0.2057\text{ V}$, $Q = 10^{-2}$):
\[ 0.2057 = 0.186 - \frac{0.059}{n} \log(10^{-2}) \] \[ 0.2057 - 0.186 = -\frac{0.059}{n} (-2) \] \[ 0.0197 = \frac{0.118}{n} \] \[ n = \frac{0.118}{0.0197} \approx 5.989 \approx 6 \] The total number of electrons transferred in the balanced cell reaction is 6.
Let's construct the balanced reaction.
Anode: $Fe \to Fe^{3+} + 3e^-$
Cathode: $M^{x+} + xe^- \to M$
To balance the electrons, we multiply the anode reaction by $x$ and the cathode reaction by $3$:
Total electrons transferred $n = 3x$.
Since we calculated $n = 6$:
$3x = 6 \implies x = 2$.
This is consistent with "M is a p-block metal" since elements like Lead (Pb) or Tin (Sn) commonly exhibit a $+2$ oxidation state.
Step 4: Final Answer:
The value of $x$ is 2.